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5 and 15 are the first

Expert replies
by BTGmoderatorDC » Wed Oct 25, 2017 4:31 pm
5 and 15 are the first two terms in a geometric sequence. What is the arithmetic difference between the 11th term and the 13th term?

A. 3*5^2
B. 5* 3^13 - 5 * 3^11
C. 5^13
D. 40 * 3^10
E. 3^12 - 3^10

How will i start solving it? Can some experts help?

OA D
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Source: — Problem Solving |

5 and 15 are the first

by EconomistGMATTutor » Wed Oct 25, 2017 5:21 pm
5 and 15 are the first two terms in a geometric sequence. What is the arithmetic difference between the 11th term and the 13th term?

A. 3*5^2
B. 5* 3^13 - 5 * 3^11
C. 5^13
D. 40 * 3^10
E. 3^12 - 3^10

How will i start solving it? Can some experts help?
Hi lheiannie07,
Let's take a look at your question.

First two terms of geometric sequence are 5 and 15.
$$a_1=5$$
$$a_2=15$$

We can find the common ratio by dividing the second term by first term.
$$r=\frac{a_2}{a_1}=\frac{15}{5}=3$$

Now, we will find 11th term and 13th term of the geometric sequence.
We know that nth term of a geometric sequence can be calculated using formula:
$$a_n=a_1r^{n-1}$$
Therefore, 11th term will be:
$$a_{11}=5.3^{11-1}$$
$$a_{11}=5.3^{10}$$
13th term will be:
$$a_{13}=5.3^{13-1}$$
$$a_{13}=5.3^{12}$$

In the question statement, we are asked to find the arithmetic difference between 11th and 13th term.
$$a_{13}-a_{11}$$
$$=5.3^{12}-5.3^{10}$$
Factor out 5.
$$=5\left(3^{12}-3^{10}\right)$$
Write 3^12 as 3^10*3^2
$$=5\left(3^{10}.3^2-3^{10}\right)$$
Factor out 3^10
$$=5.3^{10}\left(3^2-1\right)$$
$$=5.3^{10}\left(9-1\right)$$
$$=5.3^{10}\left(8\right)$$
$$=40.3^{10}$$

Therefore, Option D is correct.

Hope it helps.
I am available if you'd like any follow up.
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