Q: How do you find the common ratio in a geometric progression when the sum of the first 20 terms is 244 times the sum of its first 10 terms?

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The common ratio is the ratio of the nth term (n > 1) to the (n-1)th term. For the progression to be geometric, this ratio must be a non-zero constant.

Divide any term, except the first, by the term before it.

It is a geometric progression with common ratio 0.5

In an arithmetic progression the difference between each term (except the first) and the one before is a constant. In a geometric progression, their ratio is a constant. That is, Arithmetic progression U(n) - U(n-1) = d, where d, the common difference, is a constant and n = 2, 3, 4, ... Equivalently, U(n) = U(n-1) + d = U(1) + (n-1)*d Geometric progression U(n) / U(n-1) = r, where r, the common ratio is a non-zero constant and n = 2, 3, 4, ... Equivalently, U(n) = U(n-1)*r = U(1)*r^(n-1).

15. It's a Geometric Progression with a Common Ratio of 1/5 (or 0.2).

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The common ratio is the ratio of the nth term (n > 1) to the (n-1)th term. For the progression to be geometric, this ratio must be a non-zero constant.

For an Arithmetic Progression, Sum = 15[a + 7d].{a = first term and d = common difference} For a Geometric Progression, Sum = a[1-r^15]/(r-1).{r = common ratio }.

Divide any term, except the first, by the term before it.

It is a geometric progression with common ratio 0.5

Geology, Geography, Geometry, Gems, Gold, Gadolinium, Gallium, Germanium, Graduated Cylinder, Gametes, Gauges, Geotropism, Gigabytes, Gigapascal, Gluon, and Gravity.

The geometric series is, itself, a sum of a geometric progression. The sum of an infinite geometric sequence exists if the common ratio has an absolute value which is less than 1, and not if it is 1 or greater.

In an arithmetic progression the difference between each term (except the first) and the one before is a constant. In a geometric progression, their ratio is a constant. That is, Arithmetic progression U(n) - U(n-1) = d, where d, the common difference, is a constant and n = 2, 3, 4, ... Equivalently, U(n) = U(n-1) + d = U(1) + (n-1)*d Geometric progression U(n) / U(n-1) = r, where r, the common ratio is a non-zero constant and n = 2, 3, 4, ... Equivalently, U(n) = U(n-1)*r = U(1)*r^(n-1).

15. It's a Geometric Progression with a Common Ratio of 1/5 (or 0.2).

It's a geometric progression with the initial term 1/2 and common ratio 1/2. The infinite sum of the series is 1.

Find the 7th term of the geometric sequence whose common ratio is 1/2 and whose first turn is 5

if a number is multiplied by 1, then it does not change, it is Still the same number. A ratio of 1 is impossible . The ratio between two quantities must always be greater than 1 otherwise there is no difference between them.

ar5 = 160 ar8 = 1280 so r3 = 1280/160 = 8 and so r = 2 then, ar5 = 160 gives a = 160/32 = 5 First term = a = 5 Common ratio = r = 2