This is the question of arithmetic progression chapter in which we are to calculate the sum of the first 10 terms, if 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.

Book RS Aggarwal, Class 10, chapter 5C, question no 36.

Tn=a+(n−1)d

T4=a+(4−1)d=a+3d

Similarly

T8=a+7d,T6=a+5d,T10=a+9d

Given,

a+3d+a+7d=24

⇒2a+10d=24

⇒a+5d=12…(1)

a+5d+a+9d=44

⇒2a+14d=44

⇒a+7d=22…(2)

(2) – (1) gives

2d=10⇒d=5

From (2)

a=22−7(5)=−13

Sn=n/2[2a+(n−1)d]

S10=10/2[2(−13)+(10−1)5]

=5[−26+45]

=95