Let $$a1$$,$$a2$$,$$a3$$,…,$$a10$$ be in G.P. with $$a51=25$$ and ∑$$i=1101$$ $$ai=125$$. Then the value of ∑$$i=1101$$ $$1ai$$ is

Let the $$1st$$ term be a and common ratio be r . Then $$a1$$−$$r1011$$−$$r=125$$∑$$r=1101$$ $$1ai=1a1+1a2+$$…$$+1a101$$ $$=1a1$$−$$r1011$$−r⋅$$1r100=125ar502=125625=15$$

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$Let a_{1}, a_{2}, a_{3}, \dots, a_{10} be in G.P. with a_{51} = 25 and \sum_{i = 1}^{101} a_{i} = 125 . Then the value of \sum_{i = 1}^{101} (\frac{1}{a_{i}}) is$

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$\begin{array}{l} Let the 1^{st} term be a and common ratio be r . \\ Then \frac{a (1 - r^{101})}{1 - r} = 125 \end{array}$
$\begin{array}{l} \sum_{r = 1}^{101} \frac{1}{a_{i}} = (\frac{1}{a_{1}} + \frac{1}{a_{2}} + \dots + \frac{1}{a_{101}}) \\ = \frac{1}{a} (\frac{1 - r^{101}}{1 - r}) \cdot \frac{1}{r^{100}} = \frac{125}{{(a r^{50})}^{2}} = \frac{125}{625} = \frac{1}{5} \end{array}$

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Geometric progression

Remember concepts with our Masterclasses.

Let a1,a2,a3,…,a10 be in G.P. with a51=25 and ∑i=1101 ai=125. Then the value of ∑i=1101 1ai is

Let the 1st term be a and common ratio be r . Then a1−r1011−r=125∑r=1101 1ai=1a1+1a2+…+1a101 =1a1−r1011−r⋅1r100=125ar502=125625=15

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$Let a_{1}, a_{2}, a_{3}, \dots, a_{10} be in G.P. with a_{51} = 25 and \sum_{i = 1}^{101} a_{i} = 125 . Then the value of \sum_{i = 1}^{101} (\frac{1}{a_{i}}) is$