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 +...

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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.**