Chapter 9.6 - Sum of The Cubes of First n Natural Numbers

In the previous section, we saw the sum of squares of the first n natural numbers. In this section, we will see sum of cubes of the first n natural numbers. 

C. 1³ + 2³ + 3³ +. . . + n³

This sum can be calculated in 8 steps:
1. Consider the identity: (a+b)⁴ = a⁴+4a³b+6a²b²+4ab³+b⁴
[Recall that (a+b) can be raised to any power by using binomial theorem that we saw in the previous chapter]
Let us put a = k and b = -1. We get:

$\begin{array}{ll}
{}&{[k+(-1)]^4}
&{}={}& {[k-1]^4}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {k^4 ~+~ 4 × k^3 × -1 ~+~6 × k^2 × (-1)^2~+~4 × k × (-1)^3~+~(-1)^4}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\


{}&{}
&{}={}& {k^4 ~-~ 4 k^3~+~6k^2 ~-~4k~+~1}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

2. Subtracting (k-1)⁴ from k⁴, we get:

$\begin{array}{ll}
{}&{k^4~-~(k-1)^4}
&{}={}& {k^4~-~\left(k^4 ~-~ 4 k^3~+~6k^2 ~-~4k~+~1 \right)}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {k^4~-~k^4~+~4 k^3~-~6k^2 ~+~4k~-~1}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {4 k^3~-~6k^2 ~+~4k~-~1}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\
\end{array}$

• We can use this as an identity:
$k^4~-~(k-1)^4~=~4 k^3~-~6k^2 ~+~4k~-~1$

3. In the above identity, let us put k = 1, 2, 3, . . . , n successively. We get:

$\begin{array}{ll}
{\text{When k = 1,}}&{1^4~-~(1-1)^4}
&{}={}& {1^4~-~0^4}
&{}={}& {4 × 1^3~-~6 × 1^2 ~+~4 × 1~-~1}
&{}={}& {4(1)^3~-~6(1)^2~+~4(1)~-~1}
&{}& {}&{}& {} &{} \\

{\text{When k = 2,}}&{2^4~-~(2-1)^4}
&{}={}& {2^4~-~1^4}
&{}={}& {4 × 2^3~-~6 × 2^2 ~+~4 × 2~-~1}
&{}={}& {4(2)^3~-~6(2)^2~+~4(2)~-~1}
&{}& {}&{}& {} &{} \\

{\text{When k = 3,}}&{3^4~-~(3-1)^4}
&{}={}& {3^4~-~2^4}
&{}={}& {4 × 3^3~-~6 × 3^2 ~+~4 × 3~-~1}
&{}={}& {4(3)^3~-~6(3)^2~+~4(3)~-~1}
&{}& {}&{}& {} &{} \\

{\text{When k = 4,}}&{4^4~-~(4-1)^4}
&{}={}& {4^4~-~3^4}
&{}={}& {4 × 4^3~-~6 × 4^2 ~+~4 × 4~-~1}
&{}={}& {4(4)^3~-~6(4)^2~+~4(4)~-~1}
&{}& {}&{}& {} &{} \\


{-}&{-}
&{}& {-}
&{}& {-}
&{}& {-}
&{}& {}&{}& {} &{} \\

{-}&{-}
&{}& {-}
&{}& {-}
&{}& {-}
&{}& {}&{}& {} &{} \\

{\text{When k = n,}}&{n^4~-~(n-1)^4}
&{}& {}
&{}={}& {4 × n^3~-~6 × n^2 ~+~4 × n~-~1}
&{}={}& {4(n)^3~-~6(n)^2~+~4(n)~-~1}
&{}& {}&{}& {} &{} \\

\end{array}$

4. Picking the first and last items from each line, we get:

$\begin{array}{ll}
{}&{}
&{}& {1^4~-~0^4}
&{}& {}
&{}={}& {4(1)^3~-~6(1)^2~+~4(1)~-~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {2^4~-~1^4}
&{}& {}
&{}={}& {4(2)^3~-~6(2)^2~+~4(2)~-~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {3^4~-~2^4}
&{}& {}
&{}={}& {4(3)^3~-~6(3)^2~+~4(3)~-~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {4^4~-~3^4}
&{}& {}
&{}={}& {4(4)^3~-~6(4)^2~+~4(4)~-~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {-}
&{}& {}
&{}& {-}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {-}
&{}& {}
&{}& {-}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {n^4~-~(n-1)^4}
&{}& {}
&{}={}& {4(n)^3~-~6(n)^2~+~4(n)~-~1}
&{}& {}&{}& {} &{} \\

\end{array}$

5. In the above result in (4), let us add all terms on the left side.
• We see that diagonal elements will get cancelled:
   ♦ 1⁴ will get cancelled by -1⁴.
   ♦ 2⁴ will get cancelled by -2⁴.
   ♦ 3⁴ will get cancelled by -3⁴.
   ♦ so on . . .
• So only 0⁴ and n⁴ will remain.
• Thus the sum of all terms on the left side is: (n⁴ - 0⁴) = n⁴.

6. In the result in (4), let us add all terms on the right side. We get:
4(1³ + 2³ + 3³ +. . . + n³) - 6(1² + 2² + 3² +. . . + n²) + 4(1 + 2 + 3 +. . . + n) - (1+1+1+ . . . n times)
• This can be written in a shortened form using sigma notations:
$4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}~-~6 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}~+~4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~-~n$

7. Equating the results in (5) and (6), we get:

$\begin{array}{ll}
{}&{n^4}
&{}={}& {4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}~-~6 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}~+~4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~-~n}
&{}& {}
 \\

{\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {n^4~+~6 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}~-~4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~+~n}
&{\color {green} {\text{- - - - (a)}}}& {}
 \\

{\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {n^4~+~\left[\frac{6n(2n + 1) (n+1)}{6} \right]~-~4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~+~n}
&{\color {green} {\text{- - - - (b)}}}& {}
 \\

{\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {n^4~+~\left[\frac{6n(2n + 1) (n+1)}{6} \right]~-~\left[\frac{4n(n+1)}{2} \right]~+~n}
&{}& {}
 \\

{\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {n^4~+~n(2n + 1) (n+1)~-~2n(n+1)~+~n}
&{}& {}
 \\

{\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {n^4~+~2n^3+2n^2+n^2+n~-~2n^2 -2n~+~n}
&{}& {}
 \\

{\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {n^4 + 2n^3 + n^2 }
&{}& {}
 \\

{\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {n^2 \left(n^2 + 2n + 1 \right) }
&{\color {green} {\text{- - - - (c)}}}& {}
 \\

{\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {n^2 \left(n + 1 \right)^2 }
&{}& {}
 \\

{\Rightarrow}&{\sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {\frac{n^2 \left(n + 1 \right)^2}{4} }
&{}& {}
 \\

{\Rightarrow}&{\sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {\left[ \frac{n (n + 1)}{2}\right]^2}
&{}& {}
 \\
\end{array}$

Remarks:
(i) The line marked as (a):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}$ is in fact the sum of squares of first n natural numbers.
• So our second result B that we saw in the previous section can be used.
(ii) The line marked as (b):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k}$ is in fact the sum of first n natural numbers.
• So our first result A that we saw in the previous section can be used.
(iii) The line marked as (c):
• Using the identity (a+b)² = a² + 2ab + b²,
(n² + 2n + 1) is (n+1)²

8. We can write the result as a formula:
Sum of cubes of the first n natural numbers
= 1³ + 2³ + 3³ +. . . + n³

= $\sum\limits_{k\,=\,1}^{k\,=\,n}{k^3}~=~\left[ \frac{n (n + 1)}{2}\right]^2$

---

Let us see some solved examples:

Solved example 9.19
Find the sum to n terms of the series: 5 + 11 + 19 + 29 + 41 + . . . 
Solution:
1. Let us write:
$S_n~=~5~+~11~+~19~+~29~+~41~+~.~.~.~+~a_{n-1}~+~a_{n}$
2. The same result can be written as:
$S_n~=~5~+~11~+~19~+~29~+~41~+~.~.~.~+~a_{n-2}~+~a_{n-1}~+~a_{n}$
3. We will write the second result just below the first result. But one term to the right. We get:

$\begin{array}{ll}
{S_n}&{}={}
&{5}& {~+~}
&{11}& {~+~}
&{19}& {~+~}
&{29}& {~+~.~.~.~+~}&{a_{n-1}}& {~+~} &{a_n} &{} &{} \\

{S_n}&{}={}
&{}& {}
&{5}& {~+~}
&{11}& {~+~}
&{19}& {~+~.~.~.~+~}&{a_{n-2}}& {~+~} &{a_{n-1}} &{~+~} &{a_n} \\

\end{array}$

4. Subtracting each term in the second row, from the term directly above it, we get:

$\begin{array}{ll}
{S_n}&{}={}
&{5}& {~+~}
&{11}& {~+~}
&{19}& {~+~}
&{29}& {~+~.~.~.~+~}&{a_{n-1}}& {~+~} &{a_n} &{} &{} \\

{S_n}&{}={}
&{}& {}
&{5}& {~+~}
&{11}& {~+~}
&{19}& {~+~.~.~.~+~}&{a_{n-2}}& {~+~} &{a_{n-1}} &{~+~} &{a_n} \\

{0}&{}={}
&{5}& {~+~}
&{\left\{6 \right. }& {~+~}
&{8}& {~+~}
&{10}& {~+~.~.~.~}&{}& {} &{\left. \right \}} &{~-~} &{a_n} \\

