Question

if in an AP, p times of pth term is equal to q times of qth term, then (p + q) th term is:

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is A 0

$P[a+(p-1\left)d\right]=q[a+(q-1\left)d\right]$

$Pa+p(p-1)d+aq-q(q-1)d=0$

$a[p-q]+d\left[p\right(p-1)-q(q-1\left)\right]=0$

$a[p-q]+d[p^2-p-q^2+q]=0$

$a[p-q]+d[p^2-q^2(p-q\left)\right]=0$

$[p-q](a+d[(p+q)-1\left]d\right)=0$

$(p+q{)}^{th}$ term

$\Rightarrow (p+q{)}^{th}term=0$