In an AP, given $a_{3} = 15, S_{10} = 125$ , find d and $a_{10}$ .

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Solution

$a_{3} = 15, s_{10} = 125$

$a_{3} = a + (3 - 1) d$

$15 = a + 2 d$

$a = 15 - 2 d$

$s_{10} = \frac{10}{2} [2 a + (10 - 1) d]$

$125 = 5 [2 a + 9 d]$

$25 = 2 a + 9 d$
$25 = 2 (15 - 2 d) + 9 d$
$25 = 30 - 4 d + 9 d$
$5 d = - 5$
$d = - 1$
$a = 15 - 2 \times (- 1)$
$a = 17$
$a_{10} = a + (10 - 1) (- 1)$
$a_{10} = 17 - 9$
$a_{10} = 26$

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