Geometric Progression(G.P)

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.

nth Term

If ‘ a ‘ is the first term and ‘ r ‘ the common ratio , then G.P. can be written as a , ar , ar2 , … the nth term ‘ an ‘ is given by an = arn-1.

Sum of n Terms

The sum Sn 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 rk.

⋄ If a, a, a,….. and b, b, b,….. are two G.P.’s with common ratios r and r’ respectively then the sequence a1b1, a2b2, a3b3…..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/r3 , a/r , ar , ar3 .

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 a1 , a2 . . . . , an are in G.P., then

a1 an = a2 an-1 = a3 an−2 = ….

⋄ If a, a, a, …. is a G.P. ( each ai > 0), then

loga1 , loga2 , loga3 ….. is an A.P.

The converse is also true.

⋄ If {tn} 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 a1, a2……an are positive numbers then their G.M (G) is given by G = (a1 , a2 , a3……an)1/n.

If G1 , G2 ,…..Gn are n geometric means between a and b then a , G1, G2, …., Gn , b will be a G.P.
Here $ \displaystyle b = a r^{n+1}$

Illustration : Let a1, a2, a3 ,…… is a G.P. If 2 , 5 are two geometric means inserted between a4 and a7, find the product of first 10 terms of the G.P.

Solution: a4, 2 , 5 , a7 form a G.P.

⇒  a4a7 = 2 × 5 = 10

a1 a2 a3 …..a10 = (a1 a10)(a2 a9) (a3 a8) (a4 a7) (a5 a6) = (a4 a7)5 = 105

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 T3 = 444889 , 4 is 3 times, 8 is 2 times and 9 only once at units place always.

We have, Tn = 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 Tn as

Tn = 9+(8 × 10 + 8 × 102+ ……….+ 8 × 10n − 1)+(4 × 10n + 4 × 10n + 1 + ………..+ 4 × 102n−1)

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

= (1/9)[81 + 80 × 10n − 1 − 80 + 4 × 102n − 4 × 10n]

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

Hence each Tn is the square of an integer since 2 × 10n + 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 , ar2 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 + ar2 > a ( assuming 0 < r ≤ 1)

⇒  r2 + 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 > ar2

⇒  r2 −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 S1 , and that of the terms in the even places is S2 then find the common ratio of the progression.

(iii) If a, b, c and d are in G.P. , then show that ax2 + c divides ax3 + bx2 + cx + d.

(iv) If G1, G2 are two geometric means, and A1 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

← Back Page | Next Page →

Leave a Reply