What is the some of this geometric series: 16 + 24 + ... + 81 + 121.5?

Answer 1

The sum (to use the correct homonym) of this series, in which each term is multiplied by 3/2, is
16 + 24 + 36 + 54 + 81 + 121.5 = 332.5

There were so few terms in this that I just "cheated" and added them up individually. But if this was longer, it would make much more sense to use the handy formula for a geometric series, with a=16, r=3/2, and n=5.

a(1 - r^(n+1))/(1-r) = 16*[1 - (3/2)^6 ] / (1 - 3/2) = 332.5

Answer 2

the common ratio is 3/2 = 1.5

You can use that evil formula for the sum of a geometric series, but if you are going to use a calculator anyway, you could also do it by brute force (trying to find the missing few terms and then just adding them all up). Since r=3/2....

a(1) = 16
a(2) = 24
a(3) = 24*3/2 = 36
a(4) = 36*3/2 = 54
a(5) = 54*3/2 = 81, therefore 121.5 is the 6th term.

So, as I said before, you can just add up the 6 terms on your calculator, or you can use the following formula for the sum of the first 6 terms of this sequence:

S(6) = a1(1-r^n) / (1-r)
= 16 (1 - 1.5^6) / (1-1.5)
= -32 (1-1.5^6)

Answer 3

you want a SUM, not some (which you may obtain elsewhere than in a math book). Since you know your first and last terms, the number of terms, and the geometric series multiplier (1.5), you can substitute in the sum formula to obtain the sum or just add it up directly: (40+90+ 202.5)

Answer 4

Find the number of terms first.
common ratio=r=24/16=1.5
first term=16
general term=16(1.5)^(n-1)
last term=121.5=16(1.5)^(n-1)
(1.5)^(n-1)=121.5/16=7.59375
n-1=log(7.59375)/log(1.5)
n-1=5
n=6
sum=16(1.5^6-1)/(1.5-1)=332.5

Answer 5

a (1 - r^n) / (1-r)

a = 16
r= 1.5

For the terms you showed there,
n = 6
Answer, Sn = 16 * (1 - 1.5^6) / (1 - 1.5)
= 332.5

The sum to inftinity = infinity because r > 1