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# Find: the 10th Term of the G.P. − 3 4 , 1 2 , − 1 3 , 2 9 , . . . - Mathematics

Find:

the 10th term of the G.P.

$- \frac{3}{4}, \frac{1}{2}, - \frac{1}{3}, \frac{2}{9}, . . .$

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#### Solution

Here,

$\text { First term }, a = \frac{- 3}{4}$

$\text { Common ratio, } r = \frac{a_2}{a_1} = \frac{\frac{1}{2}}{- \frac{3}{4}} = - \frac{2}{3}$

$\therefore 10th \text { term }= a_{10} = a r^{(10 - 1)} = \left( \frac{- 3}{4} \right) \left( \frac{- 2}{3} \right)^9 = \frac{1}{2} \left( \frac{2}{3} \right)^8$

$\text { Thus, the 10th term of the given GP is } \frac{1}{2} \left( \frac{2}{3} \right)^8 .$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 20 Geometric Progression
Exercise 20.1 | Q 3.2 | Page 10
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