If a, b, c are in A.P,; b, c, d are in G.P and $$\dfrac{1}{c}$$, $$\dfrac{1}{d}$$$$\dfrac{1}{e}$$ are in A.P. prove that a, c, e are in G.P.

Asked by Abhisek | 1 year ago |  79

##### Solution :-

Given a, b, c are in A.P.

Hence, b – a = c – b  … (1)

And, given that b, c, d are in G.P.

So, c2 = bd  ..........… (2)

Also, $$\dfrac{1}{c}$$, $$\dfrac{1}{d}$$$$\dfrac{1}{e}$$ are in A.P.

So,

$$\dfrac{1}{d}- \dfrac{1}{c}= \dfrac{1}{e}- \dfrac{1}{d}$$

$$\dfrac{2}{d}= \dfrac{1}{c}+ \dfrac{1}{e}$$ ................(3)

Now, required to prove that a, c, e are in G.P. i.e., c2 = ae

From (1), we have

2b = a + c

b =$$\dfrac{(a+c)}{2}$$

And from (2), we have

d = $$\dfrac{c^2}{b}$$

On substituting these values in (3), we get

$$\dfrac{2b}{c^2}=\dfrac{1}{c}+\dfrac{1}{e}$$

$$\dfrac{2(a+c)}{2c^2}=\dfrac{1}{c}+\dfrac{1}{e}$$

= $$\dfrac{a+c}{c^2}=\dfrac{e+c}{ce}$$

$$\dfrac{a+c}{c}=\dfrac{e+c}{e}$$

= (a+c)e=(e+c)c

= ae+ce=ec+c2

c2=ae

Therefore, a, c, and e are in G.P.

Answered by Pragya Singh | 1 year ago

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