How do i find the common ratio, when the 1st, 2nd and 7th terms of a A.P with common difference 2 are the first three terms of a G.P .

The first, second, and seventh terms of your AP would be

a, a+2, and a+12

but these are supposed to be the first 3 terms of a GP, so
(x+2)/a = (a+12)/(a+2)
cross-multiply, the a^2 drops out and you get
a = 1/2

now you know your first 3 terms of the GP to be
1/2, 5/2, and 25/2

check for common ratio:
(5/2) ÷ (1/2) = 5
(25/2) ÷ (5/2) = 5

(x+2)/a = (a+12)/(a+2)

obviously should have been

(a+2)/a = (a+12)/(a+2)

Hey thanks a lot Reiny. I got it..

To find the common ratio, you can use the formula for the nth term of a geometric progression (G.P), which is given by:

An = A1 * r^(n-1)

where An is the nth term, A1 is the first term, r is the common ratio, and n is the position of the term.

In this case, we are given that the first three terms of the arithmetic progression (A.P) are the first, second, and seventh terms of a geometric progression. Let's denote them as A1, A2, and A7 respectively.

The A.P has a common difference of 2, so we can find the terms as follows:

A1 = A
A2 = A1 + d
A7 = A1 + 6 * d

Substituting these values into the formula for the G.P, we get:

A2 = A1 * r^(2-1) => A1 + 2 = A1 * r
A7 = A1 * r^(7-1) => A1 + 6 * 2 = A1 * r^6

Now we have two equations:

A1 + 2 = A1 * r
A1 + 12 = A1 * r^6

To solve these equations, we can use a method called substitution. Rearranging the first equation, we get:

r = (A1 + 2) / A1

Substituting this value into the second equation, we get:

A1 + 12 = A1 * ((A1 + 2) / A1)^6

Simplifying the equation further, we have:

A1 + 12 = A1^6 + 12A1^5 + 48A1^4 + 96A1^3 + 96A1^2 + 64A1

This is a polynomial equation of degree 6, and solving it for A1 will give us the common ratio, r.

If you prefer, you can use online equation solvers or math software to obtain the solutions numerically.