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OG 12th edition Q # 45

Expert replies
by bhumika.k.shah » Fri Feb 05, 2010 3:23 am
45. If r is a constant and an = rn for all positive integers n,
for how many values of n is an < 100 ?
(1) a50 = 500
(2) a100 + a105 = 2,050

How can i find out the sequence???
Join the discussion
Source: — Data Sufficiency |

by sanju09 » Fri Feb 05, 2010 3:33 am
bhumika.k.shah wrote:45. If r is a constant and an = rn for all positive integers n,
for how many values of n is an < 100 ?
(1) a50 = 500
(2) a100 + a105 = 2,050

How can i find out the sequence???

By an, do you mean nth term or what? Can we draw the usual meaning of rn or is it r^n? Please clarify, I don't have that OG.
Last edited by sanju09 on Fri Feb 05, 2010 3:39 am, edited 1 time in total.
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Join the discussion

by bhumika.k.shah » Fri Feb 05, 2010 3:38 am
I actually dont know ... its not raised to - for sure!

its to the right hand side below the alphabet...

i dont even know whats it called as per the "math" way :-S :-(
sanju09 wrote:
bhumika.k.shah wrote:45. If r is a constant and an = rn for all positive integers n,
for how many values of n is an < 100 ?
(1) a50 = 500
(2) a100 + a105 = 2,050

How can i find out the sequence???


Do you mean a^n = r^n? If yes, then please do the necessary edit all over.
Join the discussion

by sanju09 » Fri Feb 05, 2010 3:46 am
bhumika.k.shah wrote:I actually dont know ... its not raised to - for sure!

its to the right hand side below the alphabet...

i dont even know whats it called as per the "math" way :-S :-(
sanju09 wrote:
bhumika.k.shah wrote:45. If r is a constant and an = rn for all positive integers n,
for how many values of n is an < 100 ?
(1) a50 = 500
(2) a100 + a105 = 2,050

How can i find out the sequence???


Do you mean a^n = r^n? If yes, then please do the necessary edit all over.
OK....then it is an arithmetic progression with first term r and common difference r.

(1) a50 = r + 49 r = 500 (given), r = 10. Sufficient

(2) a100 + a105 = (r + 99 r) + (r + 104 r) = 2050 (given). r = 10. Sufficient

[spoiler]D[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Join the discussion

by sanju09 » Fri Feb 05, 2010 3:52 am
I actually dont know ... its not raised to - for sure!

its to the right hand side below the alphabet...

i dont even know whats it called as per the "math" way :-S :-(

It's called SUBSCRIPT
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Join the discussion

by bhumika.k.shah » Fri Feb 05, 2010 3:56 am
I got wht u did.but dint understand why u did so???

concept please!
sanju09 wrote:
bhumika.k.shah wrote:I actually dont know ... its not raised to - for sure!

its to the right hand side below the alphabet...

i dont even know whats it called as per the "math" way :-S :-(
sanju09 wrote:
bhumika.k.shah wrote:45. If r is a constant and an = rn for all positive integers n,
for how many values of n is an < 100 ?
(1) a50 = 500
(2) a100 + a105 = 2,050

How can i find out the sequence???


Do you mean a^n = r^n? If yes, then please do the necessary edit all over.
OK....then it is an arithmetic progression with first term r and common difference r.

(1) a50 = r + 49 r = 500 (given), r = 10. Sufficient

(2) a100 + a105 = (r + 99 r) + (r + 104 r) = 2050 (given). r = 10. Sufficient

[spoiler]D[/spoiler]
Join the discussion

by sanju09 » Fri Feb 05, 2010 4:23 am
If nth term of a sequence is given by n multiplied to a contant, then it bears the first term and common difference equal to the constant, and hence it's an AP.
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Join the discussion