I came across a difficult question from arithmetic progression chapter and it was already asked in previous year paper of 2015 in which we have to find the sum of the first n terms if in an AP, it is given that S5+S7 =167 and S10 =235, where Sn denotes the sum of its first n terms.

Book – RS Aggarwal, Class 10, chapter 5C, question no 20.

Let the first term is a and the common difference is d

By using Sn=n/2[2a+(n−1)d] we have,

S5=5/2[2a+(5−1)d]=5/2[2a+4d]

S7=7/2[2a+(7−1)d]=7/2[2a+6d]

Given: S7+S5=167

∴5/2[2a+4d]+7/2[2a+6d]=167

⇒10a+20d+14a+42d=334

⇒24a+62d=334 …(1)

S10=10/2[2a+(10−1)d]=5(2a+9d)

Given: S10=235

So 5(2a+9d)=235

⇒2a+9d=47 …(2)

Multiply equation (2) by 12, we get

24a+108d=564....(3)

Subtracting equation (3) from (1), we get

−46d=−230

∴d=5

Substing the value of d=5 in equation (1) we get

2a+9(5)=47

or 2a=2

∴a=1

Then A.P is 1,6,11,16,21,⋯