Geometric Progression (G.P) | Mathematics (Maths) Class 11 - Commerce PDF Download

C. Geometric Progression (GP)

GP is a sequence of numbers whose first term is non zero & each of the succeeding terms is equal to the proceeding terms multiplied by a constant . Thus in a GP the ratio of successive terms is constant. This constant factor is called the COMMON RATIO of the series & is obtained by dividing any term by that which immediately proceeds it . Therefore a, ar, ar², ar³, ar⁴, ...... is a GP with a as the first term & r as common ratio .

(i) n^th term = a r^n-1  

(ii) Sum of the I^st n terms i.e. 
(iii) Sum of an infinite GP when |r| < 1 when n → ∞       rⁿ → 0 if |r| < 1 therefore, S_∞ =  .

(iv) If each term of a GP be multiplied or divided by the same non-zero quantity, the resulting sequence is also a GP .

(v) Any 3 consecutive terms of a GP can be taken as a/r, a, ar ; any 4 consecutive terms of a GP can be taken as a/r³, a/r, ar, ar³ & so on.

(vi) If a, b, c are in GP ⇒ b² = ac.

Ex. 9 If the first term of G.P. is 7, its n^th term is 448 and sum of first n terms is 889, then find the fifth term of G.P.

Sol. Given a = 7 the first term t_n = ar^{n - 1} = 7(r)^{n - 1} = 44 ⇒ 7rⁿ = 448 r

Also S_n =

⇒  889 =      

⇒ r = 2. Hence T₅ = ar⁴ = 7(2)⁴ = 112

Ex.10 If the third and fourth terms of an arithmetic sequence are increased by 3 and 8 respectively, then the first four terms form a geometric sequence. Find

(i) the sum of the first four terms of A.P. 

(ii) second term of the G.P.

Sol. a, (a + d), (a + 2d), (a + 3d) in A.P.

a, a + d, (a + 2d + 3), (a + 3d + 8) are in G.P.  

hence a + d = ar

also r =  =  =

 =            

⇒  d² + 6d + 9 = d² + 5d  ⇒          d = - 9

∴ =  

⇒ a² - 18a + 81 = a² - 15a  ⇒ 3a = 81 ⇒ a = 27

hence A.P. is 27, 18, 9, 0, 

G.P. is 27, 18, 12, 8

(i) sum of the first four terms of A.P. = 54

(ii) 2^nd term of G.P. = 18

Ex.11 Three positive numbers form a G.P. If the second term is increased by 8, the resulting sequence is an A.P. In turn, if we increase the last term of this A.P. by 64, we get a G.P. Find the three numbers.

Sol. Let the numbers be a, a r, a r² where r > 0

Hence    a, (a r + 8), a r²  in  A.P.   -   (1) 

Also  a, (a r + 8),  a r² + 64 in G.P.  -   (2)

⇒   (a r + 8)² = a (a r² + 64)   a = 4/4-r  - (3)

Also(1) ⇒   2 (a r + 8) = (a + a r²) ⇒ (1 - r)² = 16/a  - (4)

From  (3)  and  (4)   r = 3  or  - 5   (rejected) . 

Hence  a = 4 numbers  are :  4, 12, 36

Ex.12 The sum of the first five terms of a geometric series is 189, the sum of the first six terms is 381, and the sum of the first seven terms is 765. What is the common ratio in this series.

Sol.

S₅ = 189; S₆ = 381; S₇ = 765 ; t₆ = S₆ – S₅ = 381 – 189 = 192

t₇ = S₇ – S₆ = 765 – 381 = 384

Ex.13 Three positive distinct numbers x, y, z are three terms of geometric progression in an order and the numbers x + y, y + z, z + x are three terms of arithmetic progression in that order. Prove that

x^x . y^y = x^y . y^z . z^x.

Sol. Let x be first term of G.P. and y and z be the m^th and n^th of same G.P. respectively

⇒ t_m = y = xr^{m - 1} and t_n = z = x r^{n - 1}, where r is a common ratio of G.P. ⇒  ...(1)

Now, we have y + z = x + y + (m - 1) d and z + x = x + y + (n - 1) d,

where id ia common difference of A.P. ⇒  From (1) and (2),

⇒  ⇒ y = z^{(z - x) /(z - y)} . x^{(x - y)/(z - y)} 

⇒ y^{(z - y)} = z^{(z - x)} . x^{(x - y)} 

⇒ x^x . y^y . z^z = x^y . y^z . z^x

Ex.14 Using G.P. express  form.

Sol. Let x =  = 0.3333............. = 0.3 + 0.03 + 0.003 + 0.0003 + .............

=

= .

Let y =  = 1.233333 = 1.2 + 0.03 + 0.003 + 0.0003 +................

= 1.2 + +................ = 1.2 + .

Ex.15 The 1^st, 2^nd and 3^rd terms of an arithmetic series are a, b and a² where 'a' is negative. The 1^st, 2^nd and 3^rd terms of a geometric series are a, a² and b find the

(a) value of a and b

(b) sum of infinite geometric series if it exists. If no then find the sum to n terms of the G.P.

(c) sum of the 40 term of the arithmetic series.

Sol. a, b, a² are in A.P. (a < 0) a, a², b are in G.P.

