Question

If the ratio of sum of $p$ terms to $q$ terms of an A.P. is $\frac{p^2+p}{q^2+q}$, then the ratio of $p^{th}$ term to $q^{th}$ term is equal to

• A. $\frac{p^2}{q^2}$
• B. $\frac{2p-1}{2q-1}$
• C. $\frac{p}{q}$
• D. $\frac{p+1}{q+1}$

Answer 1

The correct option is C $\frac{p}{q}$

Given : $\frac{S_p}{S_q}=\frac{p^2+p}{q^2+q}$
$\Rightarrow \frac{S_p}{p^2+p}=\frac{S_q}{q^2+q}=k\Rightarrow S_p=(p^2+p)k,S_q=(q^2+q)k$

Now,
$\frac{T_p}{T_q}=\frac{S_p-S_{p-1}}{S_q-S_{q-1}}\Rightarrow \frac{T_p}{T_q}=\frac{k(p^2+p)-k\left[\right(p-1{)}^2+(p-1)]}{k(q^2+q)-k\left[\right(q-1{)}^2+(q-1)]}\therefore \frac{T_p}{T_q}=\frac{p}{q}$


Alternate Solution:
Given : $\frac{S_p}{S_q}=\frac{p^2+p}{q^2+q}$
$\Rightarrow \frac{\frac{p}{2}[2a+(p-1\left)d\right]}{\frac{q}{2}[2a+(q-1\left)d\right]}=\frac{p^2+p}{q^2+q}$
$\Rightarrow \frac{a+\frac{(p-1)}{2}d}{a+\frac{(q-1)}{2}d}=\frac{p+1}{q+1}...\left(1\right)$

We know that
$\frac{T_p}{T_q}=\frac{a+(p-1)d}{a+(q-1)d}$
So, replace $p\to 2p-1$ and $q\to 2q-1$, in $\left(1\right)$ we get
$\frac{T_p}{T_q}=\frac{2p-1+1}{2q-1+1}\therefore \frac{T_p}{T_q}=\frac{p}{q}$