Question

The sum of first seven terms of an A.P is $182$. If its $4$th and the $17$th terms are in the ratio $1:5$, find the A.P

• A. $2,8,14,20,....$
• B. $2,10,18,26,....$
• C. $3,5,7,9,....$
• D. None of these

Answer 1

Answer: (B)

$2,10,18,26,....$

Given,

$S_7=\frac{7}{2}[2a+(7-1\left)d\right]=182$

$182=\frac{7}{2}[2a+6d]$........(1)

$a_n=a+(n-1)d$

$a_4=a+(4-1)d=a+3d$

$a_{17}=a+(17-1)d=a+16d$

From given, we have,

$\frac{a_4}{a_{17}}=\frac{a+3d}{a+16d}$

$\frac{1}{5}=\frac{a+3d}{a+16d}$

$5a+15d=a+16d$

solving the above equation, we get,

$4a=d$

using this value in (1), we get,

$182=\frac{7}{2}[2a+6(4a\left)\right]$

solving the above equation, we get,

$a=2$

$4\left(2\right)=8=d$

The series as follows,

$a,a+d,a+2d,.........$

$\Rightarrow 2,10,18,.......$