Question

The 16th term from the last of an AP 5, 7, 9, 11,…,165 is

• A. 130
• B. 135
• C. 140
• D. 150

Answer 1

The correct option is B 135

AP : 5, 7, 9, 11,…,165.
Here, first term (a) = 5, common difference (d) = 7 – 5 = 2.
We know that the nth term, $a_n=a+(n–1)d$.
$\Rightarrow 165=5+(n-1)\times 2$
$\Rightarrow n-1=\frac{160}{2}=80$
$\Rightarrow n=81$
The rth term from the end is $(n–r+1{)}^{th}$ term from the starting.
Therefore, 16th term from the end is $(81–16+1{)}^{th}$ = 66th term from the starting.
$\therefore a_{66}=a+65d$
$=5+65\times 2$
$=5+130$
$=135$
Therefore, the ${16}^{th}$ term from the last of the given AP is 135. Hence, the correct answer is option (2).