If a1-12- 5-a2-15 - 8-a3-18 - 11- then a1-a2-a3-a100 is -

The correct option is A 25151 a_1=12(5)=13(12 15) a_2=15(8)=13(15 18) ... a_100=1299(302)=13(1299 1302) All the terms like 15,18,...,12 ...

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Byju's Answer

Standard XII

Mathematics

nth Term of HP

If a1=12× 5...

Question

If $a_{1} = \frac{1}{2 \times 5}, a_{2} = \frac{1}{5 \times 8}, a_{3} = \frac{1}{8 \times 11}, . . .,$ then $a_{1} + a_{2} + a_{3} + . . . + a_{100}$ is :

\frac{25}{151}

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\frac{1}{2}

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\frac{1}{4}

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\frac{111}{55}

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Solution

The correct option is A $\frac{25}{151}$
$a_{1} = \frac{1}{2 (5)} = \frac{1}{3} (\frac{1}{2} - \frac{1}{5})$

$a_{2} = \frac{1}{5 (8)} = \frac{1}{3} (\frac{1}{5} - \frac{1}{8})$

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.
.
$a_{100} = \frac{1}{299 (302)} = \frac{1}{3} (\frac{1}{299} - \frac{1}{302})$

All the terms like $\frac{1}{5}, \frac{1}{8}, . . ., \frac{1}{299}$ will cancel out.

Therefore, the sum = $\frac{1}{3} (\frac{1}{2} - \frac{1}{302})$

$= \frac{1}{3} \frac{(300)}{(2) (302)} = \frac{50}{302} = \frac{25}{151}$

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Similar questions

If $a_{1} = \frac{1}{2 \times 5}, a_{2} = \frac{1}{5 \times 8}, a_{3} = \frac{1}{8 \times 11}, . . . . ., t h e n a_{1} + a_{2} + a_{3} + . . . . . + a_{100} i s$

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\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{1}{\sqrt{a_{3}} + \sqrt{a_{4}}}

is equal to

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a_{3}

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\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}} =

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If a1=12×5,a2=15×8,a3=18×11,..., then a1+a2+a3+...+a100 is :

The correct option is A 25151a1=12(5)=13(12−15)a2=15(8)=13(15−18)...a100=1299(302)=13(1299−1302)All the terms like 15,18,...,1299 will cancel out.Therefore, the sum = 13(12−1302)=13(300)(2)(302)=50302=25151

If $a_{1} = \frac{1}{2 \times 5}, a_{2} = \frac{1}{5 \times 8}, a_{3} = \frac{1}{8 \times 11}, . . .,$ then $a_{1} + a_{2} + a_{3} + . . . + a_{100}$ is :