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# In an AP, it is given that S5+S7 =167 and S10 =235, then find the AP, where Sn denotes the sum of its first n terms.

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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.

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

By using Sn=n/2[2a+(n1)d] we have,

S5=5/2[2a+(51)d]=5/2[2a+4d]

S7=7/2[2a+(71)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+(101)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,

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