find the sum of first 15 terms whose n term is 9-5n
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2
Last term = 9-5n
a+ (n-1)d = 9-5n
a+ nd - d = 9-5 n
a - d = 9 and nd = -5n⇔ d = -5
a - (-5) = 9 ⇔ a = 4
l = 9-5(15) = 9-75 = -66
sum = N/2 (a+ l) = 15/2 (4-66) = -31*15 = -465
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Answered by
1
the sum of first 15 terms whose n term is 9-5n is (-465)
Step-by-step explanation:
an = 9-5n
a1 = 9-5 = 4
a2 = 9-10 = -1
d = -5
so ,
S15 = 15/2 [2a +(15-1)d]
= 15/2 [ 2(4) + 14(-5)]
= 15/2 [ 8-70]
= 15/2 ×(-62)
= 15×(-31)
=(-465)
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