The sum of the first hundred terms of an arithmetic progression with first term a and common difference d is T

Question

The sum of the first 50 odd-numbered terms , i.e. the first, third, fifth, ..., ninety-ninth, is 1/2 T -1000. Find the value of d.

Answer 1

S100=50[2a+99d]=T
S50=25[2a+49*2d]=0.5T-1000
100a+4950d=T
100a+4900d=t-2000
subtracting
50d=2000
d=40

Answer 2

The sum of an arithmetic series is 1st term plus last term times half the number of terms. For the 1st series 1st term is a, last term is a + 99d, and so

T = 50( a + a + 99d) = 100a + 4950d

For the 2nd series, since we take every other term out of the 1st, 1st term is a, common difference is 2d, last term is a + 49(2d) = a + 98d, and so

T/2 - 1000 = 25(a + a + 98d)
T/2 - 1000 = 50a + 2450d
T - 2000 = 100a + 4900d
T = 100a + 4900d + 2000

Setting the two T equations equal,

100a + 4900d + 2000 = 100a + 4950d
2000 = 50d
40 = d.

Answer 3

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Answer 4

100a+4950d=T
50a+1225(d+1)=.5T-1000
50a+1225d+1225=50a+2475-1000
1225d=250
d=10/49