Question

15. Three numbers whose sum is 21 are in AP. If the product of the first and the third numbers exceeds the second number by 6, find the numbers.

Answer 1

Answer:

13,7,1

Step-by-step explanation:

Let the 3 numbers be (a-d), a and (a + d).

According to the question,

a-d+a+a+d=21

then 3a=21

a=7

Also,

(a-d)(a+d)=a+6

a^2-d^2=a+6

from above a=7,putting here we get

d=+-6

If d = 6, then the numbers are 7 - 6, 7, 7 + 6 

 = 1, 7, 13

If d = -6, then the numbers are 7 - (-6), 7, 7 + (-6)

 = 13, 7, 1