three numbers are in arithmetic progression and their sum is 18 and the sum of their square is 140 find the numbers
Answers
Finding the numbers in AP
Answer: Required three numbers are 2 , 6 , 10.
Explanation:
Given that three numbers are in arithematic progression.
let assume three numbers be a - d , a and a + d where first number is a - d and common difference is d .
As sum of three numbers is 18
⇒ (a - d ) + a + ( a + d ) = 18
⇒ a - d + a + a + d = 18
⇒3a = 18
⇒ a = 18/3 = 6
On substituting a = 6 , three numbers are (6 - d ) , 6 and (6 + d )
Also given sum of their square is 140
⇒ (6 - d)² + 6² + (6 + d)² = 140
=> 6² + d² -12 d + 6² + 6² + d² + 12d = 140 [ using algebraic identitites ]
⇒ 36 + d² + 36 + 36 + d² = 140
⇒ 2d² = 140 - 36 - 36 - 36
⇒ 2d² = 140 - 108 = 32
⇒ d² = 32/2
⇒ d² = 16
⇒ d = +4 or d = -4
when d = + 4
required numbers are 6 - 4 , 6 , 6 + 4 that is 2 , 6 , 10
when d = -4
required number are 6 - (-4) , 6 , 6 + (-4) thai is 10 , 6 , 2
Hence required three numbers are 2 , 6 , 10.
Answer:
Step-by-step explanation: