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.) Chalo bye gn…ab meko jana hoga ^_- Ye last wala ha About the author Valentina
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 Reply
Answer:
[tex]the \: three \: terms \: are \: 5 \: \: 12 \: \: 19[/tex]
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