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 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,
a^{1/x} = k a = k^{x} |
b^{1/y} = k b = k^{y} |
c^{1/z} = k c = k^{z} |
Since a, b and c are in G.P
b/a = c/b
b^{2} = 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)
GM^{2} = GM^{2} + 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 = 15 ab = 225 b = 225/a |
a+b/2 = 25 a + b = 50 |
a + (225/a) = 50
a^{2} + 225 = 50a
a^{2} - 50 a + 225 = 0
(a - 5)(a - 45) = 0
a = 5 and a = 45
If a = 5 b = 225/5 = 45 |
If a = 45 b = 225/45 = 5 |
Hence the required numbers are 5 and 45.
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