ARITHMETIC AND GEOMETRIC PROGRESSION QUESTION AND ANSWERS

Question 1 :

If t_k is the kth term of a GP, then show that t_n−k, t_n, t_n+k also form a GP for any positive integer k.

Solution :

t_n  =  ar^n-1 

k^th term :

t_k  =  ar^k-1

(n-k)^th term :

t_n-k  =  ar^n-k-1  -----(1)

n^th term :

t_n  =  ar^n-1  -----(2)

(n+k)^th term :

t_n+k  =  ar^n+k-1  -----(3)

In order to prove the above terms are in G.P, we have to show that the common difference are same.

(2)/(1)  ==> ar^n-1/ar^n-k-1  =  r^n-1-n+k+1 =  r^k

(3)/(2)  ==> ar^n+k-1/ar^n-1  =  r^n+k-1-n+1 =  r^k

Since the common difference are same, the above terms are in G.P.

Question 2 :

If a, b, c are in geometric progression, and if a^1/x = b^1/y = c^1/z, then prove that x, y, z are in arithmetic progression.

Solution :

a^1/x = b^1/y = c^1/z  =  k

If x, y and z are in A.P, then 2y  =  x + z.

In order to prove the above relationship,

Since a, b and c are in G.P

b/a  =  c/b

b²  =  ac

 (k^y)²  =   k^x k^z

 k^2y  =   k^(x+z)

2y  =  x + z

Hence x,y and z are in A.P

Question 3 :

The AM of two numbers exceeds their GM by 10 and HM by 16. Find the numbers.

Solution :

Let the required numbers be x and y

AM  =  GM + 10   --(1)

AM  =  HM + 16   --(2)

(1)  =  (2)

GM + 10  =  HM + 16

HM  =  GM + 10 - 16

HM  =  GM - 6 --(3)

(GM)²  =  AM (HM)

By applying the values of AM and HM in terms of G.M, we get

(GM)²  =  (GM + 10) (GM - 6)

GM²  =  GM² + 4GM - 60

4GM  =  60

GM  =  15

By applying the value o GM in (3), we get HM

HM  =  15 - 6  ==>  9

AM  =  15 + 10  =  25

a + (225/a) = 50

a² + 225  =  50a

a² - 50 a + 225  =  0

(a - 5)(a - 45)  =  0

a  =  5 and a  =  45

Hence the required numbers are 5 and 45.

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