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

Given:

The sum of first seven terms of an A.P. is 182. Its 4th and the 17th terms are in the ratio 1 : 5.

To do:

We have to find the A.P.

Solution:

Let the first term of the A.P. be $a$ and the common difference be $d$.

We know that,

nth term of an A.P. $a_n=a+(n-1)d$

Therefore,

$a_{4}=a+(4-1)d$

$=a+3d$......(i)

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

$=a+16d$......(ii)

According to the question,

$a_4 : a_{17}=(a+3d): (a+16d)$

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

$1(a+16d)=5(a+3d)$ 

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

$5a-a=16d-15d$

$4a=d$

$a=\frac{d}{4}$......(iii)

Sum of $n$ terms of an A.P. $S_n=\frac{n}{2}(2a+(n-1)d)$

$S_7=\frac{7}{2}(2a+(7-1)d)$

$182=\frac{7}{2}(2a+6d)$

$26=a+3d$

$26=\frac{d}{4}+3d$     (From (iii))

$26=\frac{d+4(3d)}{4}$

$4(26)=d+12d$

$104=13d$

$d=\frac{104}{13}$

$d=8$

Therefore,

$a=\frac{8}{4}$

$a=2$

$\Rightarrow a_{2}=a+d=2+8=10$

$a_3=a+2d=2+2(8)=2+16=18$

$a_4=a+3d=2+3(8)=2+24=26$

Hence, the required A.P. is $2, 10, 18, 26, ......$