Question

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

Answer 1

Consider the given A.P. 8, 10, 12.....

Hence the first term a is 8

And the common difference d=10-8=2 or 12-10=2

Therefore,${60}^{th}$ term is $a_{60}=8+(60-1)2$

$\Rightarrow a_{60}=8+59\times 2$

$\Rightarrow a_{60}=126$

We need to find the sum of last 10 terms

Since, sum of last 10 terms=sum of first 60 terms-sum of first 50 terms.

Sum of Last 10 terms = $\frac{60}{2}[2\times 8+(60-1\left)2\right]-\frac{50}{2}[2\times 8+(50-1\left)2\right]$

$=\frac{60}{2}\times 2[8+59]-\frac{50}{2}\times 2[8+49]$

$=60\times 67-50\times 57$

=4020-2850

=1170

Hence ,the sum of last 10 terms is 1170.