Chapter 9.7 - Miscellaneous Examples on chapter 9

In the previous section, we completed a discussion on sequences and series. In this section, we will see some miscellaneous examples.

Solved example 9.21
If p^th, q^th, r^th and s^th terms of an A.P are in G.P, then show that (p-q), (q-r), (r-s) are also in G.P.
Solution:
1. We want to prove that (p-q), (q-r), (r-s) are in G.P.
• That means, we have to prove that $\frac{q-r}{p-q}~=~\frac{r-s}{q-r}$

2. Let 'a' be the first term and 'd' the common difference of the A.P.
• Then the given terms can be written as:
    ♦ p^th term = a_p = a + (p-1)d
    ♦ q^th term = a_q = a + (q-1)d
    ♦ r^th term = a_r = a + (r-1)d
    ♦ s^th term = a_s = a + (s-1)d
3. Given that, the above four terms are in G.P. So we can write:

$\begin{array}{ll}
{}&{\frac{a_q}{a_p}}
&{~=~\frac{a_r}{a_q}}& {~=~\frac{a_s}{a_r}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{a+(q-1)d}{a+(p-1)d}}
&{~=~\frac{a+(r-1)d}{a+(q-1)d}}& {~=~\frac{a+(s-1)d}{a+(r-1)d}}
&{\color {green} {\text{- - - - (a)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{a+(q-1)d~-~[a+(p-1)d]}{a+(p-1)d}}
&{~=~\frac{a+(r-1)d~-~[a+(q-1)d]}{a+(q-1)d}}& {~=~\frac{a+(s-1)d~-~[a+(r-1)d]}{a+(r-1)d}}
&{\color {green} {\text{- - - - (b)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{a+qd-d-a-pd+d}{a+(p-1)d}}
&{~=~\frac{a+rd-d-a-qd+d}{a+(q-1)d}}& {~=~\frac{a+sd-d-a-rd+d}{a+(r-1)d}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{(q-p)d}{a+(p-1)d}}
&{~=~\frac{(r-q)d}{a+(q-1)d}}& {~=~\frac{(s-r)d}{a+(r-1)d}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

◼ Remarks:
(i) Line marked as (a):
This line is obtained by substituting the results from step (2)
(ii) Line marked as (b):
Here we apply componendo and dividendo rule:

$\begin{array}{ll}
{\text{If}}&{\frac{a}{b}~=~\frac{c}{d}}
&{\text{Then}}& {\frac{a-b}{b}~=~\frac{c-d}{d}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\text{Example:}}&{\frac{5}{2}~=~\frac{10}{4}}
&{\text{Then}}& {\frac{5-2}{2}~=~\frac{10-4}{4}~=~\frac{6}{4}~=~\frac{3}{2}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

4. Consider the above result in (3).
Take out the first two items. We get:

$\begin{array}{ll}
{}&{\frac{(q-p)d}{a+(p-1)d}}
&{~=~\frac{(r-q)d}{a+(q-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{p-q}{a+(p-1)d}}
&{~=~\frac{q-r}{a+(q-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{q-r}{p-q}}
&{~=~\frac{a+(q-1)d}{a+(p-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$ 

5. Consider again the result in (3).
Take out the last two items. We get:

$\begin{array}{ll}
{}&{\frac{(r-q)d}{a+(q-1)d}}
&{~=~\frac{(s-r)d}{a+(r-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{q-r}{a+(q-1)d}}
&{~=~\frac{r-s}{a+(r-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{r-s}{q-r}}
&{~=~\frac{a+(r-1)d}{a+(q-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

• From the line marked as (a) in step (3), we get:
$\frac{a+(r-1)d}{a+(q-1)d}~=~\frac{a+(q-1)d}{a+(p-1)d}$

• So we can write: $\frac{r-s}{q-r}~=~\frac{a+(q-1)d}{a+(p-1)d}$

6. Comparing the results in (4) and (5), we get:

$\frac{q-r}{p-q}~=~\frac{a+(q-1)d}{a+(p-1)d}~=~\frac{r-s}{q-r}$

• Hence the statement in (1) is proved.

