Question

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

Answer 1

Answer:

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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

Answer:

3,15,27,39,.....

First we need to calculate 54th term

we know that

an = a+( n-1) d

Here, a = 3

d = 15-3 = 12

n = 54

Putting values in formula

a54 = a+( n-1) d

= 3+(54-1)×12

= 3+53×12

= 3+636

= 639

So, 54th term = a54 = 639

we need to calculate term which is 132 more than 54th term

i.e. should be 639+132 = 771

Therefore an = 771

we need to find n

Putting values in formula

an = a+(n-1) d

771 = 3+(n-1)×12

771 - 3 = (n-1)×12

768 = (n-1)×12

768/12 = n-1

64 = n-1

n-1 = 64

n = 64+1

n = 65

So, 65th term is 132 more than 54th term