Question

Find the ${12}^{th}$ term from the end of the A.P. $2,6,10.........58$

• A. $22$
• B. $14$
• C. $16$
• D. $28$

Answer 1

The correct option is B $14$

The given A.P. can be written in reverse order as $58..........10,6,2$

Here $a=58,d=-4,n=12$
$a_{12}=a+(12-1)d$
$=58+11\times (-4)$
$=58-44$
$a_{12}=14$
$\therefore$ $14$ is the ${12}^{th}$ term from the end.