Question

Divide 108 into 3 parts such that these
3 are in AP and the product of the two
smaller parts is 1116.



#solve it ​

Answer 1

Answer:

ok 1st term : a-d

2nd term : a

3rd term : a+d

a-d+a+a+d = 108

3a = 108

a = 36

then,

(a-d)a = 1116

(36-d)36 = 1116

36-d = 31

-d = 31-36

d = 5

terms are : 31,36 and 41

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