In an arithmetic progression, if and , then is equal to : [2025]

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Solution

Q.
In an arithmetic progression, if $S_{40} = 1030$ and $S_{12} = 57$ , then $S_{30} - S_{10}$ is equal to : [2025]

1 505

2 525

3 510

4 515

Ans.
(4)

Given $S_{40} = 1030$ and $S_{12} = 57$

Let a be the first term and d be the common difference. We have,

$S_{40} = \frac{40}{2} [2 a + 39 d] = 1030; S_{12} = \frac{12}{2} [2 a + 11 d] = 57$

On simplifying, we get

$2 a + 39 d = \frac{103}{2}$ ... (i)

$2 a + 11 d = \frac{19}{2}$ ... (ii)

Subtracting (ii) from (i), we get

$28 d = 42 \Rightarrow d = \frac{3}{2}$

Using equation (ii),

$2 a + 11 \times \frac{3}{2} = \frac{19}{2} \Rightarrow a = \frac{- 7}{2}$

Now, we have

$S_{30} - S_{10} = \frac{30}{2} [2 a + 29 d] - \frac{10}{2} [2 a + 9 d]$

. $= 20 a + 390 d = 20 \times (- \frac{7}{2}) + 390 \times \frac{3}{2} = 515$ .

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