मराठी

Find the Sum of the Terms of an Infinite Decreasing G.P. in Which All the Terms Are Positive, the First Term is 4, and the Difference Between the Third and Fifth Term is Equal to 32/81.

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प्रश्न

Find the sum of the terms of an infinite decreasing G.P. in which all the terms are positive, the first term is 4, and the difference between the third and fifth term is equal to 32/81.

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उत्तर

\[\text{ Let r be the common ratio of the given G . P } . \]

\[ \therefore a = 4\]

\[\text { Sum of the geometric ifinite series: } \]

\[ S_\infty = 4 + 4r + 4 r^2 + . . . \infty \]

\[\text { Now, } S_\infty = \frac{4}{1 - r} . . . . . . . \left( i \right)\]

\[\text { The difference between the third and fifth term is } \frac{32}{81} . \]

\[ a_3 - a_5 = \frac{32}{81}\]

\[ \Rightarrow 4 r^2 - 4 r^4 = \frac{32}{81}\]

\[ \Rightarrow 4\left( r^2 - r^4 \right) = \frac{32}{81}\]

\[ \Rightarrow 81 r^4 - 81 r^2 + 8 = 0 . . . . . . . \left( ii \right)\]

\[\text { Now, let } r^2 = y\]

\[\text { Let us put this in } \left( ii \right) . \]

\[ \therefore 81 r^4 - 81 r^2 + 8 = 0\]

\[ \Rightarrow 81 y^2 - 81y + 8 = 0\]

\[ \Rightarrow 81 y^2 - 72y - 9y + 8 = 0\]

\[ \Rightarrow 9y\left( 9y - 1 \right) - 8\left( 9y - 1 \right) = 0\]

\[ \Rightarrow \left( 9y - 8 \right)\left( 9y - 1 \right)\]

\[ \Rightarrow y = \frac{1}{9}, \frac{8}{9}\]

\[\text { Putting y } = r^2 ,\text {  we get } r = \frac{1}{3} \text { and } \frac{2\sqrt{2}}{3}\]

\[\text { Substituting r } = \frac{1}{3} \text { and }\frac{2\sqrt{2}}{3} \text { in } \left( i \right): \]

\[ S_\infty = \frac{4}{1 - \frac{1}{3}} = \frac{12}{2} = 6\]

\[\text { Similarly }, S_\infty = \frac{4}{1 - \frac{2\sqrt{2}}{3}} = \frac{12}{3 - 2\sqrt{2}}\]

\[ \therefore S_\infty = 6, \frac{12}{3 - 2\sqrt{2}}\]

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पाठ 20: Geometric Progression - Exercise 20.4 [पृष्ठ ४०]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 20 Geometric Progression
Exercise 20.4 | Q 5 | पृष्ठ ४०

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