\end{array}$

5. In the above result, there will be (n-1) terms inside the curly brackets '{}'
• In side the curly brackets, we have the series: 6 + 8 + 10 + . . . (n-1) terms
• It is an A.P with a = 6, d = 2 and n = (n-1)
• So sum of the series = $\frac{(n-1)}{2} [2 × 6 + (n-1-1)2]~=~\frac{(n-1)}{2} [12 + (n-2)2]~=~(n-1)(6+n-2)~=~(n-1)(4+n)$

6. Now the result in (4) becomes:
0 = 5 + (n-1)(4+n) - an
⇒ a_n = 5 + (n-1)(4+n)
⇒ a_n = 5 + 4n + n² - 4 - n
⇒ a_n = 1 + 3n + n²
⇒ a_n = n² + 3n + 1

7. Thus we obtained the n^th term.
• n^th term = a_n = n² + 3n + 1.
• So the sum (S_n) of the given series can be obtained as:

$\begin{array}{ll}
{S_n}&{}={}
&{\sum\limits_{k\,=\,0}^{k\,=\,n}{a_k}}~=~\sum\limits_{k\,=\,0}^{k\,=\,n}{\left(k^2 + 3k + 1 \right)}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}~+~3\sum\limits_{k\,=\,0}^{k\,=\,n}{k}~+~n}& {}
&{\color {green} {\text{- - - - (a)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n(2n + 1) (n+1)}{6}~+~3\sum\limits_{k\,=\,0}^{k\,=\,n}{k}~+~n}& {}
&{\color {green} {\text{- - - - (b)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n(2n + 1) (n+1)}{6}~+~ \frac{3n (n+1)}{2}~+~n}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n(2n + 1) (n+1)~+~9n(n+1)~+~6n}{6}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n[(2n+1)(n+1)+9(n+1)+6]}{6}}& {}={}
&{\frac{n[2n^2+2n+n+1+9n+9+6]}{6}}& {}={}
&{\frac{n[2n^2+12n+16]}{6}}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{2n[n^2+6n+8]}{6}}& {}={}
&{\frac{n[n^2+6n+8]}{3}}& {}
&{\color {green} {\text{- - - - (c)}}}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n[(n+2)(n+4)]}{3}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

Remarks:
(i) The line marked as (a):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}$ is in fact the sum of squares of first n natural numbers.
• So our second result B that we saw in the previous section can be used.
(ii) The line marked as (b):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k}$ is in fact the sum of first n natural numbers.
• So our first result A that we saw in the previous section can be used.
(iii) The line marked as (c):
• n² + 6n + 8 = 0 can be solved as a quadratic equation.
• We will get: n = -2 and n = -4.

Solved example 9.20
Find the sum to n terms of the series whose n^th term is n(n+3)
Solution:
• n^th term = n(n+3).
• So the sum (S_n) of the given series can be obtained as:

$\begin{array}{ll}
{S_n}&{}={}
&{\sum\limits_{k\,=\,0}^{k\,=\,n}{[k(k+3)]}}~=~\sum\limits_{k\,=\,0}^{k\,=\,n}{\left[k^2 + 3k \right]}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}~+~3\sum\limits_{k\,=\,0}^{k\,=\,n}{k}}& {}
&{\color {green} {\text{- - - - (a)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n(2n + 1) (n+1)}{6}~+~3\sum\limits_{k\,=\,0}^{k\,=\,n}{k}}& {}
&{\color {green} {\text{- - - - (b)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n(2n + 1) (n+1)}{6}~+~ \frac{3n (n+1)}{2}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n(2n + 1) (n+1)~+~9n(n+1)}{6}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n[(2n+1)(n+1)+9(n+1)]}{6}}& {}={}
&{\frac{n[2n^2+2n+n+1+9n+9]}{6}}& {}={}
&{\frac{n[2n^2+12n+10]}{6}}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{2n[n^2+6n+5]}{6}}& {}={}
&{\frac{n[n^2+6n+5]}{3}}& {}
&{\color {green} {\text{- - - - (c)}}}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n[(n+1)(n+5)]}{3}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

Remarks:
(i) The line marked as (a):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}$ is in fact the sum of squares of first n natural numbers.
• So our second result B that we saw in the previous section can be used.
(ii) The line marked as (b):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k}$ is in fact the sum of first n natural numbers.
• So our first result A that we saw in the previous section can be used.
(iii) The line marked as (c):
• n² + 6n + 5 = 0 can be solved as a quadratic equation.
• We will get: n = -1 and n = -5.

---

The link below gives some more solved examples

Exercise 9.4

---

In the next section we will see some miscellaneous examples.

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Chapter 9.5 - Sum of The Squares of First n Natural Numbers

In the previous section, we completed a discussion on geometric progression and geometric mean. In this section, we will see sum to n terms of special series.

We have to find the sum of three series. They are:

A. 1 + 2 + 3 +. . . + n
   ♦ This is the sum of first n natural numbers.

B. 1² + 2² + 3² +. . . + n²
   ♦ This is the sum of squares of the first n natural numbers.

C. 1³ + 2³ + 3³ +. . . + n³
   ♦ This is the sum of cubes of the first n natural numbers.

---

A. 1 + 2 + 3 +. . . + n

This sum can be calculated in 3 steps:
1. The given series is the series related to the sequence 1, 2, 3, . . . , n
2. This sequence is an A.P with a = 1 and d = 1
• So sum to n terms will be given by:
$\begin{array}{ll}
{}&{S_n}
&{}={}& {\frac{n}{2}[2a+(n-1)d]}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {\frac{n}{2}[2 × 1+(n-1) × 1]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\


{}&{}
&{}={}& {\frac{n}{2}[2+n-1]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\


{}&{}
&{}={}& {\frac{n}{2}[n+1]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\
\end{array}$

3. We can write it as a formula:
Sum of first n natural numbers = $\frac{n}{2}[n+1]$

---

B. 1² + 2² + 3² +. . . + n²

This sum can be calculated in 8 steps:
1. Consider the identity: (a+b)³ = a³ + 3a²b + 3 ab² + b³
Let us put a = k and b = -1. We get:

$\begin{array}{ll}
{}&{[k+(-1)]^3}
&{}={}& {[k-1]^3}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {k^3 ~+~ 3 × k^2 × -1 ~+~3 × k × (-1)^2~+~(-1)^3}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\


{}&{}
&{}={}& {k^3 ~-~ 3 k^2 ~+~3k~-~1}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

2. Subtracting (k-1)³ from k³, we get:

$\begin{array}{ll}
{}&{k^3~-~(k-1)^3}
&{}={}& {k^3~-~\left(k^3 ~-~ 3 k^2 ~+~3k~-~1 \right)}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {k^3~-~k^3 ~+~ 3 k^2 ~-~3k~+~1}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {3 k^2 ~-~3k~+~1}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\
\end{array}$

• We can use this as an identity:
$k^3~-~(k-1)^3~=~3 k^2 ~-~3k~+~1$

3. In the above identity, let us put k = 1, 2, 3, . . . , n successively. We get:

$\begin{array}{ll}
{\text{When k = 1,}}&{1^3~-~(1-1)^3}
&{}={}& {1^3~-~0^3}
&{}={}& {3 × 1^2 ~-~3 × 1~+~1}
&{}={}& {3(1)^2~-~3(1)~+~1}
&{}& {}&{}& {} &{} \\

{\text{When k = 2,}}&{2^3~-~(2-1)^3}
&{}={}& {2^3~-~1^3}
&{}={}& {3 × 2^2 ~-~3 × 2~+~1}
&{}={}& {3(2)^2~-~3(2)~+~1}
&{}& {}&{}& {} &{} \\

{\text{When k = 3,}}&{3^3~-~(3-1)^3}
&{}={}& {3^3~-~2^3}
&{}={}& {3 × 3^2 ~-~3 × 3~+~1}
&{}={}& {3(3)^2~-~3(3)~+~1}
&{}& {}&{}& {} &{} \\

{\text{When k = 4,}}&{4^3~-~(4-1)^3}
&{}={}& {4^3~-~3^3}
&{}={}& {3 × 4^2 ~-~3 × 4~+~1}
&{}={}& {3(4)^2~-~3(4)~+~1}
&{}& {}&{}& {} &{} \\

{-}&{-}
&{}& {-}
&{}& {-}
&{}& {-}
&{}& {}&{}& {} &{} \\

{-}&{-}
&{}& {-}
&{}& {-}
&{}& {-}
&{}& {}&{}& {} &{} \\

{\text{When k = n,}}&{n^3~-~(n-1)^3}
&{}& {}
&{}={}& {3 × n^2 ~-~3 × n~+~1}
&{}={}& {3(n)^2~-~3(n)~+~1}
&{}& {}&{}& {} &{} \\

\end{array}$

4. Picking the first and last items from each line, we get:

$\begin{array}{ll}
{}&{}
&{}& {1^3~-~0^3}
&{}& {}
&{}={}& {3(1)^2~-~3(1)~+~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {2^3~-~1^3}
&{}& {}
&{}={}& {3(2)^2~-~3(2)~+~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {3^3~-~2^3}
&{}& {}
&{}={}& {3(3)^2~-~3(3)~+~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {4^3~-~3^3}
&{}& {}
&{}={}& {3(4)^2~-~3(4)~+~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {-}
&{}& {}
&{}& {-}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {-}
&{}& {}
&{}& {-}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {n^3~-~(n-1)^3}
&{}& {}
&{}={}& {3(n)^2~-~3(n)~+~1}
&{}& {}&{}& {} &{} \\

\end{array}$

5. In the above result in (4), let us add all terms on the left side.
• We see that diagonal elements will get cancelled:
   ♦ 1³ will get cancelled by -1³.
   ♦ 2³ will get cancelled by -2³.
   ♦ 3³ will get cancelled by -3³.
   ♦ so on . . .
• So only 0³ and n³ will remain.
• Thus the sum of all terms on the left side is: (n³ - 0³) = n³.

