Question

Find the 15th ,24th and nth (general term)term of AP given by 3,15,27,39​

Answer 1

Answer:

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Step-by-step explanation:

$a = 3 \\ d = 12 \\ a15 = a + (15 - 1)d \\ a15 = a + 14d \\ = > a15 = 3 + 14(12) = 171 \\ = > a24 = 3 + 23(12) = 279 \\ = > an = a + (n - 1)d$

an = 3+ (n-1)12

an= 12n-9

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Answer 2

The first term (a) here is 3.

And the commln difference is 15-3 = 12.

Kepping the relation in mind,
$a_{n} = a + (n - 1)d$

We will have,

$a_{15} = 3 + (14)(12) = 171$

$a_{24} = 3 + (23)(12) = 279$


So the general term will be,

$a_{n} = 3 + (n - 1)12 \\ \: \: \: \: \: \: = 3 + 12n - 12 \\ \: \: \: \: \: = 12n - 9$