If a1 = 1/(2*5), a2 = 1/(5*8), a3 = 1/(8*11), ..., then a_1 + a2 +

Question

If  a_1 = \frac{1}{2*5} ,  a_2 = \frac{1}{5*8} ,  a_3 = \frac{1}{8*11} , ..., then  a_1 + a_2 + a_3 + ...... + a_{100}  is

A. 25/151
B. 1/4
C. 1/2
D. 1
E. 111/55

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CareerGeek · Accepted Answer

a_1 = \frac{1}{2*5} ,  a_2 = \frac{1}{5*8} ,  a_3 = \frac{1}{8*11} 
-->  a_n = \frac{1}{(3n - 1)*(3n + 2)}

a_1 + a_2 + a_3 + ...... + a_{100}  =  \frac{1}{2*5}  +  \frac{1}{5*8}  +  \frac{1}{8*11}  + . . . . . .  \frac{1}{299*302}

=  \frac{1}{3} *( \frac{3}{2*5}  +  \frac{3}{5*8}  +  \frac{3}{8*11}  + . . . . . .  \frac{3}{299*302} )

=  \frac{1}{3} *( \frac{5 - 2}{2*5}  +  \frac{8 - 5}{5*8}  +  \frac{11 - 8}{8*11}  + . . . . . .  \frac{302 - 299}{299*302} )

=  \frac{1}{3} *( \frac{1}{2}  -  \frac{1}{5}  +  \frac{1}{5}  -  \frac{1}{8}  +  \frac{1}{8}  -  \frac{1}{11}  + . . . . . . . . . +  \frac{1}{299}  -  \frac{1}{302} )

=  \frac{1}{3} *( \frac{1}{2}  -  \frac{1}{302} )

=  \frac{1}{3} *( \frac{151 - 1}{302} )

=  \frac{25}{151}

Option A

exc4libur · Answer

notes: this is close to an infinite series
sum inf series: a/1-r
a=1/10
r=n_2/n_1=1/40/1/10=1/4
a/1-r=(1/10)/(3/4)=4/30=2/15~25/151

ans (A)

preetamsaha · Answer

The 100th term will be 1/299*302
The series is:
1/(2*5) + 1/(5*8) + 1/(8*11) + ……………+ 1/(299*302)
It can also be written as 
3 
3(1/2 – 1/302) = 300/(3*2*302) 
25/151
Option (A)

Vij2019 · Answer

Could someone please advise if this approach is advisable?

As the denominators are increasing over a constant numerator which means decimal is getting smaller and smaller with each next number in the sequence. And hence the solution should be in decimal and should be closer to 0.12... And the only viable choice here seems to be option A.

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Mo2men · Answer

Dear  GMATBusters   GMATGuruNY   BrentGMATPrepNow

Can you please show how to think and appraoch such a question? In above solutions, it seems there is an equation which I do not know. Can I solve the question without it?

Thanks in advance

GmatPoint · Answer

The given statement can be expressed as : 
﻿a1 =  \frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{5﻿}\right) 
﻿a2 =  \frac{1}{3}\cdot\left(\frac{1}{5}-\frac{1}{8}\right) 
﻿............................
﻿a100 =  \frac{1}{3}\cdot\left(\frac{1}{299}-\frac{1}{302}\right) 
﻿Adding the terms :
﻿a1+a2+....................... a100 =  \frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.............-\frac{1}{302}\right) 
﻿All the terms gets cancelled out other than the firs and the last last term.
 ﻿\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{302}\right)\ =\ \frac{100}{\left(2\cdot302\right)}=\ \frac{25}{151}

leoburityfilho · Answer

Can you explain how  \frac{1}{2*5}  becomes  \frac{1}{3} * (\frac{1}{2}  -  \frac{1}{5} )?

jayouh · Answer

Here is how I solved this

a1 + a2 = 1/(2*5) + 1/(5*8) = 1/8

So I want to find the 'second multiplier' in the denominator for a100.

I notice that the two multipliers in denominator increase by 3 in every term.

So the 'second multiplier' is 5 + 99 *3 = 302 which is what's going to be in the denominator for the sum.

Then for nominator, because a1+a2 equals 1, my thinking is that for the sum of 100 it's going to be 50

So 50/302 = 25/151.

I probably got lucky.

bumpbot · Answer

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If a1 = 1/(25), a2 = 1/(58), a3 = 1/(8*11), ..., then a_1 + a2 +

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