Question

If the sum of the first p terms of an A.P. is q and the sum of the first q
terms is p; then show that the sum of the first (p + q) terms is 1 - (p + 9)).​

Answer 1

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 term

Answer 2

$\bf\large\underline\blue{Answer:-}$

\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).}