Question

Which term of the ap 8,14,20,26,.... will be 72 more than its 41 th term ?

Answer 1

please mark as brainliest answer.

Answer 2

Answer:

53 rd term of given A.P

Explanation:

Given A.P:

8,14,20,26, ...

First term (a) = 8

common difference (d) =$a_{2}-a_{1}$

= 14-8

= 6

d = 6

Let n th term of A.P will be 72 more than its 41 th term.

We know that,

$\boxed {n^{th} term = a_{n}=a+(n-1)d}$

According to the problem given,

$a_{n}-a_{41}=72$

$\implies a+(n-1)d-[a+(41-1)d]=72$

$\implies a+nd-d-(a+40d)=72$

$\implies a+nd-d-a-40d=72$

$\implies nd-41d = 72$

$\implies d(n-41)=72$

$\implies n-41 = \frac{72}{6}$

$\implies n = 41 +12$

$\implies n = 53$

Therefore,

n = 53

53rd term in given A.P will be

72 more than its 41th term.

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