Question

The sum of first six terms of an arithmetic progression is $42$. The ratio of its ${10}^{th}$ term and ${30}^{th}$ term is $1:3$. Calculate the first and the thirteenth term of A.P.

Answer 1

Let $a$ be the first term and $d$ be the common difference of the given A.P. Then,
$S_6=42⟹\frac{6}{2}\{2a+(6-1\left)d\right\}=42⟹2a+5d=14$ ...(i)

It is given that

$a_{10}:a_{30}=1:3$

$⟹\frac{a+9d}{a+29d}=\frac{1}{3}$

$⟹3a+27d=a+29d$

$⟹2a-2d=0$

$⟹a=d$ ...(ii)

putting the value of $a$ in (i), we get

$2d+5d=14\Rightarrow d=2$

$\therefore a=d=2$

$\therefore a_{13}=a+12d=2+2\times 12=26$
Hence, first term $=2$ and thirteenth term $=26$