# 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.**

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**