6. In the result in (4), let us add all terms on the right side. We get:
3(1² + 2² + 3² +. . . + n²) - 3(1 + 2 + 3 +. . . + n) + (1+1+1+ . . . n times)
• This can be written in a shortened form using sigma notations:
$3 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}~-~3 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~+~n$

7. Equating the results in (5) and (6), we get:

$\begin{array}{ll}
{}&{n^3}
&{}={}& {3 \sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}~-~3 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~+~n}
&{}& {}
 \\

{\Rightarrow}&{3 \sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {n^3~+~3 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~-~n}
&{\color {green} {\text{- - - - (a)}}}& {}
 \\

{\Rightarrow}&{3 \sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {n^3~+~\frac{3n(n+1)}{2}~-~n}
&{}& {}
 \\

{\Rightarrow}&{3 \sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {\frac{2n^3 +3n(n+1) - 2n}{2}}
&{}& {}
 \\

{\Rightarrow}&{3 \sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {\frac{2n^3 + 3n^2 + 3n - 2n}{2}}
&{}& {}
 \\

{}&{}
&{}& {}
&{}& {}
 \\

{\Rightarrow}&{\sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {\frac{2n^3 + 3n^2 + n}{6}}
&{}& {}
 \\

{}&{}
&{}& {}
&{}& {}
 \\

{\Rightarrow}&{\sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {\frac{2n^3 + 3n^2 + n}{6}}
&{}& {}
 \\

{}&{}
&{}& {}
&{}& {}
 \\

{\Rightarrow}&{\sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {\frac{n(2n^2 + 3n + 1)}{6}}
&{\color {green} {\text{- - - - (b)}}}& {}
 \\

{}&{}
&{}& {}
&{}& {}
 \\

{\Rightarrow}&{\sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {\frac{n(2n + 1) (n+1)}{6}}
&{}& {}
\\

\end{array}$

Remarks:
(i) The line marked as (a):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k}$ is in fact the sum of first n natural numbers.
• So our first result A that we saw at the beginning of this section can be used.
(ii) The line marked as (b):
• (2n² + 3n +1) can be written as (2n² + 2n + n + 1)
• But (2n² + 2n + n + 1) = [2n(n+1) + (n+1)] = [(2n+1)(n+1)]
So we get: (2n² + 3n + 1) = (2n+1)(n+1)

Another method:
• Find the solutions of the quadratic equation 2n² + 3n + 1 = 0
• The solutions are: n = $-\frac{1}{2}$ and n = -1
• So we can write: $2n^2 + 3n + 1 ~=~ \left(n+ \frac{1}{2} \right) (n+1) ~=~0$
$\Rightarrow~\left(\frac{2n+1}{2} \right) (n+1) ~=~0$
$\Rightarrow~\left(2n+1 \right) (n+1) ~=~0$

8. We can write it as a formula:
Sum of squares of the first n natural numbers
= 1² + 2² + 3² +. . . + n²

= $\sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}~=~\frac{n(2n + 1) (n+1)}{6}$

---

In the next section we will see sum of the cubes of first n natural numbers.

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Chapter 9.4 - Geometric Mean

In the previous section, we saw the basic details about G.P. We saw some solved examples also. In this section, we will see two more solved examples. Later in this section, we will see geometric mean.

Solved example 9.15
Find the sum of the sequence 7, 77,777, 7777, . . . to n terms.
Solution:
1. We are asked to find the sum of the series: 7 + 77 + 777 + 7777 + . . . to n terms
2. This is not a geometric series. But we can relate it to a geometric series as follows:
$\begin{array}{ll}
{}&{S_n}
&{}={}& {7+77+777+7777+~.~.~\text{to n terms}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {7[1+11+111+1111+~.~.~\text{to n terms}]~\color{green}{\text{(Taking out common factor 7)}}}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\


{}&{}
&{}={}& {7 × \frac{9}{9}[1+11+111+1111+~.~.~\text{to n terms}]~\color{green}{\text{(Multiplying numerator and denominator by 9)}}}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\


{}&{}
&{}={}& {7 × \frac{1}{9}[9+99+999+9999+~.~.~\text{to n terms}]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {\frac{7}{9}[(10-1)+(100-1)+(1000-1)+(10000-1)+~.~.~\text{to n terms}]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {\frac{7}{9}[(10+100+1000+10000+~.~.~.~\text{to n terms})~-1-1-1-1-~.~.~\text{to n terms}]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {\frac{7}{9}[(10+100+1000+10000+~.~.~.~\text{to n terms})~-(1+1+1+1-~.~.~\text{to n terms})]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {\frac{7}{9} \left[\frac{10(10^n - 1)}{10-1}~-~n\right]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {\frac{7}{9} \left[\frac{10(10^n - 1)}{9}~-~n\right]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

Solved example 9.16
A person has 2 parents, 4 grand parents, 8 great grand parents and so on. Find the number of his ancestors during the 10 generations preceding his own.
Solution:
1. Fig.9.4 below shows the ancestors in order:

Fig.9.4

• The yellow square at the bottom represents the person under consideration.
• The cyan squares represent the parents. There are two parents.
• The green squares represent the grand parents. There are four grand parents.
• The magenta squares represent the great grand parents. There are eight great grand parents.
2. So the number of ancestors in each generation can be written as a sequence. We get: 2, 4, 8, 16, . . .
• This is a G.P with a = 2 and r = 2.
3. Sum of ten terms of this G.P will give the total number of ancestors in ten generations.
• We can write:
$S_{10}~=~\frac{a(r^n - 1)}{r-1}~=~\frac{2(2^{10} - 1)}{2-1}~=~\frac{2(2^{10} - 1)}{1}~=~2046$

---

Geometric mean

This can be written in 5 steps:
1. Suppose that, we are given two numbers a and b
• We can find a number G in such a way that, a, G, b form a G.P
2. This G can be calculated in two simple steps:
(i) Since a, G, b is a G.P, we can write: $\frac{G}{a}~=~\frac{b}{G}$
(ii) From this we get: $G^2~=~ab$
⇒ $G=\sqrt{ab}$
3. This G is called the Geometric mean of a and b.
• Geometric mean is abbreviated as G.M
4. Let us see an example:
    ♦ Let the two numbers a and b be 2 and 8
    ♦ Then the G.M of 2 and 8 = $\sqrt{2 × 8}~=~\sqrt{16}$ = 4
5. We inserted just one number G between a and b. In fact we can insert as many numbers as we like between two numbers a and b so that the resulting sequence is a G.P.
• The following solved example will demonstrate the steps.

Solved example 9.17
Insert three numbers between 1 and 256 so that the resulting sequence is a G.P
Solution:
1. The resulting G.P will be in the form: 1, G₁, G₂, G₃, 256
    ♦ The first term of this G.P is 1
    ♦ The last term is 256
    ♦ Total number of terms = 5
2. So we can write:
256 = $a r^{n-1} ~=~1 × r^{5-1} ~=~r^4$
⇒ $256~=~4^4~=~r^4$
⇒ r = 4
3. Now we can write all the intermediate terms:
2^nd term G₁ = ar = 1 × 4 = 4
3^rd term G₂ = ar² = 1 × 4² = 16
4^th term G₃ = ar³ = 1 × 4³ = 64
4. So the resulting G.P is: 1, 4, 16, 64, 256
• The three numbers to be inserted are: 4, 16 and 64

---

Relation between A.M and G.M

This can be written in 4 steps:
1. Consider two positive real numbers a and b
    ♦ Let A be the A.M of a and b
    ♦ Let G be the G.M of a and b
2. Then we can write: $A = \frac{a+b}{2}$
• Also we can write: $G=\sqrt{ab}$
3. Taking the difference, we get:
$A-G~=~ \frac{a+b}{2}~-~\sqrt{ab}~=~\frac{a+b-2 \sqrt{ab}}{2}~=~\frac{\left(\sqrt{a}- \sqrt{b} \right)^2}{2}$
• So we can write: $A-G~=~\frac{\left(\sqrt{a}- \sqrt{b} \right)^2}{2}$
4. Consider the R.H.S of the above expression. This R.H.S cannot be less than zero.
• So we can write:
    ♦ G will be either equal to A or less than A
    ♦ G can never be greater than A

---

The link below gives some more solved examples

Exercise 9.3

---

In the next section we will see special series.

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Chapter 9.3 - Geometric Progression

In the previous section, we completed a discussion on arithmetic progression. In this section, we will see geometric progression.

