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