Question

which term of the A. P. 3,15,27,39...will be 120 more than its 21st term

Answer 1

First term, a = 3
Common difference, d = 15-3 = 12
$t21 = a + (21 - 1)d \\ = 3 + 20 \times 12 = 243$

Let the required number be nth term
Then,
$tn = t21 + 120 \\ a + (n - 1)d = 243 + 120 \\ 3 + (n - 1) \times 12 = 363 \\ n - 1 = \frac{360}{12} = 30 \\ n = 31$

Therefore, 31st term is 120 more than 21st term

Answer 2

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