The sum of the first 14 terms of an AP is 1050 and its first term is 10. Find the $20^{t h}$ term. ___

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Solution

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

where $S_{n}$ is the sum of n terms of the AP,

'n' is the number of terms,

'a' is the first term,

'd' is the common difference.

Given, $S_{14}$ = 1050, n = 14, a = 10.

Substituting these values we have,

$1050 = (\frac{14}{2}) [20 + 13 d]$

$\Rightarrow 1050 = 140 + 91 d$

$\Rightarrow 910 = 91 d$

$\Rightarrow d = 10$

$t_{n} = a + (n - 1) d$

Therefore, $t_{20} = 10 + (20 - 1) \times 10 = 200$

i.e. $20^{t h}$ term is 200.

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