Find:

Solution:

In the given problem, we are given an A.P and the value of one of its term.

We need to find whether it is a term of the A.P or not.

So here we will use the formula, $a_{n}=a+(n-1) d$

(i) Here, A.P is $7,10,13, \ldots$

$a_{e}=68$

 

$a=7$

Now,

Common difference $(d)=a_{1}-a$

$=10-7$

 

$=3$

Thus, using the above mentioned formula, we get,

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

$68-7=3 n-3$

$61+3=3 n$

$n=\frac{64}{3}$

Since, the value of n is a fraction.

Thus, 68 is not the term of the given A.P

Therefore the answer is No

(ii) Here, A.P is $3,8,13, \ldots$

$a_{s}=302$

 

$a=3$

Now,

Common difference $(d)=a_{1}-a$

$=8-3$

 

$=5$

Thus, using the above mentioned formula, we get,

$302=3+(n-1) 5$

$302-3=5 n-5$

$299=5 n$

$n=\frac{299}{5}$

Since, the value of n is a fraction.

Thus, 302 is not the term of the given A.P

Therefore the answer is No

(iii) Here, A.P is $11,8,5,2, \ldots$

$a_{n}=-150$

 

$a=11$

Now,

Common difference $(d)=a_{1}-a$

$=8-11$

 

$=-3$

Thus, using the above mentioned formula

$-150=11+(n-1)(-3)$

$-150-11=-3 n+3$

$-161-3=-3 n$

$n=\frac{-164}{-3}$

Since, the value of n is a fraction.

Thus, -150 is not the term of the given A.P

Therefore, the answer is No.