Question

Find the ${12}^{th}$ term from the end of the arithmetic progression $3,5,7,9,...201?$

• A. $179$
• B. $175$
• C. $172$
• D. $174$

Answer 1

The correct option is A $179$

Clearly, the given sequence is an AP with first term $a=3$ and common difference $d=2$.
Let there are $n$ terms in this AP. Therefore,
$a_n=201\Rightarrow a+(n-1)d=201$
$\Rightarrow 3+(n-1)2=201$
$\Rightarrow (n-1)2=198$
$\Rightarrow n-1=99$
$\Rightarrow n=100$
Now ${12}^{th}$ term from end=$(100-12+1{)}^{th}$ term from beginning.
$\Rightarrow {12}^{th}$ term from end=${89}^{th}$ term from beginning=$3+(89-1)\times 2=3+176=179.$