Let S_{n} denote the sum of the first n terms | vedclass

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(D) The sum of the first $n$ terms of an arithmetic progression is given by $S_{n} = \frac{n}{2} \{2a + (n-1)d\}$.Given $S_{10} = 530$,we have $\frac{

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$1862$

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(D) The sum of the first $n$ terms of an arithmetic progression is given by $S_{n} = \frac{n}{2} \{2a + (n-1)d\}$.Given $S_{10} = 530$,we have $\frac{10}{2} \{2a + 9d\} = 530 \Rightarrow 2a + 9d = 106 \quad \dots(1)$.Given $S_{5} = 140$,we have $\frac{5}{2} \{2a + 4d\} = 140 \Rightarrow 2a + 4d = 56 \quad \dots(2)$.Subtracting equation $(2)$ from $(1)$,we get $5d = 50$,which implies $d = 10$.Substituting $d = 10$ into equation $(2)$,$2a + 4(10) = 56$ $\Rightarrow 2a = 16$ $\Rightarrow a = 8$.Now,$S_{20} - S_{6} = \frac{20}{2} \{2a + 19d\} - \frac{6}{2} \{2a + 5d\}$.$= 10(2(8) + 19(10)) - 3(2(8) + 5(10))$.$= 10(16 + 190) - 3(16 + 50)$.$= 10(206) - 3(66) = 2060 - 198 = 1862$.

Let $S_{n}$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10} = 530$ and $S_{5} = 140$,then $S_{20} - S_{6}$ is equal to:

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