The sum of 3 number in a a.p is 12 and sum of their cube is 288 .find the number
Answers
Answered by
3
Start Here
close
Sign Up
menu
X
Sign in or Register for a free account
Download the Free App

K-12 Wiki
News
Knowledge World
Exam Corner
Q & A Forum
Experts Panel
Ask
+
Academic Questions and Answers Forum, 91000+ Questions asked
View all questions
Agam answered 6 month(s) ago
If the sum of 3 numbers in a.p. is 12 and the sum of their cubes is 288. find the numbers.
Class-X Maths
person
Asked by Ishita
Feb 11
2 Likes
14401 views
editAnswer
Like
Follow
3 Answers
Top Recommend
|
Recent
person
Raghunath , SubjectMatterExpert
Member since Apr 11 2014
Consider the numbers are a , a+d , a+2d
Given that S3 =12
⇒ a + a + d + a + 2d = 12
3(a + d) =12
a + d = 4
a = 4 - d ------------(1)
And sum of their cubes is 288.
(a)3 + (a + d)3 + (a + 2d)3 = 288
a3 + a3 +d3 +3a2d +3ad2 +a3 + 8d3 + 6a2d +12ad2=288
3a3 + 9d3 +9a2d +15ad2 =288
3(4-d)3 + 9d3 +9(4-d)2d +15(4-d)d2 = 288 [using equation 1]
3(64 -d3 -48d +12d2 ) + 9d3 + 9(16 + d2 -8d) d + (60 -15d)d2 = 288
192 - 3d3 - 144d + 36d2 +9d3 + 144d + 9d3 - 72d2+60d2 - 15d3 = 288
24d2 = 288 -192
24d2 =96
d2 = 96/24
d2 = 4
d = ±2
For d = 2, a = 4 – d = 4 – 2 = 2
The numbers will be 2, 4 and 6.
For d = - 2, a = 4 - (-2) = 4 + 2 = 6
The numbers will be 6, 4 and 2.
Hence, the required numbers are 2, 4 and 6.
close
Sign Up
menu
X
Sign in or Register for a free account
Download the Free App

K-12 Wiki
News
Knowledge World
Exam Corner
Q & A Forum
Experts Panel
Ask
+
Academic Questions and Answers Forum, 91000+ Questions asked
View all questions
Agam answered 6 month(s) ago
If the sum of 3 numbers in a.p. is 12 and the sum of their cubes is 288. find the numbers.
Class-X Maths
person
Asked by Ishita
Feb 11
2 Likes
14401 views
editAnswer
Like
Follow
3 Answers
Top Recommend
|
Recent
person
Raghunath , SubjectMatterExpert
Member since Apr 11 2014
Consider the numbers are a , a+d , a+2d
Given that S3 =12
⇒ a + a + d + a + 2d = 12
3(a + d) =12
a + d = 4
a = 4 - d ------------(1)
And sum of their cubes is 288.
(a)3 + (a + d)3 + (a + 2d)3 = 288
a3 + a3 +d3 +3a2d +3ad2 +a3 + 8d3 + 6a2d +12ad2=288
3a3 + 9d3 +9a2d +15ad2 =288
3(4-d)3 + 9d3 +9(4-d)2d +15(4-d)d2 = 288 [using equation 1]
3(64 -d3 -48d +12d2 ) + 9d3 + 9(16 + d2 -8d) d + (60 -15d)d2 = 288
192 - 3d3 - 144d + 36d2 +9d3 + 144d + 9d3 - 72d2+60d2 - 15d3 = 288
24d2 = 288 -192
24d2 =96
d2 = 96/24
d2 = 4
d = ±2
For d = 2, a = 4 – d = 4 – 2 = 2
The numbers will be 2, 4 and 6.
For d = - 2, a = 4 - (-2) = 4 + 2 = 6
The numbers will be 6, 4 and 2.
Hence, the required numbers are 2, 4 and 6.
Similar questions