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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 isÂ aÂ and the common difference isÂ d

By usingÂ Snâ€‹=n/2â€‹[2a+(nâˆ’1)d]Â we have,

S5â€‹=5/2â€‹[2a+(5âˆ’1)d]=5/2[2a+4d]

S7â€‹=7/2â€‹[2a+(7âˆ’1)d]=7/2â€‹[2a+6d]

Given:Â S7â€‹+S5â€‹=167

âˆ´5/2â€‹[2a+4d]+7/2â€‹[2a+6d]=167

â‡’10a+20d+14a+42d=334

â‡’24a+62d=334Â Â  Â  Â  …(1)

S10â€‹=10/2â€‹[2a+(10âˆ’1)d]=5(2a+9d)

Given:Â S10â€‹=235

SoÂ 5(2a+9d)=235

â‡’2a+9d=47Â Â  Â  Â  …(2)

Multiply equation (2) byÂ 12, we get

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

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

âˆ’46d=âˆ’230

Â âˆ´d=5

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

2a+9(5)=47
orÂ 2a=2

Â âˆ´a=1

Then A.P isÂ 1,6,11,16,21,â‹¯

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