Some basics of geometric progression can be written in 6 steps:
1. In the case of arithmetic progression, we saw that:
We can subtract any term from the succeeding term. The difference will always be a constant.
2. In the case of geometric progression, we will see that:
We can divide any term by the preceding term. The ratio will always be a constant.
3. Let us see three examples:
Example 1:
• Consider the sequence: 2, 4, 8, 16, . . .
   ♦ Let us divide 4 by it’s preceding term 2. We will get: $\frac{4}{2}~=~2 $
   ♦ Let us divide 8 by it’s preceding term 4. We will get: $\frac{8}{4}~=~2 $
   ♦ Let us divide 16 by it’s preceding term 8. We will get: $\frac{16}{8}~=~2 $
• In this way, we can divide any term (except the first term) of this sequence by it’s preceding term. The result will be always 2
Example 2:
• Consider the sequence: $\frac{1}{9},\frac{-1}{27},\frac{1}{81},\frac{-1}{243},~.~.~.$
   ♦ Let us divide $\frac{-1}{27}$ by it’s preceding term $\frac{1}{9}$. We will get: $\frac{-1}{27} × \frac{9}{1}~=~\frac{-1}{3}$
   ♦ Let us divide $\frac{1}{81}$ by it’s preceding term $\frac{-1}{27}$. We will get: $\frac{1}{81} × \frac{27}{-1}~=~\frac{-1}{3}$
   ♦ Let us divide $\frac{-1}{243}$ by it’s preceding term $\frac{1}{81}$. We will get: $\frac{-1}{243} × \frac{81}{1}~=~\frac{-1}{3}$
• In this way, we can divide any term (except the first term) of this sequence by it’s preceding term. The result will be always $\frac{-1}{3}$
Example 3:
• Consider the sequence: 0.01, 0.0001, 0.000001, . . .
   ♦ Let us divide 0.0001 by it’s preceding term 0.01. We will get:
$\frac{0.0001}{0.01}~=~\frac{0.0001 × 100}{0.01 × 100}~=~\frac{0.01}{1}~=~ 0.01 $
   ♦ Let us divide 0.000001 by it’s preceding term 0.0001. We will get:
$\frac{0.000001}{0.0001}~=~\frac{0.000001 × 10000}{0.0001 × 10000}~=~\frac{0.01}{1}~=~ 0.01 $
• In this way, we can divide any term (except the first term) of this sequence by it’s preceding term. The result will be always 0.01
4. Note that, none of the terms in such sequences can be zero.
• The reason can be written in 5 steps:
(i) Suppose that one of the terms of the sequence is zero.
(ii) To find the constant ratio, we will want to divide the succeeding term of that zero by zero.
(iii) Division by zero will give a number that does not exist.
(iv) A number that does not exist cannot be the constant ratio.
(v) So we say that, none of the terms in such a sequence must be zero.
5. Now we can write the definition of geometric progression:
• A sequence a₁, a₂, a₃, . . . , a_n, . . . is called geometric progression, if each term is non-zero and $\frac{a_{k+1}}{a_k}~=~r~(\text{a constant})~\text{for}~k~\ge~1$
• A geometric progression is also called geometric sequence.
• Geometric progression is abbreviated as G.P
• The first term of a G.P is usually denoted by the letter a
• The constant ratio is denoted by the letter r. It is called the common ratio.
• The number of terms in a G.P is denoted by the letter n
• The last term of a G.P is denoted by the letter l
• The sum of n terms of a G.P is denoted by S_n
6. Based on the above information, a G.P will be in the form:
a, ar, ar², ar³, . . .

---

General term of a G.P

• This can be written in 6 steps:
1. We know that, first term of a G.P is a. That is., a₁ = a
• This can be written as: a₁ = a = ar⁰ = ar^1-1
2. We know that, second term of a G.P is ar. That is., a₂ = ar
• This can be written as: a₂ = ar = ar¹ = ar^2-1
3. Similarly, third term of a G.P is ar². That is., a₃ = ar²
• This can be written as: a₃ = ar² = ar^3-1
• so on . . .
4. Let us write the above results in order:
   ♦ a₁ = ar^1-1
   ♦ a₂ = ar^2-1
   ♦ a₃ = ar^3-1
   ♦ so on . . .
• We see a pattern. Based on that pattern, we can write:
   ♦ a_n = ar^n-1
• That means, n^th term of the G.P = ar^n-1
• This expression can be used as a formula to find the n^th term.
• To apply this formula, all we need to know are:
   ♦ The first term, a
   ♦ The common ratio, r
   ♦ The position of the term, n
5. Now we can write:
• A finite G.P will of the form: a, ar¹, ar², ar³, . . . , ar^n-1.
• An  infinite G.P will of the form: a, ar¹, ar², ar³, . . . , ar^n-1, . . .
6. We can write about geometric series also:
• A finite geometric series will of the form: a+ ar¹+ ar² + ar³ + . . . + ar^n-1.
• An infinite finite geometric series will of the form: a + ar¹ + ar² + ar³ + . . . + ar^n-1 + . . .

---

Sum of n terms of a G.P

Formula for sum, can be derived in 3 steps:
1. We have:
S_n = a+ ar¹+ ar² + ar³ + . . . + ar^n-1.
2. If r = 1, then the sum can be easily calculated:
S_n = a+ a(1)¹+ a(2)² + a(3)³ + . . . + a(4)^n-1.
⇒ S_n = a+ a + a + a + . . . + a. (n terms)
⇒ S_n = na
3. If r ≠ 1, then two steps are required:
(i) Multiply the expression in (1) by r. We get:
rS_n = ar + ar² + ar³ + . . . + arⁿ.
(ii) Subtract 3(i) from (1). We get:
$S_n - r S_n~=~a-a r^n$
⇒ $(1 - r) S_n~=~a(1- r^n)$
⇒ $S_n~=~\frac{a(1- r^n)}{1 - r}$
⇒ $S_n~=~\frac{a (r^n - 1)}{r-1}$
• This expression can be used as a formula to find the sum of n terms.

---

Solved example 9.9
Find the 10^th and n^th terms of the G.P. 5, 25,125, . . .
Solution:
1. We have: a_n = ar^n-1
2. In our present case:
a = 5, r = 25/5 = 5
3. So the 10^th term = 5 × 5^10-1 = 5 × 5⁹ = 5¹⁰
4. Similarly, the n^th term = 5 × 5^n-1 = 5ⁿ

Solved example 9.10
Which term of the G.P., 2,8,32, . . . up to n terms is 131072?
Solution:
1. We have: a_n = ar^n-1
2. In our present case:
a = 2, r = 8/2 = 4
3. Substituting the known values, we get:
131072 = 2 × 4^(n-1)
⇒ 65536 = 4^(n-1)
4. Factorization of 65536 is shown in fig.9.2 below:

Fig.9.2

• We can write:
$65536~=~4^5 × 8^2~=~4^5 × (4 × 2)^2 ~=~4^5 × 4^2 × 4 ~=~4^8~=~4^{n-1}$
• Thus we get: (n-1) = 8
⇒ n = 9
5. So 131072 is the 9^th term of the given G.P

Solved example 9.11
In a G.P, the 3^rd term is 24 and the 6^th term is 192. Find the 10^th term.
Solution:
1. Let a be the first term and r the common ratio.
• Then we can write: a_n = ar^n-1
2. Substituting the known values, we get:
(i) a₃ = a × r^3-1 = a × r² = 24
(ii) a₆ = a × r^6-1 = a × r⁵ = 192
3. Taking ratio, we get:
$\frac{a r^5}{a r^2}~=~\frac{192}{24}~=~8$
⇒ r³ = 8 = (2)³
⇒ r = 2
4. Substituting this value of r in 2(i), we get:
a × 2² = 24
⇒ a × 4 = 24
⇒ a = 6
5. Now we can write:
a₁₀ = a × r^10-1 = 6 × 2⁹ = 3072

Solved example 9.12
Find the sum of first n terms and the sum of first 5 terms of the geometric series:
$1+\frac{2}{3}+\frac{4}{9}+~.~.~.$
Solution:
1. In the G.P, a = 1 and r = $\frac{\frac{2}{3}}{1}~=~\frac{2}{3}$
2. We have: $S_n~=~\frac{a (r^n - 1)}{r-1}$
• Substituting the known values, we get:
$S_n~=~\frac{1 × \left[\left(\frac{2}{3} \right)^n - 1\right]}{\frac{2}{3}-1}$

⇒ $S_n~=~\frac{\frac{2^n - 3^n}{3^n}}{\frac{2-3}{3}}~=~\frac{\frac{3^n - 2^n}{3^n}}{\frac{3-2}{3}}~=~\frac{\frac{3^n - 2^n}{3^n}}{\frac{1}{3}}$

⇒ $S_n~=~\frac{3^n - 2^n}{3^n} × \frac{3}{1}~=~\frac{3^n - 2^n}{3^{n-1}}$

3. We can write the sum of first 5 terms using the above expression:
$S_5~=~\frac{3^5 - 2^5}{3^{5-1}}~=~\frac{3^5 - 2^5}{3^4}~=~\frac{211}{81}$

Solved example 9.13
How many terms of the G.P $3, \frac{3}{2}, \frac{3}{4},~.~.~.$ are needed to give the sum $\frac{3069}{512}$ ?
Solution:
1. In the G.P, a = 3 and r = $\frac{\frac{3}{2}}{3}~=~\frac{1}{2}$
2. We have: $S_n~=~\frac{a (r^n - 1)}{r-1}$
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{\frac{3069}{512}}
&{}={}& {\frac{3 × \left[\left(\frac{1}{2} \right)^n - 1\right]}{\frac{1}{2}-1}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{\Rightarrow}&{\frac{3069}{512}}
&{}={}& {\frac{\frac{3 × (1^n - 2^n)}{2^n}}{\frac{1-2}{2}}}
&{}={}& {\frac{\frac{3 × (2^n - 1^n)}{2^n}}{\frac{2-1}{2}}}
&{}={}& {\frac{\frac{3 × (2^n - 1)}{2^n}}{\frac{1}{2}}}
&{}& {}&{}& {} &{} \\


{\Rightarrow}&{\frac{3069}{512}}
&{}={}& {\frac{3 × (2^n - 1)}{2^n} × \frac{2}{1}}
&{}={}& {\frac{6 × (2^n - 1)}{2^n}}
&{}& {}
&{}& {}&{}& {} &{} \\

{\Rightarrow}&{\frac{3069}{512 × 6}}
&{}={}& {\frac{(2^n - 1)}{2^n}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{\Rightarrow}&{1023 × 2^n}
&{}={}& {1024 × 2^n~-1024}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{\Rightarrow}&{2^n}
&{}={}& {1024}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

3. Factorization of 1024 is shown in fig.9.3 below:

Fig.9.3

• We can write: $2^n~=~1024~=~4^2 × 8^2 ~=~2^4 × 2^6 ~=~2^{10}$
• Thus we get: n = 10
• That means, 10 terms are needed to obtain the given sum.

