# ARITHMETIC AND GEOMETRIC PROGRESSION QUESTION AND ANSWERS

Arithmetic and Geometric Progression Question and Answers :

Here we are going to see some practice questions on arithmetic and geometric progression question and answers.

Question 1 :

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

Solution :

tn  =  arn-1

kth term :

tk  =  ark-1

(n-k)th term :

tn-k  =  arn-k-1  -----(1)

nth term :

tn  =  arn-1  -----(2)

(n+k)th term :

tn+k  =  arn+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)  ==> arn-1/arn-k-1  =  rn-1-n+k+1 =  rk

(3)/(2)  ==> arn+k-1/arn-1  =  rn+k-1-n+1 =  rk

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 a1/x = b1/y = c1/z, then prove that x, y, z are in arithmetic progression.

Solution :

a1/x = b1/y = c1/z  =  k

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

In order to prove the above relationship,

 a1/x =  ka  =  kx b1/y =  kb  =  ky c1/z =  kc  =  kz

Since a, b and c are in G.P

b/a  =  c/b

b2  =  ac

(ky)2  =   kx kz

k2y  =   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

 √ab  =  15ab  =  225b  =  225/a a+b/2  =  25a + b  =  50

a + (225/a) = 50

a2 + 225  =  50a

a2 - 50 a + 225  =  0

(a - 5)(a - 45)  =  0

a  =  5 and a  =  45

 If a  =  5b =  225/5  =  45 If a  =  45b =  225/45  =  5

Hence the required numbers are 5 and 45.

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