we have
a=2
sum of 1st 5 terms: S5
sum of next 5 terms: S10-S5
So, we now have
S5 = (S10-S5)/4
4S5 = S10-S5
5S5 = S10
(5)(5/2)(4+4d) = (10/2)(4+9d)
d = -6
The progression is thus
2, -4, -10, -16, -22, -28, -34, -40, -46, -52 ...
S5 = -50
S10 = -250
S10-S5 = -200
T20 = 2+19(-6) = -112
S20 = (20/2)(4+19(-6)) = -1100
In an AP whose first term is 2,the sum of first five term is one fourth the sum of the next five terms.show that T20=-112. Find S20.
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