Defence Study Notes: Geometric Progression

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 r^n–1

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

t_n = ar^n–1

Sum of a geometric progression

Now, let us try to find out the sum of the first n terms of a G.P.

S_n = a + ar + ar² +…+ ar^n–1 … (6)

Multiplying both sides by r, we get,

rS_n = ar + ar² +…+ ar^n–1 + arⁿ … (7)

Subtracting (7) from (6), we have

S_n – rS_n = a – arⁿ

or, S_n = a(1–r^r)/(1–r) … (8)

From equation (8) sum of n terms of a G.P.

S_n = a(1–r^r)/(1–r)

Basically, we have to find out the value of particular case 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 rⁿ when n → ∞ will tend to zero. (take any number greater than 1 multiply it by itself several time. Are you getting convinced that it will tend to infinity. Do the same exercise with one another number less than 1).

Thus S_- = a/1–r for |r| < 1 … (9)

Example:-1 

Solution: 

Example:-2 The 1^st term of a geometric sequence is 9 and the 6^th term is 288, then find the sum of 8^th terms of the GP.

Solution:

Enquiry: What happens when any real number can be add subtract, multiply and divides 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, ……, a_nare in G.P.
   then ka₁, ka₂, ka₃, ……, ka_n and
   a₁/k, a₂/k, ... ... ... a_n/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, ……, a_n
   and b₁, b₂, b₃, ……, b_nare two G.P.
   then a₁b₁, a₂b₂, a₃b₃, ……, a_n b_n
   then a₁/b₂, a₂/b₂, ... ... ..., a_n/b_n will also be in G.P.
3. Reversing the order of a G.P.’s also results a G.P.
   Suppose a₁, a₂, a₃, ……, a_nare in G.P.
   then a_n, a_n–1, a_n–1, ……, a₃, a₂, a₁ will also be in G.P.
4. Taking the inverse of a G.P. also results in a G.P.
   Suppose a₁, a₂, a₃, ……, a_nare in G.P.
   then 1/a₁, 1/a₂, 1/a ……, 1/a_n will also be in G.P.

Note: Students are suggested to assume the known variable related to geometric progression in following way.

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

Summary of Important Notes:

• 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 r^k.
• 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/r^k, a/r^k–1, …, a, a^r, …, 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 a/r^2k–1, a/r^2k–3, ... ... ... a/r, ar, ……, ar^2k–1, in case we have to take 2k terms in a G.P.
• If a₁, a₂, ……, a_n are in G.P., then a₁a_n = a₂a_n–1 = a₃a_n–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.

t_n = a r^n–1

or b/a = r³

r =(b/a)1/3

G₁ = ar² = a (b/a)^1/3 = a^1/3 b^2/3

Example:- 3 In a G.P. if the (x+y)^th term be ‘a’ and (x- y)^th term be ‘b’ then its x^th  term is:

Solution:

Example:-4

Solution:

Example:-5 If the ratio of AM to GM of two positive numbers a and b is 5 : 3, then a : b is equal to:

Solution: Given AM : GM =5:3

AM of a, b = (a+b)/2,

GM of a, b = (ab)^1/2

So, ((a+b)/2) : (ab)^1/2 = 5: 3

a/(ab)^1/2 : b/(ab)^1/2 = 2*5/3

(a/b)^1/2 + (b/a)^1/2 = 10/3

Let (a/b)^1/2 = x, then

x+1/x = 10/3

(x² + 1)*3 = 10*x

3x² – 10x +3 =0

3x² -9x - x+3 =0

3x(x-3) -1(x-3) = 0

3x = 1 or x=3

x=1/3 or x=3

(a/b)^1/2 = 1/3 or (a/b)^1/2 = 3

a:b = 1:9 or a:b = 9:1

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