Question

if the sum of first p terms of an AP is q and the sum of first q terms is p ,then find the sum of (p+q) terms

Answer 1

$\bold{-(p+q)}$ is the sum of $\bold{(p+q)}$ terms.

Given:

Sum of first p terms of an AP = q

Sum of first q terms = p

To find:

Sum of (p + q) terms  = ?  

Solution:

The sum of first p term of AP is q, which means that the number of terms is p

Thereby, let us take the first term as A and the common difference d

The sum of first q term of AP is p, which means that the number of terms is q

Thereby, let us take the first term as A and the common difference d

Therefore, the sum $=\frac{n}{2}(2 a+(n-1) d)$

$\begin{array}{l}{q=\frac{p}{2}(2 a+(p-1) d)} \\ {p=\frac{q}{2}(2 a+(q-1) d)}\end{array}$

Subtracting the sum of the term , p and q

$\begin{array}{l}{q-p=\left(\frac{p}{2}(2 a+(p-1) d)-\frac{q}{2}(2 a+(q-1) d)\right)} \\ {q-p=a(p-q)+\frac{d}{2}\left(p^{2}-p-q^{2}+q\right)}\end{array}$

After solving the equation we get the value of d as  

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

Now with $\mathrm{d}=-\frac{2(p+q)}{p q}$, first value of the series is a and the number of terms is $p+q$

$\begin{array}{l}{\text { Sum }=\frac{n}{2}(2 a+(n-1) d)} \\ {\frac{p+q}{2}(2 a+(p+q-1) d)=\frac{p+q}{p q}(-p q)}\end{array}$

Therefore, the sum is $\bold{-(p+q).}$

Answer 2

Answer:

here is your answer

I hope it will help u