An algebra problem by Ilham Saiful Fauzi

Algebra Level pending

Let { a n } \left \{a_{n}\right \} be an arithmetic progression of real numbers such that the sum of its first p p terms is q q and sum of its first q q terms is p p . Then find sum of the first p + q p+q terms of { a n } \left \{a_{n}\right \} .

− 2 ( p + q ) -2(p+q) − ( p + q ) -(p+q) ( p + q ) − p q (p+q)-pq p + q p+q p + q − 3 p+q-3 p − q p-q

Ilham Saiful Fauzi by Ilham Saiful Fauzi , 25, Indonesia

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1 solution

The following are given, where S n S_n , a a and d d are the sum of first n n terms, first term and common difference respectively.

{ S p = p ( 2 a + ( p − 1 ) d 2 = q ⟹ 2 a + ( p − 1 ) d = 2 q p . . . ( 1 ) S q = q ( 2 a + ( q − 1 ) d 2 = p ⟹ 2 a + ( q − 1 ) d = 2 p q . . . ( 2 ) \begin{cases} S_p = \dfrac {p(2a+(p-1)d}2 = q & \implies 2a + (p-1)d = \dfrac {2q}p &...(1) \\ S_q = \dfrac {q(2a+(q-1)d}2 = p & \implies 2a + (q-1)d = \dfrac {2p}q &...(2) \end{cases}

( 1 ) − ( 2 ) : ( p − q ) d = 2 q p − 2 p q ( p − q ) d = 2 ( q 2 − p 2 p q d = − 2 ( p + q ) p q ⟹ p q d = − 2 ( p + q ) \begin{aligned} (1)-(2): \quad (p-q)d & = \frac {2q}p - \frac {2p}q \\ (p-q)d & = \frac {2(q^2-p^2}{pq} \\ d & = - \frac {2(p+q)}{pq} \\ \implies \color{#D61F06} pqd & = \color{#D61F06}-2(p+q) \end{aligned}

Now, we have:

S p + q = ( p + q ) ( 2 a + ( p + q − 1 ) d 2 = p ( 2 a + ( p − 1 + q ) d 2 + q ( 2 a + ( q − 1 + p ) d 2 = p ( 2 a + ( p − 1 ) d 2 + q ( 2 a + ( q − 1 ) d 2 + p q d = S p + S q + p q d = q + p − 2 ( p + q ) = − ( p + q ) \begin{aligned} S_{p+q} & = \frac {(p+q)(2a+(p+q-1)d}2 \\ & = {\color{#3D99F6}\frac {p(2a+(p-1+{\color{#D61F06}q})d}2} + {\color{#3D99F6}\frac {q(2a+(q-1+{\color{#D61F06}p})d}2} \\ & = {\color{#3D99F6}\frac {p(2a+(p-1)d}2} + {\color{#3D99F6}\frac {q(2a+(q-1)d}2} + \color{#D61F06}pqd \\ & = {\color{#3D99F6}S_p} + {\color{#3D99F6}S_q} + \color{#D61F06}pqd \\ & = {\color{#3D99F6}q} + {\color{#3D99F6}p} \color{#D61F06}-2(p+q) \\ & = \boxed{-(p+q)} \end{aligned}

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