If sum of n terms of an AP is $3 n^{2} + 5 n$ then which of its terms is $164$ ?

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Solution

$S_{n} = 3 n^{2} + 5 n$
$S_{1} = a_{1} = 8$
$S_{2} = a_{1} + a_{2} = 22 \to a_{2} = 14$
$S_{3} = a_{1} + a_{2} + a_{3} = 42 \to a_{3} = 20$

AP is $8, 14, 20$
$a = 8, d = 6$
$164 = a + (m - 1) d = 8 + (m - 1) 6$
$m - 1 = 26$
$m = 27$

$27^{t h}$ term is $164$

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The sum of n terms of an AP is 3n² + 5n and then which term of AP is 164.

If the sum of n terms of an AP is 3n²+5n then which of its terms is 164?

If the sum of n terms of an A.P., if $3 n^{2} + 5 n$ then which of its terms is 164 ?

n

A . P

3 n^{2} + 5 n

164

S_{n} = (3 n^{2} + 5 n)

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