Question

If the sum of first

p

terms of an A.P. is

q

and the sum of first

q

terms is

p

, then find the sum of first

(p + q)

terms.

Accepted Answer

Given,

An A.P. whose sum of first $p$ terms is $q$ and that of first $q$ terms is $p$ .

We have to find the sum of the first $(p + q)$ terms of the given A.P.

We know that the sum of first $n$ terms of an A.P. is
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

(where $a$ is the first term, $d$ is the common difference and $n$ is the number of terms).

Here, it is given, the sum of the first $p$ terms, $S_{p} = q$ .

$\Rightarrow \frac{p}{2} [2 a + (p - 1) d] = q$

$\Rightarrow p (2 a + (p - 1) d) = 2 q . . . (i)$

Again, the sum of first $q$ terms, $S_{q} = p$ .

$\Rightarrow \frac{q}{2} (2 a + (q - 1) d) = p$

$\Rightarrow q (2 a + (q - 1) d) = 2 p . . . (ii)$

Subtracting equation $(i)$ from $(ii)$ , we get $2 q a + q^{2} d - q d - 2 p a - p^{2} d + p d = 2 p - 2 q$

$\Rightarrow 2 a (q - p) + d (q^{2} - p^{2}) + d (p - q) = 2 (p - q)$

$\Rightarrow 2 a (q - p) + d (p + q) (q - p) + d (p - q) = 2 (p - q)$

$\Rightarrow (q - p) [2 a + d (p + q) - d] = - 2 (q - p)$

(Since, $p \neq q)$

$\Rightarrow 2 a + (p + q - 1) d = - 2 . . . (iii)$

Now, sum of first $p + q$ terms is,

$S_{p + q} = (\frac{p + q}{2}) [2 a + (p + q - 1) d]$

$\Rightarrow S_{p + q} = (\frac{p + q}{2}) \times (- 2) = - (p + q)$ $[$ From equation $(iii)]$

Hence, the sum of first $p + q$ terms of the A.P. is $- (p + q)$ .

Embibe Experts Solutions for Chapter: Sequences and Series, Exercise 1: Exercise-1

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Sequences and Series, Exercise 1: Exercise-1

Questions from Embibe Experts Solutions for Chapter: Sequences and Series, Exercise 1: Exercise-1 with Hints & Solutions

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