The sum of how many terms of the A.P. 14, 12, 10, ….. will be zero?
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Solution:
First term, a = 14
Common difference, d = 12 - 14 = -2
Sn = n/2 [2a + (n - 1)d]
0 = n/2 [2(14) + (n - 1)-2]
0 = n/2 (28 -2n + 2)
0 = n/2 (30 - 2n)
0 * 2 = n ( 30 - 2n)
0 = 30n - 2n^2
2n^2 - 30 = 0
2n (n - 15) = 0
2n = 0 or (n - 15) = 0
n = 0 or n = 15
Hope this will help.
First term, a = 14
Common difference, d = 12 - 14 = -2
Sn = n/2 [2a + (n - 1)d]
0 = n/2 [2(14) + (n - 1)-2]
0 = n/2 (28 -2n + 2)
0 = n/2 (30 - 2n)
0 * 2 = n ( 30 - 2n)
0 = 30n - 2n^2
2n^2 - 30 = 0
2n (n - 15) = 0
2n = 0 or (n - 15) = 0
n = 0 or n = 15
Hope this will help.
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