Question

Which term of the A.P. 6, 13, 20, 27 . . . . is 98 more than its 24th term?​

Answer 1

Answer

The 38th term of the A.P

Definition of A.P

The nth term of an A.P is $a_n=a+(n-1)d$, where $a$ is the first term and $d$ is common difference.

The method is as follows.

Given A.P: $a_n=6+7(n-1)$

Then $a_n=7n-1$ is the n-th term.

Equating, we get

$a_n=a_{24}+98$

$\implies 7n-1=(7\times 24-1)+98$

$\implies 7n=7\times 24+98$

$\implies n=24+14$

$\therefore n=38$ and hence the 38th term of the A.P is 98 more than the 24th term.

Answer 2

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•The 38th term of the A.P

Definition of A.P

The nth term of an A.P is a_n=a+(n-1)dan

=a+(n−1)d , where aa is the first term and dd is common difference.

The method is as follows:

•Given A.P: a_n=6+7(n-1)a n

•=6+7(n−1)

•Then a_n=7n-1a n

• =7n−1 is the n-th term.

•Equating, we get

•a_n=a_{24}+98a n

• =a 24+98

⟹7n−1=(7×24−1)+98

⟹7n=7×24+98

⟹n=24+14

•Therefore n=38∴n=38 and hence the 38th term of the A.P is 98 more than the 24th term.

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