Solved example 9.14
The sum of first three terms of a G.P is $\frac{13}{12}$ and their product is -1. Find the common ratio and the terms.
Solution:
1. Let the first three terms be: $\frac{a}{r}, a~\text{and}~ar$
Then we can write:
(i) $\frac{a}{r}+a+ar~=~\frac{13}{12}$
(ii) $\frac{a}{r} × a × ar~=~a^3~=~-1$
2. From 1(ii), we get: a = -1
Substituting this value of a in 1(i), we get:
$\frac{-1}{r}+(-1)+(-1)r~=~\frac{13}{12}$
⇒ $\frac{-1}{r} - 1 - r~=~\frac{13}{12}$
3. The above expression can be simplified as follows:
$\begin{array}{ll}
{}&{\frac{-1}{r} - 1 - r}
&{}={}& {\frac{13}{12}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{\Rightarrow}&{\frac{-1-r-r^2}{r}}
&{}={}& {\frac{13}{12}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{\Rightarrow}&{-12-12r-12r^2}
&{}={}& {13r}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{\Rightarrow}&{12r^2 +25r + 12}
&{}={}& {0}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

4. Solving the above quadratic equation, we get:
$r~=~\frac{-3}{4},~\frac{-4}{3}$
5. Now we can write the G.P:
(i) $\text{When}~r~=~\frac{-3}{4},~\text{the G.P is:}~\frac{-1}{\frac{-3}{4}},-1, -1 × \frac{-3}{4}$
• That is., $\frac{4}{3}, -1, \frac{3}{4}$

(ii) $\text{When}~r~=~\frac{-4}{3},~\text{the G.P is:}~\frac{-1}{\frac{-4}{3}},-1, -1 × \frac{-4}{3}$
• That is., $\frac{3}{4}, -1, \frac{4}{3}$

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In the next section we will see two more solved examples. We will also see geometric mean.

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Chapter 9.2 - Arithmetic Progression

In the previous section, we completed a discussion on the n^th term of sequences and series. In this section, we will see arithmetic progression.

We have seen the basics about arithmetic progression in our earlier classes [Details here]
Let us recall some of the formulae and properties that we studied in those classes. It can be written in steps:
1. Arithmetic progression can be written in abbreviated form as A.P
2. First term of an A.P is denoted by the letter a
• Last term of an A.P is denoted by the letter l
• Common difference of an A.P is denoted by the letter d
3. The n^th term of an A.P is given by: a_n = a+(n-1)d
4. If there is a total of m terms in an A.P, then the last term (the m^th term) will be a+(m-1)d
• So we can write l = a+(m-1)d  
5. If there are m terms in an A.P, then that A.P will be:
a, (a+d), (a+2d), (a+3d), . . . , a+(m-1)d
6. The sum of n terms of an A.P is denoted as S_n
7. Sum can be calculated using the formula: $S_n=\frac{n}{2}[2a+(n-1)d]$
8. Another formula for the sum is: $S_n=\frac{n}{2}[a+l]$
9. Since the n^th term can be calculated using an algebraic formula, an A.P is a sequence.
10. If a constant is added to each term of an A.P, the resulting sequence is also an A.P
11. If a constant is subtracted from each term of an A.P, the resulting sequence is also an A.P
12. If each term of an A.P is multiplied by a constant, the resulting sequence is also an A.P
13. If each term of an A.P is divided by a non-zero constant, the resulting sequence is also an A.P

---

Let us see some solved examples:

Solved example 9.4
In an A.P, if the m^th term is n and the n^th term is m, where m≠n, find the p^th term.
Solution:
1. The n^th term of any A.P is a+(n-1)d
• 'a' and 'd' are constants. If we can find those constants, we will be able to write the term at any position.
• To find those constants, we can use the given data.
2. Given that, m^th term is n
• So we can write: n = a+(m-1)d
⇒ a+md - d = n
3. Given that, n^th term is m
• So we can write: m = a+(n-1)d
⇒ a+nd - d = m
4. Subtracting the result in (3) from the result in (2), we get:
⇒ n-m = a+md-d - [a+nd-d]
⇒ n-m = (m-n)d
⇒ n-m = -(n-m)d
⇒ d = -1
5. Substituting this value of d in (2), we get:
a+m(-1) - (-1) = n
⇒ a-m+1 = n
⇒ a = m+n-1
6. Now we can write the p^th term:
p^th term = a+(p-1)d = (m+n-1) + (p-1) × -1 = m+n-1 - p +1 = m+n-p

Solved example 9.5
Sum of n terms of two arithmetic progressions are in the ratio (3n+8):(7n+15). Find the ratio of their 12^th terms.
Solution:
1. Let the first term and common difference of the first A.P be a and d respectively.
2. Let the first term and common difference of the second A.P be a’ and d’ respectively.
3. Then the given ration of sums can be written as:
$\frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2a'+(n-1)d1]}=\frac{3n+8}{7n+15}$
⇒ $\frac{2a+(n-1)d}{2a'+(n-1)d'}=\frac{3n+8}{7n+15}$
4. We want the ratio of the 12^th terms. That is., we want $\frac{a+11d}{a'+11d'}$
5. The left side of the result in (3) can be converted into $\frac{a+11d}{a'+11d'}$ by putting n = 23
• So the result in (3) becomes:
$\frac{2a+(23-1)d}{2a'+(23-1)d'}=\frac{3 × 23+8}{7 × 23+15}$  
⇒ $\frac{2a+22d}{2a'+22d'}=\frac{3 × 23+8}{7 × 23+15}$  
⇒ $\frac{a+11d}{a'+11d'}=\frac{77}{176}$  
⇒ $\frac{a+11d}{a'+11d'}=\frac{7}{16}$
6. Let us compare the results in (4) and (5)
• We see that, the left side is the same.
• So we can write: The ratio of the 12^th terms is $\frac{7}{16}$

Solved example 9.6
The income of a person is Rs. 3,00,000, in the first year and he receives an increase of Rs. 10,000 to his income per year for the next 19 years. Find the total amount he received in 20 years.
Solution:
1. Let us write the salaries in order:
• Salary in the first year = 300000
• Salary in the second year = (300000 + 10000) =  300000 + 1 × 10000   
• Salary in the third year = (300000 + 10000 + 10000) = 300000 + 2 × 10000
• Salary in the fourth year = (300000 + 10000 + 10000 + 10000) = 300000 + 3 × 10000
• so on . . .
2. Writing it as a sequence, we get:
300000, (300000 + 1 × 10000), (300000 + 2 × 10000), (300000 + 3 × 10000), . . .
3. Clearly, this is an A.P
• First term, a = 300000
• Common difference, d = 10000
• Number of terms, n = 20
4. Sum of all 20 terms
= $\frac{n}{2}[2a+(n-1)d]~=~\frac{20}{2}[2 × 300000+(20-1) × 10000]$        
= $10 × [600000+19 × 10000]~=~79,00,000$

---

Arithmetic mean

This can be written in 5 steps:
1. Suppose that, we are given two numbers a and b
• We can find a number A in such a way that, a, A , b form an AP
2. This A can be calculated in two simple steps:
(i) Since a, A, b is an A.P, we can write: A-a = b-A
(ii) From this we get: 2A = a+b
⇒ $A=\frac{a+b}{2}$
3. This A is called the Arithmetic mean of a and b.
• Arithmetic mean is abbreviated as A.M
4. But $\frac{a+b}{2}$ is the average of a and b
• So we can write: A.M of two numbers a and b is same as the average of a and b.
• Let us see an example:
    ♦ Let the two numbers a and b be 25 and 275
    ♦ Then the A.M of 25 and 275 = $\frac{25+275}{2}$ = 150
5. We inserted just one number A between a and b. In fact we can insert as many numbers as we like between two numbers a and b so that the resulting sequence is an A.P.
• This can be explained in steps:
(i) Let us insert n numbers A₁, A₂, A₃, A₄, . . . , A_n between a and b in such a way that,
a, A₁, A₂, A₃, A₄, . . . , A_n, b is an A.P
(ii) When n numbers are inserted, there will be a total of (n+2) terms.
• So b is the (n+2)^th term of the A.P
• We can write: b = a+[(n+2)-1]d
⇒ (n+1)d = b-a
⇒ $d=\frac{b-a}{n+1}$
(iii) So we have the first term a and the common difference d of the A.P. Using them, we can calculate the term at any position.
• That is., we can calculate A₁, A₂, A₃, A₄, . . . , A_n and insert them between a and b
• For example, A₃ is the fourth term. It will be equal to: $a+(4-1)\left(\frac{b-a}{n+1}\right)$
    ♦ Here n is number of terms inserted in between a and b

---

Let us see a solved example:

Solved example 9.7
Insert 6 numbers between 3 and 24 such that the resulting sequence is an A.P
Solution:
1. The resulting A.P will be in the form: 3, A₁, A₂, A₃, . . . A₆, 24
    ♦ The first term of this A.P is 3
    ♦ The last term is 24
    ♦ Total number of terms = 8
2. So we can write:
24 = 3+(8-1)d
⇒ d = 3
3. Now we can write all the intermediate terms:
2^nd term A₁ = 3 + (2-1) × 3 = 6
3^rd term A₂ = 3 + (3-1) × 3 = 9
4^th term A₃ = 3 + (4-1) × 3 = 12
5^th term A₄ = 3 + (5-1) × 3 = 15
6^th term A₅ = 3 + (6-1) × 3 = 18
7^th term A₆ = 3 + (7-1) × 3 = 21
4. So the resulting A.P is: 3, 6, 9, 12, 15, 18, 21, 24

---

The link below gives a few more solved examples

Exercise 9.2

---

In the next section we will see arithmetic progression.

