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Geometric Progression

A sequence is said to be in Geometric Progression, if the ratio between any two adjacent numbers in the sequence is constant (non zero). This constant is said to be the common ratio (c.r.).

        e.g. 1, 2, 4, 8 ..........……       c.r. = 2

        1, 1/2, 1/4, 1/8, …..………     c.r. = 1/2

        1/4, –1/2, 1, –2, 4, ………      c.r. = –2

In general, a geometric progression may be defined in the form

a, ar, ar2, ar3, ……, a rn–1.

By close inspection of the above series we can say the n-th term of GP will be given by

tn = arn–1

 




The value of the sum of n terms in a G.P depends on the value of r. We have different values according to whether r is equal to 1 or is not equal to 1.

Sn = a (rn – 1)/ (r-1), when r ≠1

     = na                    ,  if r =1

 

If -1 < x < 1, then lim xn = 0, as n →∞.

Hence, the sum of an infinite G.P. is

1+x+x2+…. = 1/ (1-x).

 

 

 

 


View the following video for more on geometric series

We now proceed towards some of the illustrations based on these concepts:

Illustration: An infinite G.P. has first term x and sum 5, then x belongs to what range?

Solution: we know that the sum of an infinite G.P. is

S = a/ (1-r), if |r| < 1

      = ∞,            if |r| ≥ 1

Hence, S = x/(1-r) = 5

Or 1-r = x/5

Hence, r = (5-x)/5 exists only when |r| < 1

Hence, -1 < (5-x)/5 <1

-10 < -x < 0

So this gives 0 < x < 10.

Illustration:  The third term of a geometric progression is 4. What is the product of the first five terms?

Solution: here it is given that t3 = 4.

Hence, this means ar2 =4

Now product of first five terms =

a.ar.ar2.ar3.ar4

= a5r10

= (ar2)5

= 45 

Illustration: Consider an infinite geometric series with first term ‘a’ and common ratio ‘r’. Find the values of ‘a’ and ‘r’ if its second term is ¾ and its sum is 4.

Solution: It is given in the question that the second term is ¾ and the sum is 4.

Further, the first term is ‘a’ and the common ratio is ‘r’.

Hence, we have a/ (1-r) = 4 and ar = 3/4.

This gives the value of r as 3/4a.

So, 4a2/(4a-3) = 4

This gives (a-1)(a-3) = 0

Hence, a = 1 or 3.

When a = 1 then r =3/4 and when a = 3 then r = ¼.

askIITians is an online portal which offers a platform to students where they can ask their questions on topics like sum of n terms in A.P or arithmetic progression series.

Related resources:

To read more, Buy study materials of Sequences and Series comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.

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