Question

An AP has the property that the sum of first ten terms is half the sum of next ten terms. If the second term is $13$, then the common difference is

• A. $3$
• B. $2$
• C. $5$
• D. $4$
• E. $6$

Answer 1

The correct option is B $2$

Let $a$ and $d$ be the first term and common difference, respectively.
Given, $T_2=13\Rightarrow a+d=13$ .... (i)
According to the question, we have
$(a_1+a_2+....+a_{10})=\frac{1}{2}(a_{11}+a_{12}+...+a_{20})$
$\Rightarrow 2(a_1+a_2+....+a_{10})=a_{11}+a_{12}+...+a_{20}$
$\Rightarrow 3(a_1+a_2+....+a_{10})=(a_1+a_2+...+a_{20})$
(add $a_1+a_2+...+a_{10}$ on both sides)
$\Rightarrow 3\times \frac{10}{2}[2a+9d]=\frac{20}{2}[2a+19d]$
$\Rightarrow 6a+27d=4a+38d$
$\Rightarrow 2a=11d$ .... (ii)
From Eqs. (i) and (ii), we get
$\frac{11d}{2}+d=13\Rightarrow \frac{13d}{2}=13\Rightarrow d=2$.