Question

Find the ${60}^{th}$ term of the AP $8,10,12.....$ if it has a total of $60$ terms and hence find the sum of its last $10$ terms.

Answer 1

Given,

$a=8$

$d=10-8=2$

$n=60$

Formula:

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

$T_{60}=8+(60-1)2=126$

Formula:

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Sum of last 10 terms,

$=S_{60}-S_{50}$

$=\frac{60}{2}\left[2\right(8)+(60-1\left)2\right]-\frac{50}{2}\left[2\right(8)+(50-1\left)2\right]$

$=\frac{60}{2}\left[134\right]-\frac{50}{2}\left[114\right]$

$=4020-2850$

$=1170$