Question

In an $A.P.,$ sum of $pterms=sumofqterms,$ then sum of $(p+q)$ terms equal to ?

• A. $0$
• B. $2$
• C. $-pq$
• D. $-(p+q)$

Answer 1

The correct option is A $0$

Let the first term of the A.P. is $a$ and the common difference is $d$.

Then according to the problem,

$\frac{p}{2}\{2a+(p-1\left)d\right\}=\frac{q}{2}\{2a+(q-1\left)d\right\}$

or, $2a(p-q)+\left\{\right(p^2-p)-(q^2-q\left)\right\}d=0$

or, $2a(p-q)+\left\{\right(p^2-q^2)-(p-q\left)\right\}d=0$

or, $(p-q)\{2a+(p+q-1\left)d\right\}=0$

Since $p\ne q$, $2a-(p+q-1)d=0$......(1).

Now the sum of the $p+q$ terms

$=\frac{(p+q)}{2}\{2a+(p+q-1\left)d\right\}$

$=\frac{(p+q)}{2}\times 0$

$=0$.