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(-100)+(-98)+(-96)+......+92
Asked by namupatel154 | 31 Dec, 2018, 07:38: AM
The given series is in arithmetic progression with first term a = -100 and common difference d =2.

last n-th term of the series, tn = a+(n-1)d = -100+(n-1)2 = 92 ............(1)
solving eqn.(1) for n, we get n = 97

Sum of the series = (n/2) [ 2a + (n-1)d ] = (97/2) [-200+(96×2)] = -388
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Alternate method :-

Sum of series = -100-98-96...........-2-0+2+4+6.......92

In the above series, numbers of right most terms, ( 2+4+6......92 ) are cancelled by numbers on left  ( -92-90-88  ...............-4-2 ).
hence sum of the numbers left in the series  = -100-98-96-94 = -388
Answered by Thiyagarajan K | 31 Dec, 2018, 10:04: AM

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