t8 = t5 * r^3, so
r^3 = t8/t5 = -40/5 = -8
so, r = -2
t5 = ar^4 = 16a, so a = 5/16
S7 = a(1-r^7)/(1-r)
= 5/16 (1+2^7)/(1+2) = 5/16 * 129/3 = 215/16 = 13.4375
F) t5 = 5 and t8 = -40
I'm stuck on this one!
r^3 = t8/t5 = -40/5 = -8
so, r = -2
t5 = ar^4 = 16a, so a = 5/16
S7 = a(1-r^7)/(1-r)
= 5/16 (1+2^7)/(1+2) = 5/16 * 129/3 = 215/16 = 13.4375
That was my initial thought but I couldn't figure out how...
t5 = ar^4
t8 = ar^7
r^3 = t8/t5
a = t5/r^4 = t5 / ∛(t8/t5)^4 = t5 ∛(t5/t8)^4
S7 = t5 ∛(t5/t8)^4 (1-(t8/t5)^(7/3))/(1-∛(t8/t5))
Now just plug in the numbers and let 'er rip!
Let's start by using the formula for the nth term of a geometric series:
tn = a * r^(n-1)
Where "tn" is the nth term, "a" is the first term, "r" is the common ratio, and "n" is the term number.
We are given two equations: t5 = 5 and t8 = -40.
Using the formula above, we can write these equations as:
5 = a * r^(5-1) [equation 1]
-40 = a * r^(8-1) [equation 2]
To solve this system of equations, we can divide equation 2 by equation 1 to eliminate "a":
(-40)/(5) = (a * r^(8-1))/(a * r^(5-1))
Simplifying, we get:
-8 = r^3
Now, we need to find the value of "r" to determine the common ratio. We can take the cube root of both sides of the equation:
r = ^(3√)(-8)
The cube root of -8 is -2, so:
r = -2
Now that we have the value of "r", we can substitute it back into equation 1 to find the value of "a".
5 = a * (-2)^(5-1)
Simplifying, we get:
5 = a * (-2)^4
5 = a * 16
a = 5/16
So, the first term of the series is a = 5/16 and the common ratio is r = -2.
Now, we can use the formula for the sum of a geometric series:
Sn = a * (1 - r^n) / (1 - r)
Where "Sn" is the sum of the first n terms of the series.
We want to find the sum of the first seven terms, so n = 7.
Using the values we found, we can substitute them into the formula:
S7 = (5/16) * (1 - (-2)^7) / (1 - (-2))
Calculating this expression, we get:
S7 = (5/16) * (1 - 128) / 3
S7 = (5/16) * (-127) / 3
S7 = -635/48
Therefore, the sum of the first seven terms of the given geometric series is -635/48.