how many terms of ap :24,21,18 must be taken so that their sum is 78
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Answered by
8
A=24
D=21-24=-3
Sn=78
78=n/2(2a+(n-1)d
78=n/2(2*24-3n+3)
156=n(51-3n)
156=51n-3n^2
3n^2-51n+156=0
N^2-17n+52=0
N^2-13n-4n+52
N(n-13)-4(n-13)
(N-13)(n-4)
N=13
N=4
D=21-24=-3
Sn=78
78=n/2(2a+(n-1)d
78=n/2(2*24-3n+3)
156=n(51-3n)
156=51n-3n^2
3n^2-51n+156=0
N^2-17n+52=0
N^2-13n-4n+52
N(n-13)-4(n-13)
(N-13)(n-4)
N=13
N=4
Answered by
7
AP: 24, 21, 18, ....
a = 24, d = -3, Sn = 78
Sn = (2a + (n - 1)d)n / 2
78 = (48 + (n - 1)(-3))n / 2
156 = (48 - 3n + 3)n
156 = 51n - 3n²
3n² - 51n + 156 = 0
n² - 17n + 52 = 0
n²-4n-13n+52=0
n(n-4)-13(n-4)=0
(n-13)(n-4) = 0
n = 13 or 4
therefore the sum of the first 4 terms and the sum of the first 13 terms are both equal to 78.
a = 24, d = -3, Sn = 78
Sn = (2a + (n - 1)d)n / 2
78 = (48 + (n - 1)(-3))n / 2
156 = (48 - 3n + 3)n
156 = 51n - 3n²
3n² - 51n + 156 = 0
n² - 17n + 52 = 0
n²-4n-13n+52=0
n(n-4)-13(n-4)=0
(n-13)(n-4) = 0
n = 13 or 4
therefore the sum of the first 4 terms and the sum of the first 13 terms are both equal to 78.
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