Question

If sum of three terms of an arithmetic progression is 24 and their product is 440

then find the terms.​

Answer 1

Answer:

d=±3

Step-by-step explanation:

Let the three numbers in A.P. be  a − d , a , and a + d

According to given information

Sum=( a − d ) + ( a ) + ( a + d ) = 24 ...(1)

⇒3a = 24

∴ a = 8 &  Product = ( a − d) a ( a + d ) = 440 ...(2)

⇒ ( 8 − d ) ( 8 ) ( 8 + d ) = 440

⇒ ( 8 − d ) ( 8 + d ) = 55

⇒ 64 − d2 = 55

⇒ d2 = 64 − 55 =9

⇒ d = ± 3

Therefore when d=3, the numbers are 5,8,11 and

Therefore when d=3, the numbers are 5,8,11 andwhen d=−3, the numbers are 11,8 and 5.

Therefore when d=3, the numbers are 5,8,11 andwhen d=−3, the numbers are 11,8 and 5.Thus the three numbers are 5,8 and 11.

Answer 2

Answer:

If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers. Let the three numbers in A.P. be a – d, a, and a + d. Therefore, when d = 3, the numbers are 5, 8, and 11 and when d = –3, the numbers are 11, 8, and 5. Thus, the three numbers are 5, 8, and 11.