Question

the sum of 3numbers in ap is 21 and the product of the first and the third term exceeds the second term by 6. find the number​

Answer 1

Answer:

let the 1st,2nd and 3rd no. be (x-d),x,(x+d) where d is common difference

1.

x-d+x+x+d=21

3x=21

x=7

2.

(x-d)(x+d)=x+6

7²-d²=7+6

49-13=d²

d=(+-6)

now....

the no. are 1,7,13

or 13,7,1

Answer 2

The three terms or the AP are

a+d,a,a-d

Given that,

a+d+a+a+d=21

3a=21

a=7

also given that,

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

a2-ad+da-d2=a+6

a2-d2=a+6

substituting the value of a=7

7*7-d2=7+6

49-13=d2

d2=36

d=+-6

so there will be 2 APs with a=7,d=+6 and a=7,d=-6

the APs are

7,13,19,...... and. 7,1,-5,......

HOPE THIS HELPS YOU