मराठी

In a G.P. the ratio of the sum of first 3 terms is to that of first 6 terms is 125 : 152. Find the common ratio.

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प्रश्न

In a G.P. the ratio of the sum of first 3 terms is to that of first 6 terms is 125 : 152. Find the common ratio.

बेरीज
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उत्तर

Let the first term be $$a$$ and the common ratio be $$r$$.

$$S_3 = \frac{a(1 - r^3)}{1 - r}, S_6 = \frac{a(1 - r^6)}{1 - r}$$

Since $$1 - r^6 = (1 - r^3)(1 + r^3)$$,

$$\frac{S_3}{S_6} = \frac{1 - r^3}{1 - r^6} = \frac{1}{1 + r^3}$$

The given ratio is $$125 : 152$$, so

$$\frac{1}{1 + r^3} = \frac{125}{152}$$

$$\Rightarrow 1 + r^3 = \frac{152}{125}$$

$$\Rightarrow r^3 = \frac{152 - 125}{125}$$

$$= \frac{27}{125}$$

$$= \left(\frac{3}{5}\right)^3$$

Hence, the common ratio is $$r = \dfrac{3}{5}$$.

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पाठ 11: Geometric Progression - EXERCISE 11B [पृष्ठ १८०]

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आर. एस. अग्रवाल Mathematics [English] Class 10 ICSE
पाठ 11 Geometric Progression
EXERCISE 11B | Q 11. | पृष्ठ १८०
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