Question

If the sum of the first $12$ terms of an A.P. is $468$ and its common difference is $6$, find the ${10}^{th}$ term

Answer 1

We know $d=6$ and sum of $12$ terms is $468$

Using the formula for sum of $n$ terms of an A.P.

$S_{12}=\frac{12}{2}[2a+(12-1\left)\right(6\left)\right]$

$468=6[2a+66]$

$468=12a+396$

$a=6$

Using the formula for $n^{th}$ term to find $a_{10}$:

$a_n=a+(n-1)d$

So, the ${10}^{th}$ term is:

$a_{10}=a+(10-1)d=6+9\times 6=60$.