Question

If p times the $p^{th}$ term of an A.P is equal to q times the $q^{th}$term of an A.P , then ${(p+q)}^{th}$ term is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

0

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

Given: ($p$ times $p^{th}$ term) $=$ ($q$ times $q^{th}$ term)

$p{a}_p=q{a}_q$

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

$ap+p^{2}d-pd=aq+q^{2}d-qd$

$ap-aq=-p^{2}d+q^{2}d-qd+pd$

$a(p-q)=d(q^2-p^2+p-q)$

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

$[a^2-b^2=(a+b\left)\right(a-b\left)\right]$

$a(p-q)=d(p-q)\{-1(p+q)+1\}$

$a=d(-p-q+1)$ ------ ( 1 )

$(p+q{)}^{th}$ term $=a+(n-1)d$

here , $n=(p+q)$

$(p+q{)}^{th}=a+(p+q-1)d$ ----- ( 2 )

substituting $a=d(-p-q+1)$ in eq. ( 2 )

$(p+q{)}^{th}=d(-q-p+1)+(p+q-1)d$

$(p+q{)}^{th}=-dp-dq+d+pd+qd-d$

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