In the previous section, we completed a discussion on the nth 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 nth term of an A.P is given by: an = a+(n-1)d
4. If there is a total of m terms in an A.P, then the last term (the mth 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 Sn
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 nth 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 mth term is n and the nth term is m, where m≠n, find the pth term.
Solution:
1. The nth 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, mth term is n
• So we can write: n = a+(m-1)d
⇒ a+md - d = n
3. Given that, nth 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 pth term:
pth 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 12th 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 12th 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 12th 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, . . . , An between a and b in such a way that,
a, A1, A2, A3, A4, . . . , An, 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, . . . , An 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:
2nd term A1 = 3 + (2-1) × 3 = 6
3rd term A2 = 3 + (3-1) × 3 = 9
4th term A3 = 3 + (4-1) × 3 = 12
5th term A4 = 3 + (5-1) × 3 = 15
6th term A5 = 3 + (6-1) × 3 = 18
7th 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
In the next section we will see arithmetic progression.
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