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.
You need to login to perform this action.
You will be redirected in
3 sec