Question

Which of the following can be one of the term in an Arithmetic progression $4,7,10,......$

• A. $103$
• B. $123$
• C. $171$
• D. $99$

Answer 1

The correct option is A $103$

Given: $4,7,\mathrm{10.........}$ is an A.P.
Here, first term $a=4$ and common difference $d=7-4=3$
$\therefore n^{th}$ term is

$T_n=a+(n-1)d,n\in N$
$=4+(n-1)3$
$=3n+1$
Now, $T_n=3n+1=103$
$\therefore 3n=102$
$\therefore n=34$
$\therefore$ Its $34th$ term is $103$
If $T_n=3n+1=123$
$=3n=122$
$n=\frac{122}{3}\notin N$
If $T_n=3n+1=171$
$\therefore 3n=170$
$\therefore n=\frac{170}{3}\notin N$
If $T_n=3n+1=99$
$\therefore 3n=98$
$\therefore n=\frac{98}{3}\notin N$

Hence, the answer is $103$.