Here you will learn what is geometric progression (gp) and sum of gp series formula and properties of gp.

Let’s begin –

## Sum of GP Series Formula

G.P. is a sequence of non zero numbers each of the succeeding term is equal to the preceding term multiplied by a constant. Thus in GP the ratio of successive terms is constant. This constant factor is called the COMMON RATIO of the sequence & is obtained by dividing any term by the immediately previous term. Therefore a, ar, a\(r^2\), ……. is a GP with ‘a’ as the first term and ‘r’ as common ratio.

**Note :**

**(a) nth term of GP**

\(T_n\) = a\(r^{n-1}\)

**(b) Sum of the first n terms of GP**

\(S_n\) = \(a(r^n – 1)\over {r – 1}\), if r \(\ne\) 1

**(c) Sum of infinite G.P. or sum of gp to infinity, **

\(S_n\) = \(a\over {1 – r}\); 0 < |r| < 1

Example 2 : Example : Find the sum of 7 terms of the gp 3,6, 12, ……

Solution : Here a = 3, r = 2

\(S_7\) = \(a(r^7 – 1)\over {r – 1}\)

\(T_n\) = \(3(2^7 – 1)\over {2 – 1}\)

= 3(128 – 1)

= 381

## Properties of GP

(a) If each term of a G.P. be multiplied or divided by the some non-zero quantity, then the resulting sequence is also a G.P.

(b) Three consecutive terms of a G.P. : a/r, a, ar;

Four consecutive terms of a G.P. : \(a/r^3\), a/r, ar \(ar^3\)

(c) If a, b, c are in G.P. then \(b^2\) = ac.

(d) If in a G.P, the product of two terms which are equidistant from the first and the last term, is constant and is equal to the product of first and last term. => \(T_k\).\(T_{n-k+1}\) = constant = a.l

(e) If each term of a G.P. be raised to the same power, then resulting sequence is also a G.P.

(f) In a G.P., \({T_r}^2\) = \(T_{r-k}\).\(T_{r+k}\), k < r, r \(\ne\) 1

(g) If the terms of a given G.P. are chosen at regular intervals, then the new sequence is also a G.P.

(h) If \(a_1\), \(a_2\), \(a_3\)……\(a_n\) is a G.P. of positive terms, then log \(a_1\), log \(a_2\)……log \(a_n\) is an A.P. and vice-versa.

(i) If \(a_1\), \(a_2\), \(a_3\)…… and \(b_1\), \(b_2\), \(b_3\)…… are two G.P.’s then \(a_1\)\(b_1\), \(a_2\)\(b_2\), \(a_3\)\(b_3\)……. & \(a_1 \over b_1\), \(a_2 \over b_2\), \(a_3 \over b_3\)……. is also in G.P.

Example : If a, b, c, d and p are distinct real numbers such that \((a^2 + b^2 + c^2)p^2\) – 2p(ab + bc + cd) + \((b^2 + c^2 + d^2)\) \(\leq\) 0 then a, b, c, d are in

Solution : Here, the given condition \((a^2 + b^2 + c^2)p^2\) – 2p(ab + bc + cd) + \((b^2 + c^2 + d^2)\) \(\leq\) 0

=> \((ap – b)^2\) + \((bp – c)^2\) + \((cp – d)^2\) \(\leq\) 0

\(\because\) a square cannot be negative

\(\therefore\) ap – b = 0, bp – c = 0, cp – d = 0

=> p = \(b \over a\) = \(c \over b\) = \(d \over c\) => a, b, c, d are in G.P.