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Chapter 9.1 - Solved Examples On Sequences And Series

In the previous section, we saw the basics about sequences and series. In this section, we will see some solved examples.

Solved example 9.1
(i) Write the first three terms of the sequence defined by the formula: $a_n=2n+5$
(ii) Write the first three terms of the sequence defined by the formula: $a_n=\frac{n-3}{4}$
Solution:
We can find the first three terms by substituting n = 1, 2 and 3  
Part (i):
The first three terms are:
$a_1~=~(2 × 1)+5~=~7$
$a_2~=~(2 × 2)+5~=~9$
$a_3~=~(2 × 3)+5~=~11$
Part (ii):
The first three terms are:
$a_1~=~\frac{1-3}{4}~=~\frac{-2}{4}~=~-\frac{1}{2}$
$a_2~=~\frac{2-3}{4}~=~\frac{-1}{4}~=~-\frac{1}{4}$
$a_3~=~\frac{3-3}{4}~=~\frac{0}{4}~=~0$

Solved example 9.2
What is the 20^th term of the sequence defined by $a_n=(n-1)(2-n)(3+n)$
Solution:
We can find the 20^th term by substituting n = 20
So we get: $a_{20}~=~(20-1)(2-20)(3+20)~=~19 × -18 × 23~=~-7866$

Solved example 9.3
Let the sequence a_n be defined as follows:
$a_1=1,~a_n=a_{n-1}+2~~\text{for}~n \geq 2$
Find the first five terms and write the corresponding series.
Solution:
1. The first five terms are:
$a_1~=~1$
$a_2=a_{2-1}+2~=~a_1+2~=~1+2~=~3$
$a_3=a_{3-1}+2~=~a_2+2~=~3+2~=~5$
$a_4=a_{4-1}+2~=~a_3+2~=~5+2~=~7$
$a_5=a_{5-1}+2~=~a_4+2~=~7+2~=~9$
2. So the sequence is: 1, 3, 5, 7, 9, . . .
3. So the series associated with this sequence is:
1 + 3 + 5 + 7 + 9 + . . .

---

Exercise 9.1

1. Write the first five terms of the sequence whose n^th term is $a_n=n(n+2)$
Solution:
$\begin{array}{ll}
{a_1}&{}={}
&{1 × (1+2)}& {}={}
&3& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_2}&{}={}
&{2 × (2+2)}& {}={}
&8& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_3}&{}={}
&{3 × (3+2)}& {}={}
&15& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_4}&{}={}
&{4 × (4+2)}& {}={}
&24& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_5}&{}={}
&{5 × (5+2)}& {}={}
&35& {}
&{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

2. Write the first five terms of the sequence whose n^th term is $a_n=\frac{n}{n+1}$
Solution:
$\begin{array}{ll}
{a_1}&{}={}
&{\frac{1}{1+1}}& {}={}
&\frac{1}{2}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_2}&{}={}
&{\frac{2}{2+1}}& {}={}
&\frac{2}{3}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_3}&{}={}
&{\frac{3}{3+1}}& {}={}
&\frac{3}{4}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_4}&{}={}
&{\frac{4}{4+1}}& {}={}
&\frac{4}{5}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_5}&{}={}
&{\frac{5}{5+1}}& {}={}
&\frac{5}{6}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

3. Write the first five terms of the sequence whose n^th term is $a_n=2^n$
Solution:
$\begin{array}{ll}
{a_1}&{}={}
&{2^1}& {}={}
&2& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_2}&{}={}
&{2^2}& {}={}
&4& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_3}&{}={}
&{2^3}& {}={}
&8& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_4}&{}={}
&{2^4}& {}={}
&16& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_5}&{}={}
&{2^5}& {}={}
&32& {}
&{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

4. Write the first five terms of the sequence whose n^th term is $a_n=\frac{2n-3}{6}$
Solution:
$\begin{array}{ll}
{a_1}&{}={}
&{\frac{(2 × 1)-3}{6}}& {}={}
&\frac{2-3}{6}& {}={}
&{-\frac{1}{6}}& {}
&{}& {}&{}& {} &{} \\

{a_2}&{}={}
&{\frac{(2 × 2)-3}{6}}& {}={}
&\frac{4-3}{6}& {}={}
&{\frac{1}{6}}& {}
&{}& {}&{}& {} &{} \\

{a_3}&{}={}
&{\frac{(2 × 3)-3}{6}}& {}={}
&\frac{6-3}{6}& {}={}
&{\frac{3}{6}}& {}
&{}& {}&{}& {} &{} \\

{a_4}&{}={}
&{\frac{(2 × 4)-3}{6}}& {}={}
&\frac{8-3}{6}& {}={}
&{\frac{5}{6}}& {}
&{}& {}&{}& {} &{} \\

{a_5}&{}={}
&{\frac{(2 × 5)-3}{6}}& {}={}
&\frac{10-3}{6}& {}={}
&{\frac{7}{6}}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

5. Write the first five terms of the sequence whose n^th term is $a_n=(-1)^{n-1}~5^{n+1}$
Solution:
$\begin{array}{ll}
{a_1}&{}={}
&{(-1)^{1-1} × 5^{1+1}}& {}={}
&{(-1)^{0} × 5^{2}}& {}={}
&{1 × 25}& {}={}
&{25}& {}&{}& {} &{} \\

{a_2}&{}={}
&{(-1)^{2-1} × 5^{2+1}}& {}={}
&{(-1)^{1} × 5^{3}}& {}={}
&{-1 × 125}& {}={}
&{-125}& {}&{}& {} &{} \\

{a_3}&{}={}
&{(-1)^{3-1} × 5^{3+1}}& {}={}
&{(-1)^{2} × 5^{4}}& {}={}
&{1 × 625}& {}={}
&{625}& {}&{}& {} &{} \\

{a_4}&{}={}
&{(-1)^{4-1} × 5^{4+1}}& {}={}
&{(-1)^{3} × 5^{5}}& {}={}
&{-1 × 3125}& {}={}
&{-3125}& {}&{}& {} &{} \\

{a_5}&{}={}
&{(-1)^{5-1} × 5^{5+1}}& {}={}
&{(-1)^{4} × 5^{6}}& {}={}
&{1 × 15625}& {}={}
&{15625}& {}&{}& {} &{} \\

\end{array}$

6. Write the first five terms of the sequence whose n^th term is $a_n=n\frac{n^2+5}{4}$
Solution:
$\begin{array}{ll}
{a_1}&{}={}
&{1 × \frac{1^2+5}{4}}& {}={}
&{1 × \frac{1+5}{4}}& {}={}
&{1 × \frac{6}{4}}& {}={}
&{\frac{6}{4}}& {}&{}& {} &{} \\

{a_2}&{}={}
&{2 × \frac{2^2+5}{4}}& {}={}
&{2 × \frac{4+5}{4}}& {}={}
&{2 × \frac{9}{4}}& {}={}
&{\frac{9}{2}}& {}&{}& {} &{} \\

{a_3}&{}={}
&{3 × \frac{3^2+5}{4}}& {}={}
&{3 × \frac{9+5}{4}}& {}={}
&{3 × \frac{14}{4}}& {}={}
&{\frac{42}{4}}& {}&{}& {} &{} \\

{a_4}&{}={}
&{4 × \frac{4^2+5}{4}}& {}={}
&{4 × \frac{16+5}{4}}& {}={}
&{4 × \frac{21}{4}}& {}={}
&{21}& {}&{}& {} &{} \\

{a_5}&{}={}
&{5 × \frac{5^2+5}{4}}& {}={}
&{5 × \frac{25+5}{4}}& {}={}
&{5 × \frac{30}{4}}& {}={}
&{\frac{150}{4}}& {}&{}& {} &{} \\

\end{array}$

7. Find a₁₇ and a₂₄ of the sequence whose n^th term is $a_n=4n-3$
Solution:
$\begin{array}{ll}
{a_{17}}&{}={}
&{(4 × 17)-3}& {}={}
&{68-3}& {}={}
&{65}& {}
&{}& {}&{}& {} &{} \\

{a_{24}}&{}={}
&{(4 × 24)-3}& {}={}
&{96-3}& {}={}
&{93}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

8. Find a₇ of the sequence whose n^th term is $a_n=\frac{n^2}{2^n}$
Solution:
$\begin{array}{ll}
{a_{7}}&{}={}
&{\frac{7^2}{2^7}}& {}={}
&{\frac{49}{128}}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$  

