Find $s_{30}$ for an AP, where $a_{10} = 20$ and $a_{20} = 58$ .

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Solution

As we know nth term, $a_{n} = a + (n - 1) d$
& Sum of first $n$ terms, $S_{n} = \frac{n}{2} (2 a + (n - 1) d)$ , where $a$ & $d$ are the first term & common difference of an AP.
Let $a$ be the first term and $d$ be the common difference of the AP
Given, $a_{10} = 20 = a + 9 d$
and $a_{20} = 58 = a + 19 d$
Solving above, we get
$d = 3.8$
Hence, $S_{30} = \frac{30}{2} (2 a + 29 d)$
$= 15 (a_{10} + a_{20} + d)$
$= 15 (20 + 58 + 3.8)$
$= 1227$

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