The sum of three consecutive terms of an A.P. is 36 and their product is 1140.
Find the terms. (Consider the terms to be in descending order.)
Answers
Answer:
5,12,19
Step-by-step explanation:
let a-d , a , a+d be the 3 terms of A.P such that
a-d + a + a+d = 36 (given)
=> 3a = 36
=> a = 36/3
=> a= 12
also given (a-d)* a * (a+d) = 1140
=> (12-d) * (12) * (12+d) = 1140
=> (12-d) (12+d) = 1140/12
=> 12²- d² = 95
=> d² = 144 - 95 = 49
=> d = √49 = 7
=> the 3 terms are , a-d, a, a+d
12-7, 12,12+7
5,12,19
Hope this help you
Answer:
Let the smallest Number of A.P be x.
so, the other numbers
=> x+1 and x+2
Given that,
sum of these terms = 36
According to the questions
x+2+x+1+x = 36
=> 3x+3 = 36
=> 3x = 36-3
=> 3x = 33
=> x = 33/3
=> x = 11
Therefore the terms are,
=> x+2 = 11+2 = 13
=> x+1 = 11+1 = 12
=> x = 11