Insert 6 geometric means between 27 and $$\dfrac{1}{81}$$.

Asked by Aaryan | 1 year ago |  52

##### Solution :-

Let the six terms be a1, a2, a3, a4, a5, a6.

A = 27, B = $$\dfrac{1}{81}$$

Now, these 6 terms are between A and B.

So the GP is: A, a1, a2, a3, a4, a5, a6, B.

So we now have 8 terms in GP with the first term being 27 and eighth being $$\dfrac{1}{81}$$.

We know that, Tn = arn–1

Here, Tn = $$\dfrac{1}{81}$$, a = 27 and

$$\dfrac{1}{81}$$= 27r8-1

$$\dfrac{ 1}{(81×27)}$$ = r7

r = $$\dfrac{1}{3}$$

a1 = Ar = 9

a2 = Ar2 = 3

a3 = Ar3 = 1

a4 = Ar4 = $$\dfrac{1}{3}$$

a5 = Ar5 = $$\dfrac{1}{9}$$

a6 = Ar6 = $$\dfrac{1}{27}$$

The six GM between 27 and $$\dfrac{1}{81}$$ are 9, 3, 1, $$\dfrac{1}{3}$$,$$\dfrac{1}{9}$$$$\dfrac{1}{27}$$

Answered by Aaryan | 1 year ago

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