The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P.

Given:

The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161.

To do:

We have to find the 28th 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. $a_n=a+(n-1)d$

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

Therefore,

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

$63=\frac{7}{2}(2a+6d)$

$9=a+3d$

$a=9-3d$......(i)

Sum of the next 7 terms $=161$. This implies,

Sum of the first 14 terms $=161+63=224$

$S_{14}=\frac{14}{2}(2a+(14-1)d$

$224=7(2a+13d)$

$32=2(9-3d)+13d$      (From (i))

$32=18-6d+13d$

$7d=32-18$

$7d=14$

$d=2$

This implies,

$a=9-3(2)$

$=9-6$

$=3$

$\Rightarrow a_{28}=a+(28-1)d$

$=3+27(2)$

$=3+54$

$=57$

Hence, the 28th term of the given A.P. is $57$.  

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Updated on: 10-Oct-2022

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