Question

An $A.P.$ 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

Answer 1

Answer: (B)

2

Answer 2

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