Solved example 9.22
If a, b, c are in G.P and $a^{\frac{1}{x}}~=~b^{\frac{1}{y}}~=~c^{\frac{1}{z}}$, prove that x, y, z are in A.P.
Solution:
1. We want to prove that x, y, z are in A.P.
• That means, we have to prove that y - x = z - y
2. Given that $a^{\frac{1}{x}}~=~b^{\frac{1}{y}}~=~c^{\frac{1}{z}}$
Let us assume that, the three quantities are equal to k. So we can write:
$a^{\frac{1}{x}}~=~b^{\frac{1}{y}}~=~c^{\frac{1}{z}}~=~k$
3. Consider the above result in (2). Taking the first and last items, we get:

$\begin{array}{ll}
{}&{a^{\frac{1}{x}}}
&{~=~k}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\left(a^{\frac{1}{x}} \right)^x}
&{~=~k^x}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\left(a^{\frac{x}{x}} \right)}
&{~=~k^x}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\left(a^{1} \right)}
&{~=~k^x}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{a}
&{~=~k^x}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

◼ In a similar way, we will get:
b = k^y and c = k^z

4. Now we get:
$\frac{b}{a}~=~\frac{k^y}{k^x}~=~k^{y-x}$
5. Also we get:
$\frac{c}{b}~=~\frac{k^z}{k^y}~=~k^{z-y}$
6. Given that a, b and c are in G.P. So we can write: $\frac{b}{a}~=~\frac{c}{b}$
7. So we can equate the results in (4) and (5). We get:
$\frac{b}{a}~=~\frac{c}{b}~=~k^{y-x}~=~k^{z-y}$
• Equating the powers of k, we get: y-x = z-y
• Hence the statement in (1) is proved.

Solved example 9.23
If a, b, c, d and p are different real numbers such that
$\left(a^2 + b^2 + c^2 \right)p^2~-~2(ab+bc+cd)p~+~\left(b^2 + c^2 + d^2 \right) ~\le~0$,
then show that a, b, c and d are in G.P.
Solution:
1. We want to prove that a, b, c and d are in G.P.
• That means, we have to prove that $\frac{b}{a}~=~\frac{c}{b}~=~\frac{d}{c}$.
2. The given inequality can be rearranged as follows:

$\begin{array}{ll}
{}&{\left(a^2 + b^2 + c^2 \right)p^2~-~2(ab+bc+cd)p~+~\left(b^2 + c^2 + d^2 \right)}
&{~\le~0}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\left(a^2 p^2 - 2abp + b^2 \right)~+~\left(b^2 p^2 - 2bcp + c^2 \right)~+~\left(c^2 p^2 - 2cdp + d^2 \right)}
&{~\le~0}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{(ap-b)^2~+~(bp-c)^2~+~(cp-d)^2}
&{~\ge~0}& {\color {green} {\text{- - - - (a)}}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

◼ Remarks:
Line marked as (a):
• This line is a sum of three terms. Each of the three terms is a square.
    ♦ Given that: a, b, c, d and p are real numbers.
    ♦ So (ap-b), (bp-c) and (cp-d) will also be real numbers.
    ♦ Square of real numbers will be zero or +ve.
    ♦ So the sum of the three terms will be zero or +ve.
    ♦ Thus the sign in this line is changed to ≥.

3. The given inequality is only rearranged in the above step (2).
• The rearranged result in (2) is same as the given inequality.
• But the given inequality is ≤. In step (2), it changed to ≥.
• If both ≤ and ≥ are to be satisfied, the only possibility is that, both the expressions are equal to zero. 
4. So we can write: $(ap-b)^2~+~(bp-c)^2~+~(cp-d)^2~=~0$
• If the sum of certain squares is zero, each term in that sum must be zero.
• So we get:
(i) (ap-b)² = 0
(ii) (bp-c)² = 0
(iii) (cp-d)² = 0
5. If square of a real number is zero, then that real number must be zero.
So we get:
(i) ap-b = 0
(ii) bp-c = 0
(iii) cp-d = 0
• From 5(i), we get: $p=\frac{b}{a}$
• From 5(ii), we get: $p=\frac{c}{b}$
• From 5(iiii), we get: $p=\frac{d}{c}$
6. Based on the above step (5), we can write:
$\frac{b}{a}~=~\frac{c}{b}~=~\frac{d}{c}~=~p$
• Hence the statement in (1) is proved.