(a) 2b = a + a² ⇒ b = (a + 1) and a⁴ = ab, a³ = b since a < 0 [ ]

a³ = (a + 1) ⇒ 2a² - a - 1 = 0 ⇒ (2a + 1)(a - 1) = 0, a = 1 (rejected as a < 0) or a = -

⇒ a² = ; 

  b = -  = -  

 common difference b - a

(b) common ratio r = ; r =  = - ; 

Since | r | < 1   S_∞ exists

∴ S_∞ = -  = -

(c) S₄₀ = 20 =

=  × 109 =  =

Ex.16 In an arithmetic progression, the third term is 15 and the eleventh term is 55. An infinite geometric progression can be formed beginning with the eighth term of this A.P. and followed by the fourth and second term. Find the sum of this geometric progression upto n terms. Also compute S_∞ if it exists.

Sol. Given a + 2d = 15 ....(1),  

+ 10d = 55 ....(2)

solving (1) and (2)        

 8d = 40  , d = 5  &   a = 5

∴ t₈ = 40 &  t₄ = 20     t₂ = 10

hence the G.P. is 40, 20, 10, ....... ∞ ( | r | < 1) hence S_∞ exists.

∴ S_∞ =  = 80   

S_n =  =

Ex.17 If the first 3 consecutive terms of a geometrical progression are the roots of the equation ,2x³ - 19x² + 57x - 54 = 0 find the sum to infinite number of terms of G.P.

Sol. let the roots be  , a, ar

∴ a= ....(1),      

a² = ....(2),        

a³== 27 ....(3) 

∴ a = 3

from (1)              

3(r² + r + 1) =            

⇒  6r² + 6r + 6 = 19r   

⇒   6r² - 13r + 6 = 0

⇒ (2r - 3)(3r - 2) = 0  

⇒  r =  or  (rejected)

 Numbers are , a, ar 

⇒ , 3, 2                   

S_∞ =  =  =

Ex.18 Let P =  then find log_0.01(P).

Sol. P = =

=  = 10² = 100  log_0.01(P) = - 1

Ex.19 A positive number is such that its fractional part, its integral part and the number itself constitute the first three terms of a geometrical progression . Show that the n^th term and the sum of the first n terms of the G.P. are,  respectively.

Sol. x  ⇒ y, m, m + y in G.P. m² = y (m + y)

⇒ y² + my - m² = 0 

If y =  put m = 1 ; y =  or

y = (not possible) 

G.P. , 1 ,  .... Now proceed

Ex.20 One of the roots of the equation 2000x⁶ + 100x⁵ + 10x³ + x - 2 = 0 is of the form , where m is non zero integer and n and r are relatively prime natural numbers. Find the value of m + n + r.

Sol. 

hence, m = – 1;  n = 161;  r = 40

m + n + r = 200

Ex.21 Suppose that a₁, a₂,........,a_n,......... is an A.P. Let S_k = a_{(k – 1) n + 1} + a_{(k – 1)n + 2} + ...... + a_kn . Prove that S₁, S₂,.... are in A.P. having common difference equal to n2 times the common difference of the A.P. a₁, a₂,.....

Sol.  We have S_k = a_{(k – 1) n + 1} + a_{(k – 1)n + 2} + .... + a_{(k – 1)n + n}

Now we have a_{(k – 1) n + 1} = a₁ + (k – 1) nd;

a_{(k – 1) n + 2} = a₁ + (k – 1) nd + d

a_{(k – 1) n + 3} = a₁ + (k – 1) nd + 2d

-----------------------

  and 

[putting k + 1 in place of k] Hence, we have S_{k + 1}–S_k = n²d 

which is a constant independent of k. This proves that the sequence (S^k) is an A.P. having common difference n²d.

Ex.22 In a G.P.  if  S₁ , S₂ & S₃  denote respectively the sums of the first n terms ,  first 2 n terms and first 3 n terms ,  then prove that 

Sol. 

T P T S₁ (S₁ - S₃)  =  S₂ (S₁ - S₂)

Ex. 23 How many increasing 3-term geometric progressions can be obtained from the sequence 1, 2, 2², 2³,....,2ⁿ ? (e.g., {2², 2⁵ ,2⁸} is a 3-term geometric progression for n ≥ 8.)

Sol. Let us start counting 3–term G.P.’s with common ratios 2, 2², 2³,...

The 3–term G.P.’s with common ratio 2 are 1, 2,2²; 2, 2², 2³;.......; 2^{n – 2}, 2^{n – 1}, 2ⁿ.

Theyy are (n – 1) in number. The 3–term GP’s with common ratio 2² are

1, 2², 2⁴; 2, 2³, 2⁵; ....... ; 2^{n – 4}, 2^{n – 2}, 2ⁿ

They are (n – 3) in number. Similarly we see that the 3–term GP’s with common ratio 2³ are (n – 5) in number and so on. Thus the number of 3–term GP’s which can be formed from the sequence 1, 2, 2², 2³,.....,2ⁿ is equal to S = (n – 1) + (n – 3) + (n – 5) +......... 

Here the last term is 2 or 1 according as n is odd or even. If n is odd, then

S = (n – 1) + (n – 3) + (n – 5) + .... + 2 

If n is even, then S = (n – 1) + (n – 3) + .... + 1 = n²/4

Hence the required number is (n² – 1)/4 or n²/4 according as n is odd or even.

Ex.24 N = 2^{n – 1} (2ⁿ – 1) and (2ⁿ – 1) is a prime number.

1 < d₁ < d₂ < ..... < d_k = N are the divisors of N. Show that 

Sol. Let 2ⁿ – 1 = q

1, d₁ d₂ .... d_k are divisors      

1, 2, 2², ...., 2^n–1, 2q, ...., 2^{n – 1} q respectively.