question_answer

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.

Answer:

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

Hence the first term is 8

And the common difference

\[d=10-8=2\]

or                \[12-10=2\]

Therefore, 60th 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.

\[{{S}_{10}}=\frac{60}{2}[2\times 8+(60-1)2]-\frac{50}{2}[2\times 8+(50-1)2]\]

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

\[=60\times 67-50\times 57\]

\[=4020-2850\]

\[=1170\]

Hence, the sum of last 10 terms is 1170.