A geometric sequence of positive terms has a first term of 9 and a third term of 4. What i… — ACT

A geometric sequence of positive terms has a first term of $9$ and a third term of $4$ . What is the sum of the first $4$ terms of this sequence?

The correct answer is

\frac{65}{3}

To find the sum of the first 4 terms of a geometric sequence with

a_1 = 9

and

a_3 = 4

, we first determine the common ratio

r

. Since

a_3 = a_1 \cdot r^2

, we have

4 = 9 \cdot r^2

, which gives

r^2 = \frac{4}{9}

. Because the terms are positive,

r = \frac{2}{3}

. The first 4 terms are

9

6

4

, and

\frac{8}{3}

. Summing these terms gives

9 + 6 + 4 + \frac{8}{3} = 19 + \frac{8}{3} = \frac{65}{3}

Find the common ratio

r

of the geometric sequence.

r = \frac{2}{3}

Since the sequence is geometric, the third term is related to the first term by

a_3 = a_1 \cdot r^2

. Substituting the given values yields

4 = 9 \cdot r^2

, which simplifies to

r^2 = \frac{4}{9}

. Because all terms in the sequence are positive,

r

must be positive, so

r = \sqrt{\frac{4}{9}} = \frac{2}{3}

Calculate the first 4 terms of the sequence.

a_1 = 9

a_2 = 6

a_3 = 4

a_4 = \frac{8}{3}

Multiply each term by the common ratio

r = \frac{2}{3}

to find the subsequent term:

a_1 = 9

a_2 = 9 \cdot \frac{2}{3} = 6

a_3 = 6 \cdot \frac{2}{3} = 4

, and

a_4 = 4 \cdot \frac{2}{3} = \frac{8}{3}

Sum the first 4 terms of the sequence.

Sum =

\frac{65}{3}

Add the four terms:

9 + 6 + 4 + \frac{8}{3} = 19 + \frac{8}{3} = \frac{57}{3} + \frac{8}{3} = \frac{65}{3}

Calculating the sum of the first

n

terms of a geometric sequence given its first and third terms.

Question