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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.
Given: A.P. 8, 10, 12, ……., if it has a total of 60 terms
To do: Find the 60th term of the A.P. and find the sum of its last 10 terms.
Solution:
The given A.P. 8, 10, 12, …
Here the first term $a= 8$ and the common difference $d=10 ‐ 8 = 2$
As known
$n^{th}$ term of an A.P., $a_{n}=a+(n-1)d$
Its $60^{th}$ term $a_{60}=8+(60-1)2$
$=124$
As known sum of n terms in an A.P., $S_{n}=\frac{n}{2}( 2a+( n-1) d)$
We need to find the sum of the last 10 terms.
Thus,
Sum of last 10 terms $=$Sum of first 60 terms$ ‐$ Sum of first 50 terms
Thus the sum of last 10 terms $=S_{60} -S_{50}=\frac{60}{2}( 2\times 8+59\times 2) -\frac{50}{2}( 2\times 8+49\times 2)$
$=30( 134)-25( 114)$
$=4020-2850$
$=1170$
$\boxed{Therefore,\ Sum\ of\ last\ 10\ terms\ of\ the\ given\ A.P.\ is\ 1170.}$
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