prepfortests1

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simba12123: Senior | Next Rank: 100 Posts; Posts: 94; Joined: Tue Oct 14, 2008 1:05 pm

by simba12123 » Wed Oct 29, 2008 8:37 am

What is the sum of the 20th and 21st terms in the sequence 1, 3, 9, 27... where each term is 3 times the preceding term?

1. 4×3^19
2. 10×3^19
3. 4×3^20
4. 10×3^20
5. 3^39

qa is1

i counted out and got to the point of 3^19+3^20 and fell for trap of qa is 5. what did i do wrong?

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Source: Beat The GMAT — Problem Solving | Wed Oct 29, 2008 8:37 am

logitech: Legendary Member; Posts: 2134; Joined: Mon Oct 20, 2008 11:26 pmFollowed by: 25 members; Thanked: 237 times; GMAT Score:730

Re: prepfortests1

by logitech » Wed Oct 29, 2008 8:54 am

simba12123 wrote:What is the sum of the 20th and 21st terms in the sequence 1, 3, 9, 27... where each term is 3 times the preceding term?

1. 4×3^19
2. 10×3^19
3. 4×3^20
4. 10×3^20
5. 3^39

qa is1

i counted out and got to the point of 3^19+3^20 and fell for trap of qa is 5. what did i do wrong?

Simba,

First of all, counting will take some time, especially if you face a question that asks to find 5326th term. So it is always a good idea to look for a rule

1st term is 3^0=1
2nd term is 3^1=3
.....
......
20th term is 3^(20-1) = 3^19
21st term is 3^(21-1) = 3^20

So --> 3^19+3^20 = 3^19 ( 1 + 3 ) = 4 x 3^19

Hence, A

LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

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vittalgmat: Legendary Member; Posts: 621; Joined: Wed Apr 09, 2008 7:13 pmFollowed by: 4 members; Thanked: 33 times

by vittalgmat » Wed Oct 29, 2008 8:55 am

U almost got to the last step!!!

3^19 + 3^20 = 3^19 + 3^19 *3.

Taking the common factor 3^19,

3^19(1+3)

= 4* 3^19

HtHelps

-V

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tkhalid: Newbie | Next Rank: 10 Posts; Posts: 7; Joined: Sun Oct 19, 2008 12:09 am; Location: Faisalabad, Pakistan

by tkhalid » Wed Oct 29, 2008 9:05 am

I just want to add a simple formula

Tn = a * r^(n-1)

Where, Tn= nth term
a= first term
r= common ratio (in this case it was 3)

I hope this will help in future to determine geometric terms

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

Re: prepfortests1

by Stuart@KaplanGMAT » Wed Oct 29, 2008 11:23 am

simba12123 wrote:What is the sum of the 20th and 21st terms in the sequence 1, 3, 9, 27... where each term is 3 times the preceding term?

1. 4×3^19
2. 10×3^19
3. 4×3^20
4. 10×3^20
5. 3^39

i counted out and got to the point of 3^19+3^20 and fell for trap of qa is 5. what did i do wrong?

There's no simple way to add exponents unless they have the same base AND the same power. For example, if we have:

x^4 + x^7,

there's no easy way to add them together.

However,

3(x^4) + 4(x^4) = 7(x^4).

You applied the "multiplication of exponents" rule:

(x^a) * (x^b) = x^(a+b)

For addition, we need to do what vittalgmat did so well - get a common denominator to equalize the powers.

To do so, we almost always factor out from the bigger power to reduce it to the lower power.

Using the multiplication of exponents rule noted above, we can rewrite 3^20 as:

3^1 * 3^19 = 3(3^19)

Once we substitute that into your sum, we get:

3^19 + 3^20 = 1(3^19) + 3(3^19) = 4(3^19)

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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dhanda.arun: Senior | Next Rank: 100 Posts; Posts: 52; Joined: Thu Jul 31, 2008 12:30 pm; Thanked: 5 times

by dhanda.arun » Wed Oct 29, 2008 5:15 pm

Hi,
This is simply a Geometric Progeression where a= 1 and r = 3.
So Tn = a.r^(n-1)
T20 = 3^19
T21 = 3^20
Adding ->
3^19(1+3)
= 4.3^19

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mbaapplicant2008: Senior | Next Rank: 100 Posts; Posts: 42; Joined: Thu Oct 30, 2008 1:21 am

by mbaapplicant2008 » Thu Oct 30, 2008 1:50 am

Great, it is choice A: 4×3^19

MBA Applicant

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vivek.kapoor83: Legendary Member; Posts: 833; Joined: Mon Aug 04, 2008 1:56 am; Thanked: 13 times

by vivek.kapoor83 » Thu Oct 30, 2008 2:52 am

it has to be A only.

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simba12123: Senior | Next Rank: 100 Posts; Posts: 94; Joined: Tue Oct 14, 2008 1:05 pm

geomteric progression

by simba12123 » Sat Nov 01, 2008 9:43 am

This is a final weakspot for me. I am struggling to understand Stuart Kovinsky. However, I have one last question about this. By following the logical steps, doesn't this all boil down to ---

3^20 + 3^21 as 3^20 as best factor to pull out of? Hence I see no real reason to go to 3^19.

second question : a.) 3(3^19)= 3^20
b.) (3^2)(3^19) = 3^21

Unforutunately, I am lost on this one!

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

Re: geomteric progression

by Stuart@KaplanGMAT » Wed Nov 05, 2008 1:19 pm

simba12123 wrote:This is a final weakspot for me. I am struggling to understand Stuart Kovinsky. However, I have one last question about this. By following the logical steps, doesn't this all boil down to ---

3^20 + 3^21 as 3^20 as best factor to pull out of? Hence I see no real reason to go to 3^19.

We're asked to add the 20th and 21st terms in the sequence. Since the first term is 1 = 3^0, the 20th and 21st terms are actually 3^19 and 3^20.

In other words, the nth term in the sequence is 3^(n-1), NOT 3^n.

second question : a.) 3(3^19)= 3^20
b.) (3^2)(3^19) = 3^21

Unforutunately, I am lost on this one!

a) 3(3^19) = (3^1)(3^19) = 3^(1+19) = 3^20
b) (3^2)(3^19) = 3^(2+19) = 3^21

Both rely on the multiplication rule for exponents: when you mutliply two terms with the same base, you ADD the exponents, i.e.:

(x^a)(x^b) = x^(a+b)

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