A Weird Arithmetic Progression

Algebra Level 2

If the sum of the first p p terms of an arithmetic progression is equal to the sum of the first q q terms of the same arithmetic progression, where p ≠ q p \ne q , find the sum for the first p + q p+q terms of the arithmetic progression.


The answer is 0.

Karthick Shiva by Karthick Shiva , 24, India

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1 solution

Let the first term and the common difference of the arithmetic progression be a 1 a_1 and d d respectively. Then we have:

p ( 2 a 1 + ( p − 1 ) d ) 2 = q ( 2 a 1 + ( q − 1 ) d ) 2 2 a 1 ( p − q ) = ( q 2 − q − p 2 + p ) d = ( q − p ) ( q + p − 1 ) d ⟹ 2 a 1 = ( 1 − p − q ) d \begin{aligned} \frac {p(2a_1+(p-1)d)}2 & = \frac {q(2a_1+(q-1)d)}2 \\ 2a_1(p-q) & = (q^2-q-p^2+p)d = (q-p)(q+p-1)d \\ \implies 2a_1 & = (1-p-q)d \end{aligned}

Then the sum of the first p + q p+q terms,

S p + q = p + q 2 ( 2 a 1 + ( p + q − 1 ) d ) = p + q 2 ( ( 1 − p − q ) d + ( p + q − 1 ) d ) = 0 \begin{aligned} S_{p+q} & = \frac {p+q}2 (2a_1 + (p+q-1)d) \\ & = \frac {p+q}2 ((1-p-q)d + (p+q-1)d) \\ & = \boxed 0 \end{aligned}

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