Question

Which term of the A.P. 3,15,27,39,.... Will be 132more than it's 54th term

Answer 1

Given,

First term ( a ) = 3

Common difference ( d ) = Difference of two consecutive terms

= 15 - 3 = 12

Now,

Nth term = a + ( N - 1 ) d

54th term = 3 + ( 54 - 1 ) 12

54th term =3 + 53 * 12

54th term = 3 + 636

54th term = 639.

The term which is 132 more than 639.

The value of that term = 132 + 639 = 771.

Now,

Nth term = a +( N - 1 ) d

771 = 3 + ( N - 1 ) 12

771 = 3 + 12N - 12

771 = 12N - 9

12N = 771 + 9

12N = 780

N = 780 ÷ 12

N = 65.

The required term is 65th.


Hope it helps !!

Answer 2

Hey there!

Given A.P : 3,15,27, 39....

a = 3

d = 15-3 = 12

a54 = a + 53d
= 3 + 53 × 12
= 639 is 54th term.

132 more than it's 54th term

132 + 639 = 771

an = a + (n-1) d
771 = 3 + (n-1) 12

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

(n-1) = 64
n = 65

So, 65th term will be 132 more than it's 54th term

Glad if helped. :D