Find the 60th 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`