the sum of three consecutive terms of an A.P. is 36 and their products is 1140. find in the terms
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Answered by
16
Answer:
The sum of three constraints equal 36
Answered by
1
Answer:
The terms are : 5,12,19 or 19,12,5
Step-by-step explanation:
Let a-d,a,a+d be the three consecutive term of A.P.
Given: a-d+a+a+d=36 , (a-d)a(a+d)=1140
a-d+a+a+d=36
3a=36
a =36/3
a=12
(a-d)a(a+d)=1140
(12-d)12(12+d)=1140
12(12²-d²)=1140
12(144-d²)=1140
144-d²=1140/12
= 95
-d² = 95-144
-d²=-49
d²=49
d=7 or -7
The terms are : 5,12,19 or 19,12,5
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