Let S_n denote the sum of first n-terms of an arithmetic progression. If S_10=530,S_5=140, then S_20-S_6 is equal to:

We know that the sum of $n$n terms of an arithmetic progression with its first term as $a$a and common difference $d$d is ${S}_{n}=\frac{n}{2}[2a+(n-1)d].$Sn​=2n​[2a+(n−1)d].

Given, ${S}_{10}=530$S10​=530

$$\Rightarrow \frac{10}{2}[2a+9d]=530$$⇒210​[2a+9d]=530

$$\Rightarrow 2a+9d=106\ldots (1)$$⇒2a+9d=106…(1)

And ${S}_{5}=140$S5​=140

$$\Rightarrow \frac{5}{2}[2a+4d]=140$$⇒25​[2a+4d]=140

$$\Rightarrow 2a+4d=56\ldots (2)$$⇒2a+4d=56…(2)

Subtracting the two equation, we get, $5d=50$5d=50

$$\Rightarrow d=10$$⇒d=10

On putting the value of $d$d in the equation $(2),$(2), we get

$$2a+4\times 10=56$$2a+4×10=56

$$\Rightarrow 2a=16$$⇒2a=16

$$\Rightarrow a=8.$$⇒a=8.

Now, ${S}_{20}-{S}_{6}=\frac{20}{2}[2a+19d]-\frac{6}{2}[2a+5d]$S20​−S6​=220​[2a+19d]−26​[2a+5d]

$$=14a+175d$$=14a+175d

$$=(14\times 8)+(175\times 10)$$=(14×8)+(175×10)

$$=1862.$$=1862.