Question

three numbers are in ap whose sum is 12 if we add 1,2 and 6 to these numbers respectively , the numbers become in gp. find the numbers. With explanation

Answer 1

let the three numbers be a-d, a, a+d

therefore

a-d+a+a+d=12

3a=12

a=4

a-d+1= 4+1-d = 5-d

a+2= 4+2= 6

a+d+6 = 4+6+d = 10+d

In gp the common ratio is same

I am sorry I could solve till here only

hope this helps you to solve the rest

Answer 2

Step-by-step explanation:

Let the terms of A.P. be:

a−d, a, a+d

We can write a−d + a + a+d=12

3a = 12

a = 4

if we add 1,2 and 6 to these numbers respectively

a−d + 1, a+2, a+d+6 are in G.P.

Now, we need to substitute the "a" value into the GP series.

5-d, 6, 10+d are in G.P

6/(5-d) = (10+d)/6

Now, we need to cross multiply the terms.

36 = (5-d)(10+d)

36 = 50−10d+5d−d^2

36 = 50−5d−d^2  

Move all terms to one side.

d ^2+5d-14=0

(d-2)(d+7) = 0

The d values are 2, -7

when d=2, a=4 the series is, 2, 4, 6

when d=−7, a=4 the series is, 11, 4, -3