In a geomatic sequence, the seventh term exceeds the fifth term by 1920. Find the sum of the first eleven terms if the common ratio of the sequence is 2.

ar^6 - ar^4=1920

64a-16a=1920
48a=1920
a = 40

sum(11) = 40(2^11 - 1)/(2-1) = 81880

that's geometric sequence

T7 = T5*r^2 = T5+1920
r = 2, so

4*T5 = T5 + 1920
3T5 = 1920
T5 = 640

640 = 2^4 * 40

So, a=40 and the sequence is

40 80 160 320 640 1280 2560 5120 10240 20480 40960 ...

Note that 2560 = 640 + 1920

oops - good catch, Reiny. You actually answered the question :-)

find the fifth term in the sequence a1=-4a2=8a3=-16 can you guys help me

find the first term if the 6th and 7th term are 320 and 1920 respectively.

Why did the geomatic sequence go see a doctor? Because the seventh term had an acute case of "I'm 1920 units bigger than the fifth term"!

But fear not! I'm here to help you compute the sum of the first eleven terms.

In a geometric sequence, the n-th term can be found using the formula:

𝑎𝑛 = 𝑎1 × 𝑟^(𝑛−1)

Where 𝑎𝑛 is the n-th term, 𝑎1 is the first term, 𝑟 is the common ratio, and 𝑛 is the term number.

From the given information:

𝑎7 = 𝑎5 + 1920

Substituting the equation for the n-th term:

𝑎1 × 𝑟^6 = 𝑎1 × 𝑟^4 + 1920

Since the common ratio is given as 2, we can simplify the equation:

𝑎1 × 2^6 = 𝑎1 × 2^4 + 1920

64𝑎1 = 16𝑎1 + 1920

48𝑎1 = 1920

𝑎1 = 40

Now that we know the first term (𝑎1), we can find the sum of the first eleven terms using the formula for the sum of a geometric sequence:

𝑆𝑛 = (𝑎1 × (1 − 𝑟^𝑛)) / (1 − 𝑟)

Substituting the values:

𝑆11 = (40 × (1 − 2^11)) / (1 − 2)

Calculating this expression, we find that the sum of the first eleven terms is -30720.

So, the sum of the first eleven terms of the geometric sequence with a common ratio of 2 is -30720 (which is definitely not a laughing matter).

To find the sum of the first eleven terms in a geometric sequence, we first need to determine the first term and the common ratio.

Let's suppose the first term of the geometric sequence is represented by 'a', and the common ratio is represented by 'r'.

According to the problem, the seventh term exceeds the fifth term by 1920. Using this information, we can write the following equation:

a * r^6 = a * r^4 + 1920

Now, we are given that the common ratio (r) is 2. Substituting this value into the equation, we have:

a * 2^6 = a * 2^4 + 1920

64a = 16a + 1920

Simplifying further:

48a = 1920

Dividing both sides of the equation by 48:

a = 40

So, we now know that the first term of the geometric sequence (a) is 40, and the common ratio (r) is 2.

To find the sum of the first eleven terms, we can use the formula for the sum of a geometric series:

Sum = a * (1 - r^n) / (1 - r)

where:
- a is the first term,
- r is the common ratio, and
- n is the number of terms.

Substituting the given values into the formula:

Sum = 40 * (1 - 2^11) / (1 - 2)

Simplifying further:

Sum = 40 * (1 - 2048) / (1 - 2)

Sum = 40 * (-2047) / (-1)

Sum = 40 * 2047

Sum = 81,880

Therefore, the sum of the first eleven terms in the geometric sequence is 81,880.