What is the sum of the 12 terms of AP if a1+a5+a8+a12=24 ?
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For an AP the nth terms can be written as a + (n-1)*d, where a is the first term and d is the common difference between consecutive terms. The sum of the first n terms is (t1 + tn)*(n/2)
In the problem we have a1+a5+a8+a12=24
=> a1 + a1 + 4d + a1 + 7d + a1 + 11d = 24
=> 4*a1 + 22d = 24
=> 2*a1 + 11d = 12
The sum we have to find is (a1 + a12)*6
=> (a1 + a1 + 11d)*6
=> (2a1 + 11d)*6
=> 12*6
=> 72
The sum is 72.
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We'll write the sum of 12 terms of an arithmetical progression:
S12 = (a1 + a12)*12/2
We'll have to determine the sum a1 + a12, for finding S12.
We also know, from enunciation, that a1+a5+a8+a12=24.
According to the rule:
(a1 + a8)/2 = (a5 + a12)/2
Since the ratios are equal, we'll interchange the terms:
a1 + a12 = a5 + a8
We'll substitute a5 + a12 into the sum:
a1 + a12 + a1 + a12 = 24
2(a1 + a12) = 24
a1 + a12 = 12
S12 = 12*12/2
The sum is: S12 = 72
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