If the sum of first p terms of an AP is same as the sum of first q terms of an AP, then find the summation of first p+q terms.

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Solution

Formula,

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

Consider sum of first p terms,

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

Consider sum of first q terms,

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

Given,

$S_{p} = S_{q}$

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

$p [2 a + (p - 1) d] = q [2 a + (q - 1) d]$

$2 a (p - q) = d [(q - 1) q - (p - 1) p]$

$2 a (p - q) = - d (p - q) [p + q - 1]$

$2 a (p - q) + d (p - q) [p + q - 1] = 0$

$(p - q) [2 a + (p + q - 1) d] = 0$

$p - q \neq 0 \Rightarrow [2 a + (p + q - 1) d] = 0$

Now,

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

$= \frac{p + q}{2} \times 0$

$= 0$

Therefore the sum of (p+q) terms is 0.

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