If the first term of AP is 2 and the sum of first 5 terms is equal to one fourth of the sum of the next five terms find the sum of the 20 terms
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It is given that a = 2
And S5 = 1/4(S10-S5)
n/2(2a+(n-1)d) = 1/4(n/2(2a+(n-1)d-n/2(2a+(n-1)d)
5/2(4+4d) = 1/4(5(4+9d)-5/2(4+4d)
10+10d =1/4 (20+45d-10-10d)
40+40d = 10+35d
30 = -5d
d = -6
So, S20 = 20/2(2(2)+(20-1)-6)
=-1100
And S5 = 1/4(S10-S5)
n/2(2a+(n-1)d) = 1/4(n/2(2a+(n-1)d-n/2(2a+(n-1)d)
5/2(4+4d) = 1/4(5(4+9d)-5/2(4+4d)
10+10d =1/4 (20+45d-10-10d)
40+40d = 10+35d
30 = -5d
d = -6
So, S20 = 20/2(2(2)+(20-1)-6)
=-1100
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