Question

The sum of the 4th and the 8th terms of an AP is 24 and the sum of its 6th and 10th terms is 44. Find the sum of its first 10 terms.

Answer 1

Let the first term of an A.P = a
and the common difference of the given A.P = d
As we know that
$a_n=a+(n-1)d$
$a_4=a+(4-1)d$
$a_4=a+3d$
Similarly ,
$a_8=a+7d$
$a_6=a+5d$
$a_{10}=a+9d$
Sum of 4 th and 8th terms of an A.P = 24 ( given )
$a_4+a_8=24$
$a+3d+a+7d=24$
$2a+10d=24$
$a+5d=12$ .....................(i)
Sum of 6 th and 10 th term of an A.P = 44 ( given )
$a_6+a_{10}=44$
$a+5d+a+9d=44$
$2a+14=44$
$a+7d=22$ .....................(ii)
Solving (i) & (ii)

$2d=10$
$d=5$
From equation (i) ,
$a+5d=12$
$a+5\left(5\right)=12$
$a+25=12$
$a=-13$

Sum of n terms, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]S_{10}=\frac{10}{2}\left[2\right(-13)+(10-1\left)5\right]=5[-26+9\times 5]=5[-26+45]=5\times 19=95$

Sum of first 10 terms = 95