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,⋯