This Sum Is That; That Sum Is This

Algebra Level 3

For distinct numbers p p and q q , consider an arithmetic progression such that

  • the sum of the first p p terms is equal to q q , and
  • the sum of the first q q terms is equal to p p .

What is the sum of the first p + q p+ q terms?

0 0 p + q p+q p − q p-q − ( p + q ) -(p+q)

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Sunny Dhondkar by Sunny Dhondkar , 20, India

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3 solutions

Viki Zeta
Oct 12, 2016

Relevant wiki: Arithmetic Progression Sum

Recall that the n th n^\text{th} of an arithmetic progression is denoted by T n = a + ( n − 1 ) d T_n = a + (n-1)d , where a a is the first term and d d is the common difference. And the sum of the first n th n^\text{th} terms of this progression is S n = n 2 [ 2 a + ( n − 1 ) d ] S_n = \dfrac n2 [ 2a +(n-1) d ] .

We have

S p = q p 2 ( 2 a + ( p − 1 ) d ) = q 2 a + ( p − 1 ) d = 2 q p S q = p 2 a + ( q − 1 ) d = 2 p q ( − ) 2 a + ( p − 1 ) d = 2 q p ⟹ ( q − 1 ) d − ( p − 1 ) d = 2 p q − 2 q p d ( q − p ) = 2 ( p 2 − q 2 ) p q − d ( p − q ) = 2 ( p + q ) ( p − q ) p q − d = 2 ( p + q ) p q d = − 2 ( p + q ) p q S p + q = ( p + q ) 2 ( 2 a + ( ( p − 1 ) + q ) d ) = ( p + q ) 2 ( 2 a + ( p − 1 ) d + q d ) = ( p + q ) 2 ( 2 q p + q − 2 ( p + q ) p q ) = ( p + q ) 2 ( 2 q p − 2 p + 2 q p ) = p + q 2 ( − 2 p p ) = − ( p + q ) S_p = q \\ \dfrac{p}{2} \left(2a + (p-1)d\right) = q \\ \boxed{2a + \left(p-1\right)d = \dfrac{2q}{p} }\\ S_q = p \\ \boxed{2a + \left(q-1\right)d = \dfrac{2p}{q}} \\ (-)2a + \left(p-1\right)d = \dfrac{2q}{p} \implies \left(q-1\right)d - \left(p-1\right)d = \dfrac{2p}{q} - \dfrac{2q}{p} \\ d(q-p) = \dfrac{2\left(p^2 - q^2\right)}{pq} \\ -d(p-q) = 2\dfrac{\left(p+q\right)\left(p-q\right)}{pq} \\ -d = \dfrac{2\left(p+q\right)}{pq} \\ d = \dfrac{-2\left(p+q\right)}{pq} \\ S_{p+q} = \dfrac{\left(p+q\right)}{2}\left(2a + \left(\left(p-1\right)+q\right)d\right) \\ = \dfrac{\left(p+q\right)}{2}\left(2a + \left(p-1\right)d +qd\right) \\ = \dfrac{\left(p+q\right)}{2}\left(\dfrac{2q}{p} + q\dfrac{-2\left(p+q\right)}{pq} \right) \\ = \dfrac{\left(p+q\right)}{2}\left(\dfrac{2q}{p} - \dfrac{2p+2q}{p}\right) \\ = \dfrac{p+q}{2} \left(\dfrac{-2p}{p}\right) \\ = -(p+q)

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how did you get the first term 'a'?

Sunny Dhondkar - 4 years, 8 months ago

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First term is always a a or a 1 a_1 . It's the general form of an AP

Viki Zeta - 4 years, 8 months ago

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i meant that you didn't show how you get the first term.. (procedure)

Sunny Dhondkar - 4 years, 8 months ago

Mistake: − d ( p − q ) = 2 ( p + q ) ( p − q ) p q ⟹ − d = 2 ( p + q ) p q -d(p-q) = 2\dfrac{(p+q)(p-q)}{pq} \implies -d = \dfrac{2(p+q)}{pq} , iff p ≠ q p \neq q , since division by 0 0 is not allowed.

If p = q p = q we have a counterexample, which you can see from my report.

Jesse Nieminen - 4 years, 8 months ago

Total of terms of this progression is : S n = a n + d 2 ∗ ( n 2 − n ) S_n =an+\dfrac d 2 *(n^2 - n) .

We have

S p = q ⟹ a p + d 2 ∗ ( p 2 − p ) = q . . . . . ( 1 ) S q = p ⟹ a q + d 2 ∗ ( q 2 − q ) = p . . . . . ( 2 ) F r o m ( 1 ) a n d ( 2 ) , e l e m i n a t i n g a w e g e t ( q − p ) d = 2 p q − 2 q p = 2 ∗ ( p + q ) ∗ ( p − q ) p q . S i n c e p ≠ q , d p q = − 2 ∗ ( p + q ) . . . . . ( 3 ) A l s o ( 1 ) + ( 2 ) g i v e s S q + S p = q + p = a ( p + q ) + d 2 ∗ ( p 2 + q 2 ) − d 2 ∗ ( p + q ) = p + q . . . . . ( 4 ) S p + q = a ( p + q ) + d 2 ∗ ( p 2 + 2 p q + q 2 ) − d 2 ∗ ( p + q ) = p + q S o f r o m ( 4 ) a n d ( 3 ) , w e g e t , S p + q = a ( p + q ) + d 2 ∗ ( p 2 + q 2 ) − d 2 ∗ ( p + q ) + p q d = p + q + p q d ∴ S p + q = p + q − 2 ( p + q ) = − ( p + q ) S_p = q \implies\ \ \ \ ap+ \dfrac d 2 * (p^2 - p) = q.....(1) \\ S_q = p \implies\ \ \ \ \ aq+ \dfrac d 2 * (q^2 - q)= p.....(2) \\ From\ (1)\ and\ (2),\ \ eleminating\ a\ we\ get\\ (q -p)d = \dfrac{2p}{q} - \dfrac{2q}{p}=2*\dfrac{(p+q)*(p-q)}{pq}.\\ Since\ p\neq q,\ \ \ dpq = - 2*(p+q) .....(3)\\ Also\ (1)+(2)\ gives\ S_q+S_p=q+p=a(p+q)+{ \color{#3D99F6} {\dfrac d 2 *(p^2+q^2) } } \ - \ \dfrac d 2 * (p+q) = p+q.....(4) \\ S_{p+q} = a(p+q)+{ \color{#3D99F6} {\dfrac d 2 *(p^2+2pq+q^2) } } \ - \ \dfrac d 2 * (p+q) = p+q \\ So\ from \ (4)\ and\ (3), we\ get,\\ S_{p+q}=a(p+q)+{ \color{#3D99F6} {\dfrac d 2 *(p^2+q^2) } } \ - \ \dfrac d 2 * (p+q) {\color{#3D99F6}{+pqd}}= p+q+pqd\\ \therefore\ S_{p+q}=p+q -2(p+q)=-(p+q)

Niranjan Khanderia - 4 years, 7 months ago
Sunny Dhondkar
Oct 11, 2016

Here is a special hint

1 Helpful 0 Interesting 0 Brilliant 0 Confused

This is a very overrated question.

Thomas Jacob - 4 years, 8 months ago
Ankush Gogoi
Oct 16, 2016

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