Question

FROM CHAPTER ARITHEMETIC PROGRESSIONS.

CLASS 10

which term of the AP:3, 15, 27, 39, ........will be 132 more than its 54th terms? ​

Answer 1

a = 3

d = 15 - 3 = 12

T54 = 3 + (54 - 1)*12

= 3 + 53*12

= 3 + 636

= 639

let nth term will be equal to T54 + 132

= 639 + 132 = 771

a + (n - 1)d = 771

3 + (n - 1) 12 = 771

(n - 1) 12 = 771 - 3 = 768

n - 1 = 768/12 = 64

n = 64 + 1

n = 65

Answer 2

Given ,

The AP is 3 , 15 , 27 , 39 , ...

First term (a) = 3

Common difference (d) = 12

We know that ,

The nth term of an AP is given by

$\boxed{ \tt{a + (n - 1)d}}$

According to the question ,

nth term of AP is 132 more than its 54th term

Thus ,

a + (n - 1)d = 132 + 54th term

3 + (n - 1)12 = 132 + 3 + (54 - 1)12

3 + 12n - 12 = 135 + 636

12n - 9 = 771

12n = 780

n = 780/12

n = 65

Therefore , 65th term of given AP is 132 more than its 54th term

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