Question

the sum of three terms of AP is 36 and their product 1620 find the three terms of AP should take the three terms as a minus b, a, a + b

Answer 1

(a-b)+(a+b)+(a)=36
3a=36
a=12

now
(a)(a-b)(a+b) =1620
(12)(144-b²)=1620
1728-1620= b²
b=√108

Answer 2

Answer:

let the 3 terms of the ap's be :- a, a+b, a-b

given that :- sum of the 3 terms=36

a+(a+b)+(a-b)=36

3a=36

a=12

given that :- product of the 3 terms=1620

a(a+b)(a-b)=1620

12(12^2- b^2)=1620

12^2-b^2=1620/12

144-b^2=135

-b^2=135-144= -9

b^2= root9

b=3

therefore, sub value of a&b in a, a-b, a+b to get the 3 terms=

a=12

a+b= 15

a-b= 9

hence the 3 terms of the given ap are 12, 15 , 9.

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