in an AP whose 1st term is 2 the sum of 1st 5 terms is one fourth the sum of the next 5 terms ....show that T20 = -112...find S20.. plz help me
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S20 = -1100
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Sn = n/2[2a + (n – 1)d]
S5 = 5/2[4 + 4d] = 10 + 10d
S10 = 10/2[4 + 9d] = 20 + 45d
difference
S10 – S5 = 20 + 45d – 10 – 10d
= 10 + 35d
S5 = 1/4(10 + 35d)
10 + 10d = 2.5 + 8.75d
1.25d = – 7.5
d = – 6
S30 = 30/2[4 + 29(–6)]
= 15[–170]
= – 2550
S5 = 5/2[4 + 4d] = 10 + 10d
S10 = 10/2[4 + 9d] = 20 + 45d
difference
S10 – S5 = 20 + 45d – 10 – 10d
= 10 + 35d
S5 = 1/4(10 + 35d)
10 + 10d = 2.5 + 8.75d
1.25d = – 7.5
d = – 6
S30 = 30/2[4 + 29(–6)]
= 15[–170]
= – 2550
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