Question #3e258

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Answer 1 · 2017-02-19T09:17:53

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

Here

$a_{3} = 10$ $a_3=10$

$a_{5} = 26$ $a_5=26$

On solving, we get

$a + 4 d = 26$ $a+4d=26$

$a + 3 d = 10$ $a+3d=10$

On substituting $a = 1$ $a=1$ , we get

$d = 3$ $d=3$

Therefore,

$a_{n} = 1 + 3 (n - 1)$ $a_n = 1 + 3(n-1)$