Which term of the A.P. 8,14,20,26,...will …
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Posted by Abhishek Singh 5 years, 11 months ago
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Yogita Ingle 5 years, 11 months ago
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,
{tex}\boxed {n^{th} term = a_{n}=a+(n-1)d}{/tex}
According to the problem given,
{tex}a_{n}-a_{41}=72{/tex}
{tex}\implies a+(n-1)d-[a+(41-1)d]=72{/tex}
{tex}\implies a+nd-d-(a+40d)=72{/tex}
{tex}\implies a+nd-d-a-40d=72{/tex}
{tex}\implies nd-41d = 72{/tex}
{tex}\implies d(n-41)=72{/tex}
{tex}\implies n-41 = \frac{72}{6}{/tex}
{tex}\implies n = 41 +12{/tex}
{tex}\implies n = 53{/tex}
Therefore,
n = 53
53rd term in given A.P will be 72 more than its 41th term.
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