9. Find a₉ of the sequence whose n^th term is $a_n=(-1)^{n-1} n^3$
Solution:
$\begin{array}{ll}
{a_{9}}&{}={}
&{(-1)^{9-1} × 9^3}& {}={}
&{(-1)^{8} × 729}& {}={}
&{1 × 729}& {}={}
&{729}& {}&{}& {} &{} \\

\end{array}$

10. Find a₂₀ of the sequence whose n^th term is $a_n=\frac{n(n-2)}{n+3}$
Solution:
$\begin{array}{ll}
{a_{20}}&{}={}
&{\frac{20(20-2)}{20+3}}& {}={}
&{\frac{20 × 18}{23}}& {}={}
&{\frac{360}{23}}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

11. Write the first five terms of the following sequence and obtain the corresponding series.
$a_1=3,~a_n=3a_{n-1}+2~~\text{for all}~n \gt 1$
Solution:
1. The first five terms are:
$\begin{array}{ll}
{a_{1}}&{}={}
&{3}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_{2}}&{}={}
&{3 × a_{2-1}+2}& {}={}
&{3 × a_{1}+2}& {}={}
&{(3 × 3)+2}& {}={}
&{9+2}& {}={}&{11}& {} &{} \\

{a_{3}}&{}={}
&{3 × a_{3-1}+2}& {}={}
&{3 × a_{2}+2}& {}={}
&{(3 × 11)+2}& {}={}
&{33+2}& {}={}&{35}& {} &{} \\

{a_{4}}&{}={}
&{3 × a_{4-1}+2}& {}={}
&{3 × a_{4}+2}& {}={}
&{(3 × 35)+2}& {}={}
&{105+2}& {}={}&{107}& {} &{} \\

{a_{5}}&{}={}
&{3 × a_{5-1}+2}& {}={}
&{3 × a_{4}+2}& {}={}
&{(3 × 107)+2}& {}={}
&{321+2}& {}={}&{323}& {} &{} \\

\end{array}$

2. So the sequence is: 3, 11, 35, 107, 323, . . .
3. So the series associated with this sequence is:
3 + 11 + 35 + 107 + 323, . . .

12. Write the first five terms of the following sequence and obtain the corresponding series.
$a_1=-1,~a_n=\frac{a_{n-1}}{n}~~\text{for all}~n \ge 2$
Solution:
1. The first five terms are:
$\begin{array}{ll}
{a_{1}}&{}={}
&{-1}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_{2}}&{}={}
&{\frac{a_{2-1}}{2}}& {}={}
&{\frac{a_{1}}{2}}& {}={}
&{\frac{-1}{2}}& {}
&{}& {}&{}& {} &{} \\

{a_{3}}&{}={}
&{\frac{a_{3-1}}{3}}& {}={}
&{\frac{a_{2}}{3}}& {}={}
&{\frac{\frac{-1}{2}}{3}}& {}={}
&{\frac{-1}{6}}& {}&{}& {} &{} \\

{a_{4}}&{}={}
&{\frac{a_{4-1}}{4}}& {}={}
&{\frac{a_{3}}{4}}& {}={}
&{\frac{\frac{-1}{6}}{4}}& {}={}
&{\frac{-1}{24}}& {}&{}& {} &{} \\

{a_{5}}&{}={}
&{\frac{a_{5-1}}{5}}& {}={}
&{\frac{a_{4}}{5}}& {}={}
&{\frac{\frac{-1}{24}}{5}}& {}={}
&{\frac{-1}{120}}& {}&{}& {} &{} \\

\end{array}$

2. So the sequence is: $-1, \frac{-1}{2}, \frac{-1}{6}, \frac{-1}{24}, \frac{-1}{120}, ~.~.~.$
3. So the series associated with this sequence is:
$-1 + \frac{-1}{2} + \frac{-1}{6} + \frac{-1}{24} + \frac{-1}{120} ~.~.~.$

13. Write the first five terms of the following sequence and obtain the corresponding series.
$a_1=a_2=2,~a_n=a_{n-1}-1~~\text{for all}~n \gt 2$
Solution:
1. The first five terms are:
$\begin{array}{ll}
{a_{1}}&{}={}
&{2}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_{2}}&{}={}
&{2}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_{3}}&{}={}
&{a_{3-1}-1}& {}={}
&{a_{2}-1}& {}={}
&{2-1}& {}={}
&{1}& {}&{}& {} &{} \\

{a_{4}}&{}={}
&{a_{4-1}-1}& {}={}
&{a_{3}-1}& {}={}
&{1-1}& {}={}
&{0}& {}&{}& {} &{} \\

{a_{5}}&{}={}
&{a_{5-1}-1}& {}={}
&{a_{4}-1}& {}={}
&{0-1}& {}={}
&{-1}& {}&{}& {} &{} \\

\end{array}$

2. So the sequence is: 2, 2, 1, 0, -1, . . .
3. So the series associated with this sequence is:
2 + 2 + 1 + 0 + (-1), . . .

14. The Fibonacci sequence is defined by:
$a_1=a_2=1,~a_n=a_{n-1}~+~a_{n-2}~~\text{for all}~n \gt 2$
Find $\frac{a_{n+1}}{a_n}$ for n = 1, 2, 3, 4, 5
Solution:
1. The first six terms are:
$\begin{array}{ll}
{a_{1}}&{}={}
&{1}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_{2}}&{}={}
&{1}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_{3}}&{}={}
&{a_{3-1}~+~a_{3-2}}& {}={}
&{a_{2}~+~a_{1}}& {}={}
&{1+1}& {}={}
&{2}& {}&{}& {} &{} \\

{a_{4}}&{}={}
&{a_{4-1}~+~a_{4-2}}& {}={}
&{a_{3}~+~a_{2}}& {}={}
&{2+1}& {}={}
&{3}& {}&{}& {} &{} \\

{a_{5}}&{}={}
&{a_{5-1}~+~a_{5-2}}& {}={}
&{a_{4}~+~a_{3}}& {}={}
&{3+2}& {}={}
&{5}& {}&{}& {} &{} \\

{a_{6}}&{}={}
&{a_{6-1}~+~a_{6-2}}& {}={}
&{a_{5}~+~a_{4}}& {}={}
&{5+3}& {}={}
&{8}& {}&{}& {} &{} \\

\end{array}$

2. So the required ratios are:
$\begin{array}{ll}
{\text{When n = 1,}~\frac{a_{n+1}}{a_n}}&{}={}
&{\frac{a_{1+1}}{a_1}}& {}={}
&{\frac{a_{2}}{a_1}}& {}={}
&{\frac{1}{1}}& {}
&{}& {}&{}& {} &{} \\

{\text{When n = 2,}~\frac{a_{n+1}}{a_n}}&{}={}
&{\frac{a_{2+1}}{a_2}}& {}={}
&{\frac{a_{3}}{a_2}}& {}={}
&{\frac{2}{1}}& {}
&{}& {}&{}& {} &{} \\

{\text{When n = 3,}~\frac{a_{n+1}}{a_n}}&{}={}
&{\frac{a_{3+1}}{a_3}}& {}={}
&{\frac{a_{4}}{a_3}}& {}={}
&{\frac{3}{2}}& {}
&{}& {}&{}& {} &{} \\

{\text{When n = 4,}~\frac{a_{n+1}}{a_n}}&{}={}
&{\frac{a_{4+1}}{a_4}}& {}={}
&{\frac{a_{5}}{a_4}}& {}={}
&{\frac{5}{3}}& {}
&{}& {}&{}& {} &{} \\

{\text{When n = 5,}~\frac{a_{n+1}}{a_n}}&{}={}
&{\frac{a_{5+1}}{a_5}}& {}={}
&{\frac{a_{6}}{a_5}}& {}={}
&{\frac{8}{5}}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

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In the next section we will see arithmetic progression.

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Chapter 9 - Sequences And Series

In the previous section, we completed a discussion on binomial theorem. In this  chapter, we will see sequences and series.

The word sequence can be explained using three examples:
Example 1:
This can be written in 4 steps:
1. Consider a bundle of freshly printed currency notes.
2. Let there be a hundred notes in the bundle.
• Each note in that bundle will have a unique number.
3. Suppose that, the 14^th note from the top has the number 175214
    ♦ Then the 15^th note from the top will have the number 175215
    ♦ The 16^th note from the top will have the number 175216
    ♦ so on . . .
4. We say that, the currency notes in the bundle are arranged in a sequential order.
• The sequence can be written as:
175201, 175202, 175203, . . . , 175214, 175215, 175216, 175217, . . . , 175299, 175300.

Example 2
This can be written in 3 steps:
1. Suppose that, we want to make equilateral triangles with identical spheres. Then the arrangement will be as shown in fig.9.1 below:

Fig.9.1

2. The first triangle will contain 3 spheres
    ♦ The second triangle will contain 6 spheres
    ♦ The third triangle will contain 10 spheres
    ♦ The fourth triangle will contain 15 spheres
    ♦ so on . . .
3. This is the only way to form equilateral triangles using identical spheres.
• So the number of spheres in the triangles form a sequence.
• The sequence can be written as: 3, 6, 10, 15, . . .

