Definition: A G.P. is a sequence whose first term is non-zero and each of whose succeeding terms is r times the preceding term, where r is some fixed non-zero number, known as the common ratio of the G.P.

__n__^{th} Term

^{th}Term

If ‘ a ‘ is the first term and ‘ r ‘ the common ratio , then G.P. can be written as a , ar , ar^{2} , … the n^{th} term ‘ a_{n} ‘ is given by a_{n} = ar^{n-1}.

__ Sum of n Terms __

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

$ \displaystyle S_n = \frac{a(r^n – 1)}{r-1} , \, r\ne 1 $

$ \displaystyle S_n = na , \, r = 1 $

If − 1 < r < 1 , then

The sum of the infinite G.P. is

$ \displaystyle S_\infty = a + ar + ar^2 + ….\infty = \frac{a}{1-r} $

Notes:

⋄ If each term of a G.P. is multiplied (divided) by a fixed non-zero constant, then the resulting sequence is also a G.P. with same common ratio as that of the given G.P.

⋄ If each term of a G.P. (with common ratio r) is raised to the power k , then the resulting sequence is also a G.P. with common ratio r^{k}.

⋄ If a_{1 }, a_{2 }, a_{3 },….. and b_{1 }, b_{2 }, b_{3 },….. are two G.P.’s with common ratios r and r’ respectively then the sequence a_{1}b_{1}, a_{2}b_{2}, a_{3}b_{3}…..is also a G.P. with common ratio r r’ .

⋄ If we have to take three terms in a G.P., it is convenient to take them as a/r , a , ar . In general, we take

$ \displaystyle \frac{a}{r^k} ,\frac{a}{r^{k-1}} …. a , ar …. ar^k $ in case we have to take (2k+1) terms in a G.P.

⋄ If we have to take four terms in a G.P., it is convenient to take them as a/r^{3} , a/r , ar , ar^{3} .

In general, we take

$ \displaystyle \frac{a}{r^{2k-1}} ,\frac{a}{r^{2k-3}} …. \frac{a}{r} , ar …. ar^{2k-1} $ in case we have to take 2k terms in a G.P.

⋄ If a_{1 }, a_{2} . . . . , a_{n} are in G.P., then

a_{1} a_{n} = a_{2} a_{n-1} = a_{3} a_{n−2} = ….

⋄ If a_{1 }, a_{2 }, a_{3 }, …. is a G.P. ( each a_{i} > 0), then

loga_{1} , loga_{2} , loga_{3} ….. is an A.P.

The converse is also true.

⋄ If {t_{n}} is a G.P. then the common ratio r , is given by

$ \displaystyle r = (\frac{t_p}{t_q})^{1/p-q} $

__Geometric Means:__

⋄ 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 $ \displaystyle b = \sqrt{ac} $ is the geometric mean of a and c.

⋄ If a_{1}, a_{2}……a_{n} are positive numbers then their G.M (G) is given by G = (a_{1} , a_{2} , a_{3}……a_{n})^{1/n}.

If G_{1} , G_{2} ,…..G_{n} are n geometric means between a and b then a , G_{1}, G_{2}, …., G_{n} , b will be a G.P.

Here $ \displaystyle b = a r^{n+1}$

Illustration : Let a_{1}, a_{2}, a_{3} ,…… is a G.P. If 2 , 5 are two geometric means inserted between a_{4} and a_{7}, find the product of first 10 terms of the G.P.

Solution: a_{4}, 2 , 5 , a_{7} form a G.P.

⇒ a_{4}a_{7} = 2 × 5 = 10

a_{1} a_{2} a_{3} …..a_{10} = (a_{1} a_{10})(a_{2} a_{9}) (a_{3} a_{8}) (a_{4} a_{7}) (a_{5} a_{6}) = (a_{4} a_{7})^{5} = 10^{5}

Illustration : Prove that the numbers 49, 4489, 444889, …. obtained by inserting 48 into the middle of the preceding numbers are squares of integers

Solution: As we observe in T_{3} = 444889 , 4 is 3 times, 8 is 2 times and 9 only once at units place always.

We have, T_{n} = 444 ….4888…..89

It is obvious that 4 is repeated n times and 8 is repeated (n − 1) times and 9 is in units place. Now, we can write T_{n} as

T_{n} = 9+(8 × 10 + 8 × 10^{2}+ ……….+ 8 × 10^{n − 1})+(4 × 10^{n} + 4 × 10^{n + 1} + ………..+ 4 × 10^{2n−1})

$ \displaystyle 9 + \frac{80(10^{n-1}-1)}{10-1} + \frac{4\times10^n(10^n – 1)}{10-1} $

= (1/9)[81 + 80 × 10^{n − 1} − 80 + 4 × 10^{2n} − 4 × 10^{n}]

$ \displaystyle = (\frac{1 + 2\times10^n}{3})^2 $

Hence each T_{n} is the square of an integer since 2 × 10^{n} + 1 is divisible by 3 as the sum of digits in numerator is always divisible by 3.

e.g. for n = 1 , 2 , 3 , ….., Numerator = 21 , 201 , 2001 , 20001 , ….. each divisible by 3.

Illustration : If three successive terms of a G.P form the sides of a triangle then show that common ratio ‘ r ‘ satisfies the inequality .

$ \displaystyle \frac{1}{2}(\sqrt 5 -1) < r < \frac{1}{2}(\sqrt 5 +1) $

Solution: Let a , ar , ar^{2} be the terms. For triangle formation the necessary and sufficient condition is the sum of any two sides be larger than the third side.

Hence ar + ar^{2} > a ( assuming 0 < r ≤ 1)

⇒ r^{2} + r − 1 > 0 ( since a > 0 )

$ \displaystyle (r-\frac{-1-\sqrt5}{2})(r-\frac{-1+\sqrt5}{2}) > 0 $

$ \displaystyle \frac{\sqrt5 -1}{2} < r \le 1 $ .. . . . (1)

Consider r ≥ 1 then a + ar > ar^{2}

⇒ r^{2} −r −1 < 0

$ \displaystyle (r-\frac{1+\sqrt5}{2})(r-\frac{1-\sqrt5}{2}) < 0 $

$ \displaystyle 1\le r < \frac{1+\sqrt5}{2} $ . . . . . (ii)

Hence the result.

Exercise :

(i) If the pth, qth, rth terms of an A.P are in G.P, then find the common ratio of the G.P.

(ii) A G.P. consists of 2n terms. If the sum of the terms occupying the odd places is S_{1} , and that of the terms in the even places is S_{2} then find the common ratio of the progression.

(iii) If a, b, c and d are in G.P. , then show that ax^{2} + c divides ax^{3} + bx^{2} + cx + d.

(iv) If G_{1}, G_{2} are two geometric means, and A_{1} is the arithmetic mean between two positive numbers then show that $ \displaystyle \frac{G_1^2}{G_2} + \frac{G_2^2}{G_1} = 2A_1 $

(v) Show that : $ \large \left| \begin{array}{ccc} a & b & a\alpha + b \\ b & c & b\alpha + c \\ a\alpha +b & b\alpha +c & 0 \end{array} \right| = 0 $ , if a, b, c are in G. P.

### Also Read :

∗ Arithmetic Progression(G.P) ∗ Arithmetico Geometric Progression(G.P) ∗ Harmonic Progression(H.P) ∗ Miscellaneous Progression ∗ INEQUALITIES ∗ Solved Problems : Progression & Series |