CBSE Class 10 Mathematics Arithmetic Progression Notes Set 06

Sum of the First n Terms of an A.P.

Let the A.P. be a, a + d, a + 2d, ..., a + (n − 1)d and let S denote the sum of its first n terms. Then

\[ S=a+(a+d)+(a+2d)+\dots+[a+(n-2)d]+[a+(n-1)d]\qquad\dots(i) \]

Writing the terms in the reverse order,

\[ S=[a+(n-1)d]+[a+(n-2)d]+\dots+(a+d)+a\qquad\dots(ii) \]

Adding (i) and (ii), the sum of each pair of terms (the first term of (i) with the first of (ii), and so on) is 2a + (n − 1)d, and there are n such pairs:

\[ 2S=n[2a+(n-1)d]\ \Rightarrow\ S=\frac{n}{2}[2a+(n-1)d] \]

Therefore the sum of the first n terms of an A.P. is

\[ S_n=\frac{n}{2}[2a+(n-1)d] \]where a is the first term, n is the number of terms and d is the common difference. We also use Sₙ in place of S.

Since a + (n − 1)d is the last term l of the A.P.,

\[ S_n=\frac{n}{2}(a+l) \]If there are only n terms in the A.P., then aₙ = l, the last term.

Example 1. Find the sum of the first 20 terms of an A.P. if the 5th and 12th terms are 30 and 65 respectively.
Answer:
Let a be the first term and d the common difference. a + 4d = 30 ...(i) and a + 11d = 65 ...(ii).
(ii) − (i): 7d = 35 ⇒ d = 5. From (i), a = 30 − 20 = 10.
S₂₀ = 10[2 × 10 + 19 × 5] = 10 × 115 = 1150.

Example 2. If the ratio of the 11th term of an A.P. to its 18th term is 2 : 3, find the ratio of the sum of the first five terms to the sum of the first 10 terms.
Answer:
\( \frac{a_{11}}{a_{18}}=\frac{a+10d}{a+17d}=\frac{2}{3} \) ⇒ 3a + 30d = 2a + 34d ⇒ a = 4d.
S₅ = \( \frac{5}{2}(2a+4d)=\frac{5}{2}\times12d=30d \) and S₁₀ = \( 5(2a+9d)=5\times17d=85d \).
S₅ : S₁₀ = 30d : 85d = 6 : 17.

Example 3. A sum of Rs. 1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests form an A.P.? Find the interest after 30 years.
Answer:
Simple interest after n years = \( \frac{1000\times8\times n}{100}=80n \). For n = 1, 2, 3, ... the interests are Rs. 80, 160, 240, ..., which form an A.P. with a = 80 and d = 80.
Interest after 30 years = 80 × 30 = Rs. 2400.

Example 4. Find the sum of all three-digit numbers which leave remainder 3 when divided by 5.
Answer:
The smallest such number is 103 and the largest is 998, so the numbers 103, 108, 113, ..., 998 form an A.P. with a = 103 and d = 5.
998 = 103 + (n − 1) × 5 ⇒ n = 180.
S = \( \frac{180}{2}(103+998)=90\times1101 \) = 99090.

Example 5. If in an A.P. the sum of the first m terms is zero, show that the sum of its first n terms is \( \frac{an(m-n)}{m-1} \), where a is the first term.
Answer:
Sₘ = 0 ⇒ \( \frac{m}{2}[2a+(m-1)d]=0 \) ⇒ 2a + (m − 1)d = 0 ⇒ d = \( \frac{-2a}{m-1} \).
Sₙ = \( \frac{n}{2}\left[2a-\frac{2a(n-1)}{m-1}\right]=an-\frac{an(n-1)}{m-1}=\frac{an(m-n)}{m-1} \). Hence proved.

Summary

1. A sequence is a set of numbers arranged in some definite order according to some definite rule.
2. An arithmetic progression (A.P.) is a sequence of numbers in which the difference between consecutive terms is constant.
3. The general form of an A.P. is a, a + d, a + 2d, ..., where a is the first term and d is the common difference.
4. The nth term of an A.P. is aₙ = a + (n − 1)d = l (for the last term).
5. Selection of terms: take the middle term as a and the common difference as d when an odd number of terms is required; take a − d and a + d as the two middle terms and 2d as the common difference when an even number of terms is required.
6. Three numbers are in A.P. when a₂ − a₁ = a₃ − a₂ = ...
7. The sum of the first n terms is Sₙ = n/2 [2a + (n − 1)d] or Sₙ = n/2 (a + l).
8. The nth term from the end of an A.P. is l − (n − 1)d, where l is the last term. Also the (n − m + 1)th term from the beginning is the mth term from the end.