Question

which term of Ap 8 , 14 , 20, 26,.... will be 72 more than it's 41st term?​

Answer 1

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}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}

n

th

term=a

n

=a+(n−1)d

According to the problem given,

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

n

−a

41

=72

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

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

\implies a+nd-d-a-40d=72⟹a+nd−d−a−40d=72

\implies nd-41d = 72⟹nd−41d=72

\implies d(n-41)=72⟹d(n−41)=72

\implies n-41 = \frac{72}{6}⟹n−41=

6

72

\implies n = 41 +12⟹n=41+12

\implies n = 53⟹n=53

Therefore,

n = 53

53rd term in given A.P will be

72 more than its 41th term.

Answer 2

Answer:

Here a=8

d= 6

Let the term be An

Then An = 72+ 41th term

= 72 + (8 + ( 40 × 6 ))

=72 + (8 + 240)

=320

Hence 320= 8+ (n - 1)6

n - 1 = 52

n=53

Hope it helped ❤