Question

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

Answer 1

Given,

$S_6=42$

$\frac{6}{2}[2a+(6-1\left)d\right]=42$

$3[2a+5d]=42$

$2a+5d=14$.......(1)

$\frac{a+9d}{a+12d}=\frac{1}{3}$

$3a+27d=a+12d$

$2a+15d=0$.......(2)

(2)-(1) gives,

$d=-\frac{14}{10}$

From (2)

$a=\frac{-15d}{2}=10.5$

$T_{13}=10.5+(13-1)(-\frac{14}{10})$

$=10.5-16.8$

$=-6.3$