Question

Is 68 a term of the A.P 7, 10, 13.....?

Answer 1

use formula of nth term=a+(n-1) d
68=7+(n-1) 3
61/3+1=n
here we look n is fractional but n can't be fractional so 68 is not a term of above series

Answer 2

Step-by-step explanation:

Given A.P:

7,10,13, ...

First term (a) = 7

$Common\: difference (d) = a_{2}-a_{1}\\=10-7\\=3$

Let n th term of given A.P is 68.

$a_{n} = 68$

$\implies a+(n-1)d = 68$

$\implies 7+(n-1)3=68$

$\implies (n-1)3 = 68-7$

$\implies (n-1)3 = 61$

$\implies (n-1)= \frac{61}{3}$

$\implies n = \frac{61}{3}+1\\=\frac{61+3}{3}\\=\frac{64}{3}$

But ,

n should not be a fraction.

Therefore,

68 is not a term in given A.P.

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