If a is the first term and r is the common ratio of the geometric progression,

then its n^{th} term is given by a_{n} = ar^{n-1}

The sum S_{n} of the first n terms of the G.P. is given by

S_{n} = a (r^{n} – 1)/ (r-1), when r ≠1; = na if r =1

If -1 < x < 1, then limx^{n} = 0, as n →∞. Hence, the sum of an infinite G.P. is 1+x+x^{2}+ ….. = 1/(1-x)

If -1 < r< 1, then the sum of the infinite G.P. is a +ar+ ar^{2}+ ….. = a/(1-r)

If each term of the G.P is multiplied or divided by a non-zero fixed constant, the resulting sequence is again a G.P.

If a_{1}, a_{2}, a_{3}, …. andb_{1}, b_{2}, b_{3}, … are two geometric progressions, then a_{1}b_{1}, a_{2}b_{2}, a_{3}b_{3}, …… is also a geometric progression and a_{1}/b_{1}, a_{2}/b_{2}, ... ... ..., a_{n}/b_{n} will also be in G.P.

Suppose a_{1}, a_{2}, a_{3}, ……,a_{n} are in G.P. then a_{n}, a_{n–1}, a_{n–2}, ……, a_{3}, a_{2}, a_{1} will also be in G.P.

Taking the inverse of a G.P. also results a G.P. Suppose a_{1}, a_{2}, a_{3}, ……,a_{n} are in G.P then 1/a_{1}, 1/a_{2}, 1/a_{3} ……, 1/a_{n} will also be in G.P

If we need to assume three numbers in G.P. then they should be assumed as a/b, a, ab (here common ratio is b)

Four numbers in G.P. should be assumed as a/b^{3}, a/b, ab, ab^{3} (here common ratio is b^{2})

Five numbers in G.P. a/b^{2}, a/b, a, ab, ab^{2} (here common ratio is b)

If a_{1}, a_{2}, a_{3},… ,a_{n} is a G.P (a_{i}> 0 ∀i), then log a_{1}, log a_{2}, log a_{3}, ……, log a_{n} is an A.P. In this case, the converse of the statement also holds good.

If three terms are in G.P., then the middle term is called the geometric mean (G.M.) between the two. So if a, b, c are in G.P., then b

= √ac is the geometric mean of a and c.