What is the sum of the last 7terms of the geometric series with 𝑡3 = 3, 𝑡7 = 768 and the last term being the 10th term?

Question

What is the sum of the last 7terms of the geometric series with 𝑡3 = 3, 𝑡7 = 768 and the last term being the 10th term?

Answer 1

wrong again t7/t3 = r^4, not r^6

so, r^4 = 768/3 = 256
so r = ±4
now, t1 = 3/4^2 = 3/16

S3 = 3/16 (r^3-1)/(r-1) = 3/16 (63/3) = 63/16
S10 = 3/16(r^10-1)/(r-1) = 3/16 (1048575/3) = 1048575/16
So, the last 7 terms add up to
S10-S3 = 65532

Answer 2

I apologize for the mistake. You are correct. The common ratio should be r^4, not r^6.

Using the correct common ratio, we have:
(t7/t3) = (ar^(7-1))/(ar^(3-1))
(t7/t3) = r^(4)

Substituting the given values, we have:
768/3 = r^4
256 = r^4

Taking the 4th root of both sides, we get:
r = ±4

Since the common ratio of a geometric series cannot be negative, we take r = 4.

Now, let's find the value of the first term (a) using the first equation:
3 = a * 4^(3-1)
3 = 16a
a = 3/16

So, the first term (a) is 3/16 and the common ratio (r) is 4.

Now, let's find the last term of the series, which is the 10th term:
tn = ar^(n-1)
t10 = (3/16) * 4^(10-1)
t10 = (3/16) * 4^9
t10 = (3/16) * 262144
t10 = 49152

To find the sum of the last 7 terms, we need to find the sum of the 10 terms and subtract the sum of the first three terms.

Sum of the 10 terms = (3/16 * (4^10 - 1)) / (4 - 1)
Sum of the 10 terms = (3/16 * 1048575) / 3
Sum of the 10 terms = 1048575/16

Sum of the first three terms = (3/16 * (4^3 - 1)) / (4 - 1)
Sum of the first three terms = (3/16 * 63) / 3
Sum of the first three terms = 63/16

Sum of the last 7 terms = Sum of the 10 terms - Sum of the first three terms
Sum of the last 7 terms = 1048575/16 - 63/16
Sum of the last 7 terms = (1048575 - 63) / 16
Sum of the last 7 terms = 1048512/16
Sum of the last 7 terms = 65532

Answer 3

To find the sum of the last 7 terms of a geometric series, we need to find the common ratio (r) and the first term (a).

Given that 𝑡3 = 3 and 𝑡7 = 768, we can use these two terms to find the common ratio (r) and the first term (a).

Step 1: Finding the common ratio (r)
Using the formula for the nth term of a geometric series:
𝑡ₙ = 𝑎 * 𝑟^(𝑛−1)

For 𝑡₇, we have:
𝑡₇ = 𝑎 * 𝑟^(7−1)
768 = 𝑎 * 𝑟^6 --------(1)

For 𝑡₃, we have:
𝑡₃ = 𝑎 * 𝑟^(3−1)
3 = 𝑎 * 𝑟^2 --------(2)

By dividing equation (1) by equation (2), we can eliminate the first term 'a':
768/3 = (𝑎 * 𝑟^6) / (𝑎 * 𝑟^2)
256 = 𝑟^4

Taking the fourth root of both sides gives us:
𝑟 = ∛256
𝑟 = 4

Step 2: Finding the first term (a)
Using equation (2), we can rearrange it to solve for a:
𝑎 = 3 / 𝑟^2
𝑎 = 3 / 4^2
𝑎 = 3 / 16
𝑎 = 3/16

Now that we have the common ratio (r) and the first term (a), we can find the sum of the last 7 terms of the geometric series.

Step 3: Finding the sum of the last 7 terms
The sum of a geometric series can be calculated using the formula:

𝑆𝑛 = 𝑎 * (1 − 𝑟^𝑛) / (1 − 𝑟)

For the last 7 terms, n = 7. Plugging in the values:
𝑆₇ = 𝑎 * (1 − 𝑟^7) / (1 − 𝑟)
𝑆₇ = (3/16) * (1 − 4^7) / (1 − 4)

Calculating the values inside the brackets:
𝑆₇ = (3/16) * (-16383) / (-3)

Simplifying:
𝑆₇ = 17139/16
𝑆₇ = 1071.19

Therefore, the sum of the last 7 terms of the geometric series is approximately 1071.19.

Answer 4

Let's denote the first term of the geometric series as "a" and the common ratio as "r". We are given that t3 = 3 and t7 = 768.

The formula for the nth term of a geometric series is given by tn = ar^(n-1).

Using this formula, we can write equations based on the given information:
t3 = ar^(3-1) = 3
t7 = ar^(7-1) = 768

Dividing the second equation by the first equation, we get:
(t7/t3) = (ar^(7-1))/(ar^(3-1))
(t7/t3) = r^(6)

Substituting the given values, we have:
768/3 = r^6
256 = r^6

Now, let's find the common ratio (r):
Taking the 6th root of both sides, we get:
r = 2

Now, we can find the value of the first term (a) using the first equation:
3 = a * 2^(3-1)
3 = 4a
a = 3/4

So, the first term (a) is 3/4 and the common ratio (r) is 2.

Now, let's find the last term of the series, which is the 10th term:
tn = ar^(n-1)
t10 = (3/4) * 2^(10-1)
t10 = (3/4) * 2^9
t10 = (3/4) * 512
t10 = 384

To find the sum of the last 7 terms, we need to find the sum of the 10 terms and subtract the sum of the first three terms:
Sum = (a * (r^n - 1)) / (r - 1)

Sum of the 10 terms = (3/4 * (2^10 - 1)) / (2 - 1)
Sum of the 10 terms = (3/4 * 1023) / 1
Sum of the 10 terms = 767.25

Sum of the first three terms = (a * (r^3 - 1)) / (r - 1)
Sum of the first three terms = (3/4 * (2^3 - 1)) / (2 - 1)
Sum of the first three terms = (3/4 * 7) / 1
Sum of the first three terms = 21/4

Sum of the last 7 terms = Sum of the 10 terms - Sum of the first three terms
Sum of the last 7 terms = 767.25 - 21/4
Sum of the last 7 terms = (3070 - 21) / 4
Sum of the last 7 terms = 3049/4
Sum of the last 7 terms = 762.25