Question

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

Answer 1

Answer:

AP - 3, 15, 27, 39...........

a = 3

d = 15-3 = 12

54th term = t54 = a + 53d

=> 3 + 53(12)

=> 3+ 636

=> 639

132 more than 54th term = 132 + 639 = 771

=> an = a + (n-1)d

=> 771 = 3 + (n-1)12

=> 771 - 3 = 12n - 12

=> 768 + 12 = 12n

=> 780/12 = n

=> n = 65

therefore the term which is 132 more than 54th term is 65th term

i think this is the answer...........I hope this helps u

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Assume,

First term be a

Common difference be d

Here,

a = 3,

d = 15 - 3 = 12

n = 54

Using Formula :-

a(n) = a + (n - 1)d

a(54) = 3 + (54 - 1) × 12

a(54) = 3 + 53 × 12

a(54) = 3 + 636

a(54) = 639

Now,

132 more than its 54th term :-

= 132 + 639

= 771

Now,

a(n) = a + (n - 1)d

a + (n - 1)d = 771

3 + (n - 1)12 = 771

3 + 12n - 12 = 771

12n = 771 + 12 - 3

12n = 780

n = 780/12

n = 65

Therefore,

We get,

65th term is 132 more than 54th term