Question

can 68 be a term of AP 7 10 13? Justify your answer

Answer 1

Answer:

Let us assume that 68 is a term in the given AP sequence.

Then, 68 will be some nth term of the AP, here n is a natural number.

Therefore we know the formula that,

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

$\implies a_n = 68$

$\implies a = 7$

$\implies d = 3$

$\implies n = ?$

Substituting in the formula we get,

$\implies 68 = 7 + ( n - 1 ) 3\\\\\implies 68 - 7 = ( n - 1 ) 3\\\\\implies 61 = ( n - 1 )3\\\\\implies \dfrac{61}{3} = n - 1 \\\\\implies n = \dfrac{61}{3} +1 \implies \dfrac{ 61 + 3 }{3} \implies \dfrac{64}{3}$

Here we get n as a fraction. So our assumption was wrong.

Hence 68 is not a term of the AP.

Hence Justified !!