In an arithmetic progression,if S_{40} = 1030 | vedclass

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(B) Let $a$ be the first term and $d$ be the common difference of the $AP$.$S_{n} = \frac{n}{2}[2a + (n-1)d]$For $S_{40} = 1030$: $\frac{40}{2}[2a + 3

Vedclass · Accepted Answer

$515$

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(B) Let $a$ be the first term and $d$ be the common difference of the $AP$.$S_{n} = \frac{n}{2}[2a + (n-1)d]$For $S_{40} = 1030$: $\frac{40}{2}[2a + 39d] = 1030 \implies 20(2a + 39d) = 1030 \implies 2a + 39d = 51.5$  $(1)$For $S_{12} = 57$: $\frac{12}{2}[2a + 11d] = 57 \implies 6(2a + 11d) = 57 \implies 2a + 11d = 9.5$  $(2)$Subtracting $(2)$ from $(1)$: $(2a + 39d) - (2a + 11d) = 51.5 - 9.5 \implies 28d = 42 \implies d = 1.5$Substituting $d = 1.5$ in $(2)$: $2a + 11(1.5) = 9.5 \implies 2a + 16.5 = 9.5 \implies 2a = -7 \implies a = -3.5$Now,$S_{30} - S_{10} = \frac{30}{2}[2a + 29d] - \frac{10}{2}[2a + 9d] = 15[2a + 29d] - 5[2a + 9d] = 30a + 435d - 10a - 45d = 20a + 390d$Substituting values: $20(-3.5) + 390(1.5) = -70 + 585 = 515$.

In an arithmetic progression,if $S_{40} = 1030$ and $S_{12} = 57$,then $S_{30} - S_{10}$ is equal to:

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