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# The sum of first 9 terms of an A.P. is 162. The ratio of its 6th term to its 13th term is $1 : 2$. Find the first and 15th term of the A.P.

Given:

The sum of first 9 terms of an A.P. is 162. The ratio of its 6th term to its 13th term is $1 : 2$.

To do:

We have to find the first and 15th term of 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,

$a_{6}=a+(6-1)d$

$=a+5d$......(i)

$a_{13}=a+(13-1)d$

$=a+12d$......(ii)

According to the question,

$a_6 : a_{13}=(a+5d): (a+12d)$

$\frac{1}{2}=\frac{a+5d}{a+12d}$

$1(a+12d)=2(a+5d)$

$a+12d=2a+10d$

$2a-a=12d-10d$

$a=2d$......(iii)

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

$S_9=\frac{9}{2}(2a+(9-1)d)$

$162=\frac{9}{2}(2a+8d)$

$18=a+4d$

$18=2d+4d$ (From (iii))

$6d=18$

$d=\frac{18}{6}$

$d=3$

Therefore,

$a=2(3)$

$a=6$

$\Rightarrow a_{15}=a+(15-1)d$

$=6+14(3)$

$=6+42$

$=48$

Hence, the first term and the 15th term of the given A.P. are $6$ and $48$ respectively.