Question

The 1th ,
3th and 11th term of an arithmetic progression are the first three terms of a

geometric progression.

It is further given that the sum of the first 13 terms of the arithmetic progression is

260 .



Find, in any order, the common ratio of the geometric progression and the first term

and common difference of the arithmetic progression.​

Answer 1

For the arithmetic progression,

let the first term be a and common difference be d

so the terms are a, a+d, a+2d.. a+5d...

2nd term is a+d

3rd term is a+2d

5th term is a+5d

now if these are in GP,

(a+2d) / (a+d) = (a+5d) / (a+2d)

(a+2d)^2 = (a+d)(a+5d)

a^2 + 4ad + 4d^2 = a^2 + 6ad + 5d^2

d^2 = -2ad (Assuming d is not equal to 0 else we wont have an AP initially)

or d = -2a ... (1)

Now, using this to determine the ratio

(a+2d) / (a+d) = (a-4a) / (a-2a) = -3a/-a = 3

So,

common ratio is 3.

Answer 2

$\huge \mathfrak \orange{answer}$

Let the 3 terms of the AP be (a−d),a,(a+d)

Terms of the GP: a,(a−d),(a+d) in that order.

In a GP, terms next to each other have the same ratio.

So,

a

(a−d)

=

(a−d)

(a+d)

(a−d)

2

=a(a+d)

d

2

−2ad=ad

d

2

−3ad=0

d(d−3a)=0

We know that d is not 0 from the question. So d=3a

Common ratio=

a

(a−d)

=

a

(a−3a)

=−2