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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 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, ar, ar2, ar3, ……, a rn–1

By having a close look we can say the n-th term of GP will be given by tn = arn–1

Sum of a geometric progression

Sn = a + ar + ar2 +…+ arn–1                                  … (6)

Multiplying both sides by r, we get,

rSn = ar + ar2 +…+ arn–1 + arn                               … (7)

Subtracting (7) from (6), we have

Sn – rSn = a – arn

or, Sn = a(1–rr)/(1–r)                                           … (8)


         Sn = a(1–rn)/(1–r)


What happens when n tends to infinity.

If we take any value of |r| greater than 1 then value of rn when n ∞  will tend to infinite. Hence value of S will also tend to infinite. If we take of |r| less than 1 then rn will tend to 0 when n  ∞ .

Thus S = a/1–r for |r| < 1                                               … (9)


When any real number is added, subtracted, multiplied or divided to each term of a geometric series?

  1.     Multiplication/Division by a constant number to each term of a G.P. also results a G.P.

        Suppose a1, a2, a3, ……, an are in G.P.

        then ka1, ka2, ka3, ……, kan and

        a1/k, a2/k, ... ... ... an/k will also be in G.P.

        Where k ? R and k ≠ 0.


2.     Multiplication/Division of two G.P.’s also results a G.P.

        Suppose a1, a2, a3, ……, an

        and b1, b2, b3, ……, bn are two G.P.

        then a1b1, a2b2, a3b3, ……, an bn

        then a1/b2, a2/b2, ... ... ..., an/bn will also be in G.P.


3.     Reversing the order of a G.P.’s also results a G.P.

        Suppose a1, a2, a3, ……, an are in G.P.

        then an, an–1, an–1, ……, a3, a2, a1 will also be in G.P.


4.     Taking the inverse of a G.P. also results a G.P.

        Suppose a1, a2, a3, ……, an are in G.P.

        then 1/a1, 1/a2, 1/a ……, 1/an will also be in G.P.

 To solve problems numbers should be taken intelligently eg.


Three number in G.P.  α/ß, α, αß                       c.r = ß

Four number in G.P.  α/ß3, α ß, αß, αß3            c.r. = ß2

Five numbers in G.P.   α/ß2, α/ß, α . αß . αß2   c.r. = ß


• If each term of a G.P. is multiplied (or divided) by a fixed non-zero constant, then the resulting sequence is also a G.P. with same 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 a1, a2, a3, ……, b1, b2, b3, …… 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 rr’.

• 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 a/rk, a/rk–1, …, a, ar, …, ark 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 a/r2k–1, a/r2k–3, ... ... ... a/r, ar, ……, ar2k–1, in case we have to take 2k terms in a G.P.

• If a1, a2, ……, an are in G.P., then a1an = a2an–1 = a3an–2 = ……

• If a1, a2, a3, …… is a G.P. (each a1 > 0), then loga1, loga2, loga3 …… is an A.P. The converse is also true.

Geometric Mean between two numbers

If three numbers are in G.P. then middle one is said to be geometric mean (GM) of two others.

Let, G be the geometric mean between two number a and b

So, a G b are in G.P.

         G/a = b/G.

or,    G2 = ab

         G =√ab

Similarly we can find two geometric means between two given numbers a and b.

Let a, G1, G2, b are in G.P.

tn = a rn–1

or b/a = r3

r =(b/a)1/3

G1 = ar2 = a (b/a)1/3 = a1/3 b2/3

Geometric Mean(s)

• 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 b = √ac is the geometric mean of a and c.

• If a1, a2, ……, an are non-zero positive numbers, then their G.M.(G) is given by G = (a1a2a3, ……, an)1/n. If G1, G2, …… Gn are n geometric means between and a and b then a, G1, G2, ……, Gn b will be a G.P. Here b = arn+1.

r = n+1√b/a  G1 = an+1√b/a, G2 = a(n+1√b/a)2,…, Gn = a(n+1√b/a)n.



The 7th term of a G.P. is 8 times the 4th term. Find the G.P. when its 5th term is 48.


Given that t7 = 8t4 ⇒ ar6 = 8ar3
⇒ r3 = 8 = 23 ⇒ r = 2.
Also t5 = 48 ⇒ ar4 = 48 or 16a = 48 ⇒ a = 3.
Hence the required G.P. is 3, 6, 12, 24 ……



Does there exists a G.P. containing 27, 8 and 12 as three of its terms? If it exists, how many such progressions are possible?


Let 8 be the mth, 12 the nth and 27 be the tth terms of a G.P. whose first term is A and common ratio is R.
Then 8 = ARm–1, 12 = ARn–1, 27 = ARt–1
⇒ 8/12 = Rm–n = 2/3,          12/27 = Rn–t = (2/3)2,          8/27 = Rm–t =(2/3)3
⇒ 2m – 2n = n – t                        and                         3m – 3n = m – t
⇒ 2m + t = 3n                              and                         2m + t = 3n
⇒ 2m+t/3 = n.

​There are infinite sets of values of m, n, t which satisfy this relation. For example, take m = 1, then 2+t/3 = n = k ⇒ n = k, t = 3k – 2. By giving different values to k we get integral values of n and t. Hence there are infinite numbers of G.P.’s whose terms may be 27, 8, 12 (not consecutive).



In a four term series if first three are in G.P. and last three are in A.P. with common different 6 and last terms is equal to the first term then find all four terms in series.


This is very tricky question. If you read question carefully then it is clear that we have to start with A.P. because common difference is given.
Let the numbers be a + 6, a–6, a, a+6 now first three are in G.P. is (a–6)2 = a(a+6) or, a2 – 12a + 36 = a2 therefore numbers are 8, –4, 2, 8.

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