prepfortests1

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

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

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

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)

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

Great, it is choice A: 4×3^19

it has to be A only.

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!

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)