The sum of the first (n) terms of the sequence is given by the formula (given below) . If this sequence is a geometric progression, find q. If this sequence is an Arithmetic progression, find d

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.

Step-by-step explanation:

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)