How many terms of the AP : 24, 21, 18, .... must be taken so that the sum is 78 ? Explain the
double answer.
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t1= 24 = a
d = t2 - t1 = 21 - 24 = -3
Sn = n/2 [2a+(n-1)d]
78=n/2[2(24)+(n-1)-3]
156=n(48-3n+3)
156=n(51-3n)
156=51n-3n^2
3n^2-51n+156=0
Dividing by 3 throughout,
n^2-17n+52=0
n^2-14n-3n+52=0
n(n-14)-3(n-14)=0
(n-3)(n-14)=0
n=3 or n=14
We get double answers because the given AP is in descending order and the common difference is negative.
d = t2 - t1 = 21 - 24 = -3
Sn = n/2 [2a+(n-1)d]
78=n/2[2(24)+(n-1)-3]
156=n(48-3n+3)
156=n(51-3n)
156=51n-3n^2
3n^2-51n+156=0
Dividing by 3 throughout,
n^2-17n+52=0
n^2-14n-3n+52=0
n(n-14)-3(n-14)=0
(n-3)(n-14)=0
n=3 or n=14
We get double answers because the given AP is in descending order and the common difference is negative.
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