If a 1-1-2 - 5- a 2-1-5 - 8- a 3-1-8 - 11- - - then a 1- a 2- a 3- a 100 isA- 25-151B-CD- 411-55

The correct option is A 25/151 a_1=1/2 × 5; a_2=1/5 × 8; a_3=1/8 × 11, .....,a_n=1/3 ( 1/2+(n 1)3 1/5+(n 1)3 ) Let, S=a_1+a_2+a_3........+a_100 S=1/3 ( 1/2 1/ ...

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If a 1=1/2 × ...

Question

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

$\frac{25}{151}$

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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
$\frac{25}{151}$

$a_{1} = \frac{1}{2 \times 5}; a_{2} = \frac{1}{5 \times 8}; a_{3} = \frac{1}{8 \times 11}, . . . . ., a_{n} = \frac{1}{3} (\frac{1}{2 + (n - 1) 3} - \frac{1}{5 + (n - 1) 3})$
Let, $S = a_{1} + a_{2} + a_{3} . . . . . . . . + a_{100}$
$S = \frac{1}{3} (\frac{1}{2} - \frac{1}{5}) + \frac{1}{3} (\frac{1}{5} - \frac{1}{8}) + \frac{1}{3} (\frac{1}{8} - \frac{1}{11}) + \frac{1}{3} (\frac{1}{299} - \frac{1}{302}) S = \frac{1}{3} (\frac{1}{2} - \frac{1}{302}) S = \frac{25}{151}$

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Similar questions

Q. 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 :

Q. If

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

then

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

Q. Show that

\frac{a_{1}}{a_{1}} + \frac{a_{2}}{a_{2}} + \frac{a_{3}}{a_{3}} + \dots \frac{a_{n}}{a_{n}} = \frac{1}{1} + \frac{1}{a_{1}} + \frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}} + \dots \frac{a_{n - 2}}{a_{n - 1}}

Q. If

a_{1}, a_{2}, a_{3}, . . . . a_{n}

are in H.P, then

\frac{a_{1}}{a_{2} + a_{3} + . . . . + a_{n}}, \frac{a_{2}}{a_{1} + a_{3} + . . . . + a_{n}}, . . . ., \frac{a_{n}}{a_{1} + a_{2} + . . . . + a_{n - 1}}

are in

Q. If

a_{1}, a_{2}, a_{3}, . . . . . a_{n}

are in H.P., then

\frac{a_{1}}{a_{2} + a_{3} + . . . a_{n}}, \frac{a_{2}}{a_{1} + a_{3} + . . . . + a_{n}}, \frac{a_{n}}{a_{1} + a_{2} + . . . + a_{n - 1}}

are in

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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 25151 a1=12×5; a2=15×8; a3=18×11,.....,an=13(12+(n−1)3−15+(n−1)3) Let, S=a1+a2+a3........+a100 S=13(12−15)+13(15−18)+13(18−111)+13(1299−1302)S=13(12−1302)S=25151

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