Question

Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54th term?​

Answer 1

$\huge\underline\mathfrak\purple{Solution}:-$

65th term of the AP will be 132 more than its 54th term.

$\bold \pink{\underline{Step\:by\:step\: explanation:-}}$

Given A.P. is 3, 15, 27, 39, …

first term, a = 3

common difference, d = a2 − a1 = 15 − 3 = 12

We know that,

an = a + (n − 1) d

Therefore,

a54 = a + (54 − 1) d

⇒ 3 + (53) (12)

⇒ 3 + 636 = 639

a54 = 639

We have to find the term of this A.P. which is 132 more than a54, i.e. 771.

Let nth term be 771.

= an = a + (n − 1) d

= 771 = 3 + (n − 1) 12

= 768 = (n − 1) 12

= (n − 1) = 64

= n = 65

Therefore, 65th term was 132 more than 54th term.

Answer 2

Step-by-step explanation:

54 term = 3+ 53*12

= 3+636

=639

132 more is 771

771= 3+12 n-12

12 n -12 = 768

12 n= 780

n= 65

verification

a65= 3+ 64*12

= 771