Example 3
• Suppose that in a society, the generation gap is 30 years. That is., a person will have children when he/she is 30 years old.
• We want to find the total number of members of a family in each generation for 300 years.
• This can be calculated in 13 steps:
1. Let the total number of members in 300 years be x.
2. Consider a person whose age at present is 30.
• Then 30 years ago, he was not born. His family had upto his father and mother only. So 30 years ago, the number of members will be (x-1)
3. 60 years ago, his father and mother were not born.
• So 60 years ago, the number of members would be [(x-1)-2] = (x-3)
4. 90 years ago, the parents of the persons deducted in (3) were not born.
• Two persons were deducted in (3).
• There would be four parents for those two persons.
• So 90 years ago, the number of members would be [(x-3)-4] = (x-7)
5. 120 years ago, the parents of the persons deducted in (4) were not born.
• 4 persons were deducted in (4).
• There would be 8 parents for those 4 persons.
• So 120 years ago, the number of members would be [(x-7)-8] = (x-15)  
6. 150 years ago, the parents of the persons deducted in (5) were not born.
• 8 persons were deducted in (5).
• There would be 16 parents for those 8 persons.
• So 150 years ago, the number of members would be [(x-15)-16] = (x-31) 
7. 180 years ago, the parents of the persons deducted in (6) were not born.
• 16 persons were deducted in (5).
• There would be 32 parents for those 16 persons.
• So 180 years ago, the number of members would be [(x-31)-32] = (x-63) 
8. 210 years ago, the parents of the persons deducted in (7) were not born.
• 32 persons were deducted in (7).
• There would be 64 parents for those 32 persons.
• So 210 years ago, the number of members would be [(x-63)-64] = (x-127)
9. 240 years ago, the parents of the persons deducted in (8) were not born.
• 64 persons were deducted in (8).
• There would be 128 parents for those 64 persons.
• So 240 years ago, the number of members would be [(x-127)-128] = (x-255)
10. 270 years ago, the parents of the persons deducted in (9) were not born.
• 128 persons were deducted in (9).
• There would be 256 parents for those 128 persons.
• So 270 years ago, the number of members would be [(x-255)-256] = (x-511)
11. 300 years ago, the parents of the persons deducted in (10) were not born.
• 256 persons were deducted in (10).
• There would be 512 parents for those 256 persons.
• So 300 years ago, the number of members would be [(x-511)-512] = (x-1023)
12. But 300 years ago, there will be just two members. A father and mother.
• So we can write: x-1023 = 2
• So x = 1025
• That is., total numbers of members in 300 years = 1025
• We can write: The present person under consideration has 1024 ancestors.
13. Now consider the number of parents, grand parents, great grand parents etc.,
• We get:
    ♦ Number of parents = 2
    ♦ Number of grand parents = 4
    ♦ Number of great grand parents = 8
    ♦ so on . . .
• These numbers form a sequence. The sequence is:
2, 4, 8, 16, 32, . . .

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• Let us see some features related to sequences. They can be written in steps:
1. The various numbers occurring in a sequence are called terms of the sequence.
• For example, we can write:
‘8’ is a term of the sequence 2, 4, 8, 16, 32, . . .
2. The terms are denoted by the letter a.
• A subscript is also given to ‘a’. The subscript will indicate the position of the term.
• For example, we can write:
In the sequence 2, 4, 8, 16, 32, . . . , a₄ = 16
3. a_n denotes the term at the n^th position in the sequence. This term is also called the general term of the sequence.
4. If the number of terms in a sequence is finite, then that sequence is called a finite sequence.
• For example, the sequence in example 1 that we saw above, is a finite sequence. This is because, the number of terms in that sequence is 100, which is a finite number. (Something that is finite has a definite fixed size or extent)
• The sequence in example 3 is also finite because, we are considering the number of generations within 300 years.
5. If the number of terms in a sequence is infinite, then that sequence is called a infinite sequence.
• For example, the sequence in example 2 that we saw above, is an infinite sequence. This is because, infinite number of such triangles are possible and correspondingly, infinite number of terms will be present in that sequence. (Something that is infinite, is limitless or endless in space, extent, or size. It is impossible to count, measure or calculate an infinite quantity)

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Algebraic formula for the n^th term of a sequence

This can be explained with the help of some examples.
Example 1:
This can be written in 4 steps:
1. Consider the sequence 2, 4, 6, 8, . . .
• It is the sequence of even natural numbers.
2. We have:
• a₁ = 2
    ♦ 2 can be obtained by multiplying the ‘position number 1’ by 2
• a₂ = 4
    ♦ 4 can be obtained by multiplying the ‘position number 2’ by 2
• a₃ = 6
    ♦ 6 can be obtained by multiplying the ‘position number 3’ by 2
• a₄ = 8
    ♦ 8 can be obtained by multiplying the ‘position number 4’ by 2
so on . . .
3. So it is clear that, any term in this sequence can be obtained by multiplying it’s ‘position number’ by 2
That is., a_n = n × 2
4. We can write:
The algebraic formula for the n^th term of the sequence 2, 4, 6, 8, . . . is: a_n = 2n

Example 2:
This can be written in 4 steps:
1. Consider the sequence 1, 3, 5, 7, . . .
• It is the sequence of odd natural numbers.
2. We have:
• a₁ = 1
    ♦ 1 can be obtained by multiplying the ‘position number 1’ by 2 and then subtracting 1
    ♦ That is., 1 = (1 × 2) - 1
• a₂ = 3
    ♦ 3 can be obtained by multiplying the ‘position number 2’ by 2 and then subtracting 1
    ♦ That is., 3 = (2 × 2) - 1
• a₃ = 5
    ♦ 1 can be obtained by multiplying the ‘position number 3’ by 2 and then subtracting 1
    ♦ That is., 5 = (3 × 2) - 1
• a₄ = 7
    ♦ 7 can be obtained by multiplying the ‘position number 4’ by 2 and then subtracting 1
    ♦ That is., 7 = (4 × 2) - 1
so on . . .
3. So it is clear that, any term in this sequence can be obtained by multiplying it’s ‘position number’ by 2 and then subtracting 1
That is., a_n = n × 2 - 1
4. We can write:
The algebraic formula for the n^th term of the sequence 1, 3, 5, 7, . . . is:
a_n = 2n - 1  

Example 3:
This can be written in 5 steps:
1. Consider the sequence 1, 1, 2, 3, 5, 8, . . .
• Here we do not see a visible pattern.
2. We have:
• a₁ = 1
• a₂ = 1
• a₃ = 2
    ♦ 2 can be obtained by adding a₂ and a₁.
    ♦ a₂ and a₁ are the two terms coming just before a₃
• a₄ = 3
    ♦ 3 can be obtained by adding a₃ and a₂.
    ♦ a₃ and a₂ are the two terms coming just before a₄.
• a₅ = 5
    ♦ 5 can be obtained by adding a₄ and a₃.
    ♦ a₄ and a₃ are the two terms coming just before a₅.
• a₆ = 8
    ♦ 8 can be obtained by adding a₅ and a₄.
    ♦ a₅ and a₄ are the two terms coming just before a₆.
so on . . .
3. So it is clear that, any term in this sequence can be obtained by adding the two terms coming just before it.
That is., a_n = a_n-1 + a_n-2.
4. We can write:
The algebraic formula for the n^th term of the sequence 1, 1, 2, 3, 5, 8, . . . is:
a_n = a_n-1 + a_n-2, n > 2
• Note that, n must be greater than 2. If we put n = 2, then a_n-2 will denote the zero^th term, which is not available.
5. This sequence is called Fibonacci sequence.

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• Now we have a basic idea about the algebraic formula for the n^th term.
• In the formula, we put natural numbers starting from 1 on the left side. We get corresponding values on the right side.
• Based on this information, we can consider a sequence as a function of the form a(n)
• In some cases, the domain will be the set of natural numbers. {1, 2, 3, 4, . . .}
    ♦ Then it will be an infinite sequence.
• In some cases, the domain will be a subset of natural numbers. {1, 2, 3, 4, . . . k}
    ♦ Then it will be a finite sequence.
• The range will be the terms of the sequence.

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• There are some sequences which do not have an algebraic formula for the n^th term.
• For example, consider the sequence 2, 3, 5, 7, 11, 13, 17, . . .
• It is the sequence of prime numbers. There is no algebraic formula to obtain the terms of this sequence.
• In such cases, we describe the sequence by a verbal description. This verbal description is called the rule for generating the terms of the sequence.

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Series

Basics about series can be written in steps:
1. Let a₁, a₂, a₃, a₄, . . . a_n be a given sequence.
• Then the expression a₁ + a₂ + a₃ + a₄ + . . . a_n + . . . is called the series associated with the given sequence.
2. Note that the expression a₁ + a₂ + a₃ + a₄ + . . . a_n + . . . is the series.
• We must not consider the actual sum as the series.
• The actual sum is referred to as sum of the series.
3. We can write about finite and infinite series:
• If the given sequence is finite, then the associated series is a finite series.
• If the given sequence is infinite, then the associated series is an infinite series.
4. Since the series involves a summation, we can use the sigma notation. We have seen the details about sigma notation in a previous chapter. [see section 8.1]
• So the series can be abbreviated as: $\sum\limits_{k\,=\,1}^{k\,=\,n}{a_k}$
• 'k=1' at the bottom of '∑' indicates that, the value of k starts from 1. In other words, the value of k in the first term is 1.
• 'k=n' at the top of '∑' indicates that, the value of k ends at n. In other words, the value of k in the last term is n.
• So the values of k are: 1, 2, 3, . . . n
• It is known as the summation from k = 1 to k = n.
• Let us write an example:
The series associated with the sequence of odd natural numbers can be written in two forms:
(i) 1 + 3 + 5 + 7 + . . .
(ii) $\sum\limits_{k\,=\,1}^{k\,=\,\infty}{2k-1}$

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In the next section we will see some solved examples on sequences and series.

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