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 its last 10 terms.

#### Solution

Consider the given A.P. 8, 10, 12, …

Here the initial term is 8 and the common difference is 10 - 8 = 2 and 12 - 10 = 2

General term of an A.P. is tn and formula to tn is

`t_n= a +(n-1) d`

`t_60=8+(60-1)xx2`

`t_60=8+59xx2`

`t_60=8+118`

`t_60=126`

We need to find the sum of last 10 terms.

Thus,

Sum of last 10 terms = Sum of first 60 terms - Sum of first 50 terms

`S_n=n/2[2a+(n-1)d]`

`S_60=60/2[2xx8+(60-1)xx2]`

`S_60=30[16+118]`

`S_60=30[134]`

`S_60=4020`

Similarly,

`S_50=50/2[2xx8+(50-1)xx2]`

`S_50=25[16+98]`

`S_50=25[114]`

`S_50=2850`

Therefore,

Sum of last 10 terms = Sum of first 60 terms Sum of first 50 terms

Thus the sum of last 10 terms`=S_60 - S_50= 4020-2850= 1170`