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Question 1
The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms of this AP is 235, find the sum of its first twenty terms.

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Solution

Let the first term, common differences and the number of terms of an AP are a, d and n, respectively.
Sum of first n terms of an AP, Sn=n2[2a+(n1)d] (i)
Sum of first five terms of an AP, S5=52[2a+(51)d] [from Eq.(i)]
=52(2a+4d)=5(a+2d)
S5=5a+10d
And, sum of first seven terms of an AP, S7=72[2a+(71)d]
=72[2a+6d]=7(a+3d)
S7=7a+21d
Now, by given condition,
S5+S7=167
5a+10d+7a+21d=167
12a+31d=167 ...(ii)
Given that, sum of first ten terms of this AP is 235.
S10=235
102[2a+(101)d]=235
5(2a+9d)=235
2a+9d=47 ...(iii)
Solving (ii) and (iii) gives,
12a+54d=28212a+31d=167 ––––––––––––––– 23d=115 d=5
Now, put the value of d in Eq. (iii), we get,
2a+9(5)=472a+45=47
2a=4745=2a=1
Sum of first twenty terms of this AP, S20=202[2a+(201)d]
=10[2×(1)+19×(5)]=10(2+95)
=10×97=970

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