# Sum of GP Series Formula | Properties of GP

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.

### (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 : 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.

### Related Questions

The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ……. , is

If $$(10)^9$$ + $$2(11)^1(10)^8$$ + $$3(11)^2(10)^7$$ + …… + $$10(11)^9$$ = $$K(10)^9$$, then k is equal to