Question

cOULD YOU HELP ME PLEASE :( In a geometric series of 8 terms, whose ratio is 2, the sum of all the terms is 1020, what is the sum of the fourth and fifth term?

Answer 1

Step-by-step explanation:

If a sequence is geometric there are ways to find the sum of the first n terms, denoted Sn, without actually adding all of the terms.

To find the sum of the first Sn terms of a geometric sequence use the formula

Sn=a1(1−rn)1−r,r≠1,

where n is the number of terms, a1 is the first term and r is the common ratio.

The sum of the first n terms of a geometric sequence is called geometric series.

Example 1:

Find the sum of the first 8 terms of the geometric series if a1=1 and r=2.

S8=1(1−28)1−2=255

Example 2:

Find S10 of the geometric sequence 24,12,6,⋯.

First, find r.  

r=r2r1=1224=12

Now, find the sum:

S10=24(1−(12)10)1−12=306964

Example 3:

Evaluate.

∑n=1103(−2)n−1

(You are finding S10 for the series 3−6+12−24+⋯, whose common ratio is −2.)

Sn=a1(1−rn)1−rS10=3[1−(−2)10]1−(−2)=3(1−1024)3=−1023  

In order for an infinite geometric series to have a sum, the common ratio r must be between −1 and 1.  Then as n increases, rn gets closer and closer to 0.  To find the sum of an infinite geometric series having ratios with an absolute value less than one, use the formula, S=a11−r, where a1 is the first term and r is the common ratio.

Example 4:

Find the sum of the infinite geometric sequence

27,18,12,8,⋯.

First find r:  

r=a2a1=1827=23

Then find the sum:

S=a11−r

S=271−23=81

Example 5:

Find the sum of the infinite geometric sequence

8,12,18,27,⋯ if it exists.

First find r:  

r=a2a1=128=32

Since r=32 is not less than one the series has no sum.

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