Question

If the sum of the first $p$ terms of an AP is the same as the sum of its first $q$ terms (where $p\ne q$) then what is the sum of its first $(p+q)$ terms ?

Answer 1

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

Now the sum of first $p$ terms is $\frac{p}{2}(2a+(p-1\left)d\right)$.....(1) and that of first $q$ terms is $\frac{q}{2}(2a+(q-1\left)d\right)$......(2).

According to the problem,

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

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

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

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

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

Now the sum of first $(p+q)$ terms is $\frac{p+q}{2}\{2a+(p+q-1\left)d\right\}=0$ [ Using (1)].