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प्रश्न
Q 2
बेरीज
उत्तर
Let G1, G2, G3, be three geometric means between a = `1/3` and b = 432
Then, `1/3` G1, G2, G3, 432 is a G.P.
Thus, we have
First term = a =`1/3`
5th term of the G.P. = ar4 = 432
`=>1/3xxr^4=432`
⇒ r4 = 1296
⇒ r4 = 64
⇒ r = 6
∴ G1 = ar = `1/3` x 6 = 2
G2 = ar2 = `1/3` x 6 x 6 = 12
G3 = ar3 = `1/3` x 6 x 6 x 6 = 72
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Simple Applications - Geometric Progression
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