Question

If second, sixth and eighteenth terms of an A.P. are the consecutive terms of a G.P., find the common ratio.
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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

The common ratio is 3

Given :

Second , sixth and eighteenth terms of an AP are the consecutive terms of a GP

To find :

The common ratio

Concept :

1. The nth term of an AP is

aₙ = a + (n - 1 )d.

a = first term

aₙ = nth term

d = common difference.

2. If a , b , c are in GP then b² = ac

Solution :

Step 1 of 2 :

Form the equation

Let for the AP , first term = a and common difference = d

Second term = a + d

Sixth term = a + 5d

Eighteenth term = a + 17d

∴ a + d , a + 5d , a + 17d are in GP

We know that if a , b , c are in GP then b² = ac

So by the given condition

$\displaystyle \sf \therefore {(a + 5d)}^{2} = (a + d)(a + 17d)$

Step 2 of 2 :

Find common ratio

$\displaystyle \sf {(a + 5d)}^{2} = (a + d)(a + 17d)$

$\displaystyle \sf{ \implies } {a}^{2} + 10ad + 25{d}^{2} = {a}^{2} + 17ad + ad + 17 {d}^{2}$

$\displaystyle \sf{ \implies } {a}^{2} + 10ad + 25{d}^{2} = {a}^{2} + 18ad + 17 {d}^{2}$

$\displaystyle \sf{ \implies } 8{d}^{2} = 8ad$

$\displaystyle \sf{ \implies }d = a$

So terms of the GP are 2d , 6d , 18d

Hence common ratio of the GP

$\displaystyle \sf = \frac{ Second \: term}{ First \: term }$

$\displaystyle \sf = \frac{6d}{2d}$

$\displaystyle \sf = 3$

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