Solved example 9.24
If p, q, r are in G.P and the equations px² + 2qx + r = 0 and dx² + 2ex + f = 0 have a common root, then show that $\frac{d}{p}, \frac{e}{q}, \frac{f}{r}$ are in A.P.
Solution:
1. We want to prove that $\frac{d}{p}, \frac{e}{q}, \frac{f}{r}$ are in A.P.
• That means, we have to prove that:

$\begin{array}{ll}
{}&{\frac{e}{q} - \frac{d}{p}}
&{~=~\frac{f}{r} - \frac{e}{q}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{e}{q} + \frac{e}{q}}
&{~=~\frac{f}{r} + \frac{d}{p}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{2e}{q}}
&{~=~\frac{f}{r} + \frac{d}{p}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

2. Given that, p, q, r are in G.P. So we can write: q² = pr
3. Also given that, px² + 2qx + r = 0 and dx² + 2ex + f = 0 have a common root.
• Let us write the roots of the first equation. We get:

$\begin{array}{ll}
{}&{\frac{-2q \pm \sqrt{(2q)^2 ~-~4pr}}{2p}}
&{~=~\frac{-2q \pm \sqrt{4q^2 ~-~4pr}}{2p}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{~=~}&{\frac{-2q \pm \sqrt{4pr ~-~4pr}}{2p}}
&{~=~\frac{-2q \pm \sqrt{0}}{2p}}& {\color {green} {\text{- - - - (a)}}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{~=~}&{\frac{-2q}{2p}}
&{~=~\frac{-q}{p}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

◼ Remarks:
Line marked as (a): From (2), we have: q² = pr

4. So the first equation has only one root, which is: $\frac{-q}{p}$
• Given that, the two equations have a common root. So $\frac{-q}{p}$must be a root of the second equation also.
• Since $\frac{-q}{p}$ is a root of the second equation, we can write:

$\begin{array}{ll}
{}&{d \left(\frac{-q}{p} \right)^2~+~2e\left(\frac{-q}{p} \right)~+~f}
&{~=~0}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{d q^2}{p^2}~-~\frac{2eq}{p}~+~f}
&{~=~0}& {}
&{\color {green} {\text{- - - - (a)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{d q^2}{p^2 q^2}~-~\frac{2eq}{p q^2}~+~\frac{f}{q^2}}
&{~=~\frac{d}{p^2}~-~\frac{2e}{p q}~+~\frac{f}{q^2}}& {~=~0}
&{\color {green} {\text{- - - - (b)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{p d}{p^2}~-~\frac{2ep}{p q}~+~\frac{f p}{q^2}}
&{~=~\frac{d}{p}~-~\frac{2e}{q}~+~\frac{f p}{q^2}}& {~=~0}
&{\color {green} {\text{- - - - (c)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{d}{p}~-~\frac{2e}{q}~+~\frac{f p}{pr}}
&{~=~\frac{d}{p}~-~\frac{2e}{q}~+~\frac{f}{r}}& {~=~0}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{2e}{q}}
&{~=~\frac{d}{p}~+~\frac{f}{r}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

◼ Remarks:
(i) Line marked as (a):
We want to transform the statement in (a) to the staement in (1). For that, we divide by q²
(ii) Line marked as (b):
We want to further transform the statement in (b) to the statement in (1). For that, we multiply by p.
(iii) Line marked as (c):
From (2), we have: q² = pr
• Hence the statement in (1) is proved.

---

The link below gives some more solved examples

Miscellaneous Exercise on chapter 9

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In the next chapter we will see straight lines.

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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 9.2 - 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
= $\frac{n}{2}[2a+(n-1)d]~=~\frac{20}{2}[2 × 300000+(20-1) × 10000]$        
= $10 × [600000+19 × 10000]~=~79,00,000$

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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

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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

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The link below gives a few more solved examples

Exercise 9.2

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In the next section we will see arithmetic progression.

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