We know that the sum of n terms of an arithmetic progression with its first term as a and common difference d is Sn=2n[2a+(n−1)d].
Given, S10=530
⇒210[2a+9d]=530
⇒2a+9d=106…(1)
And S5=140
⇒25[2a+4d]=140
⇒2a+4d=56…(2)
Subtracting the two equation, we get, 5d=50
⇒d=10
On putting the value of d in the equation (2), we get
2a+4×10=56
⇒2a=16
⇒a=8.
Now, S20−S6=220[2a+19d]−26[2a+5d]
=14a+175d
=(14×8)+(175×10)
=1862.