Question

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

Answer 1

Answer:A=3

D=15-3=12

an=a+(n-1).d

132=3+(n-1).12

132=3+12n-12

132-3=12n-12

129=12n-12

129+12=12n

141=12n

n=11.5

Step-by-step explanation:

Answer 2

Solution

65th term will be 132 more than the 54th term

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