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+(n−1)d] ⋯(i)
∴ Sum of first five terms of an AP, S5=52[2a+(5−1)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+(7−1)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+(10−1)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)=47⇒2a+45=47
⇒2a=47−45=2⇒a=1
Sum of first twenty terms of this AP, S20=202[2a+(20−1)d]
=10[2×(1)+19×(5)]=10(2+95)
=10×97=970