Question

Let $ {I_n} = \int_0^{\dfrac{\pi }{4}} {{{\tan }^n}xdx,} $ then $ \dfrac{1}{{{I_2} + {I_4}}},\dfrac{1}{{{I_3} + {I_5}}},\dfrac{1}{{{I_4} + {I_6}}},\dfrac{1}{{{I_5} + {I_7}}}, $ form an:
A) AP with common difference \[2\]
B) GP with common ratio \[2\]
C) AP with common difference \[1\]
D) HP

Answer 1

Hint: To solve this question, we will integrate the expression given to us, i.e., $ {I_n} = \int_0^{\dfrac{\pi }{4}} {{{\tan }^n}xdx,} $ then after solving it, we will get an equation in the form, $ \dfrac{1}{{{I_n} + {I_{n - 2}}}} = (n - 1). $ Now, on substituting the values of n = 4,5,6,7…in this equation, we will check whether we will get AP, GP or HP.

Complete step-by-step answer:
n a sequence of numbers when the difference between any two consecutive terms is same, then the sequence is an arithmetic progression, while if the sequence of numbers have the ratio of any two consecutive terms is same, then the sequence is a geometric progression, if in the sequence of numbers, the reciprocal of the terms are in AP, then it is a harmonic progression.
We have been given that, $ {I_n} = \int_0^{\dfrac{\pi }{4}} {{{\tan }^n}xdx,} $ we need to check the expression $ \dfrac{1}{{{I_2} + {I_4}}},\dfrac{1}{{{I_3} + {I_5}}},\dfrac{1}{{{I_4} + {I_6}}},\dfrac{1}{{{I_5} + {I_7}}}, $ will form an AP, GP or HP.
 $ {I_n} = \int_0^{\dfrac{\pi }{4}} {{{\tan }^n}xdx} $
 $ {I_n} = \int_0^{\dfrac{\pi }{4}} {{{\tan }^{n - 2}}x.{{\tan }^2}xdx} $
 $ {I_n} = \int_0^{\dfrac{\pi }{4}} {{{\tan }^{n - 2}}x({{\sec }^2}x - 1)dx} $ $ [\because {\tan ^2}x = {\sec ^2}x - 1] $
 $ {I_n} = \int_0^{\dfrac{\pi }{4}} {({{\tan }^{n - 2}}x.{{\sec }^2}x)dx - \int_0^{\dfrac{\pi }{4}} {{{\tan }^{n - 2}}xdx} } $
 $ {I_n} = \int_0^{\dfrac{\pi }{4}} {({{\tan }^{n - 2}}x.{{\sec }^2}x)dx - } {I_{n - 2}} $ $ [\because {I_n} = \int_0^{\dfrac{\pi }{4}} {{{\tan }^n}xdx} ,\therefore {I_{n - 2}} = \int_0^{\dfrac{\pi }{4}} {{{\tan }^{n - 2}}xdx} ] $
 $ {I_n} + {I_{n - 2}} = \int_0^{\dfrac{\pi }{4}} {({{\tan }^{n - 2}}x.{{\sec }^2}x)dx} $
 $ {I_n} + {I_{n - 2}} = (\dfrac{{{t^{n - 1}}}}{{n - 1}})_0^ 1 $
 $ {I_n} + {I_{n - 2}} = (\dfrac{1}{{n - 1}}) $
 $ \Rightarrow \dfrac{1}{{{I_n} + {I_{n - 2}}}} = (n - 1) $
Now, let us check this expression, $ \dfrac{1}{{{I_2} + {I_4}}},\dfrac{1}{{{I_3} + {I_5}}},\dfrac{1}{{{I_4} + {I_6}}},\dfrac{1}{{{I_5} + {I_7}}}. $
On putting n=4 in $ \dfrac{1}{{{I_n} + {I_{n - 2}}}} = (n - 1), $ we get
\[\dfrac{1}{{{I_4} + {I_{4 - 2}}}} = \dfrac{1}{{{I_2} + {I_4}}} = (4 - 1) = 3\]
On putting n=5 in $ \dfrac{1}{{{I_n} + {I_{n - 2}}}} = (n - 1), $ we get
\[\dfrac{1}{{{I_5} + {I_{5 - 2}}}} = \dfrac{1}{{{I_3} + {I_5}}} = (5 - 1) = 4\]
On putting n=6 in $ \dfrac{1}{{{I_n} + {I_{n - 2}}}} = (n - 1), $ we get
\[\dfrac{1}{{{I_6} + {I_{6 - 4}}}} = \dfrac{1}{{{I_4} + {I_6}}} = (6 - 1) = 5\]
On putting n=7 in $ \dfrac{1}{{{I_n} + {I_{n - 2}}}} = (n - 1), $ we get
\[\dfrac{1}{{{I_7} + {I_{7 - 2}}}} = \dfrac{1}{{{I_5} + {I_7}}} = (7 - 1) = 6\]
So, we get 3,4,5,6, we can see the common difference between two consecutive terms is 1. Hence, 3,4,5,6 are in AP, with a common difference 1.
Thus, option (C) AP with common difference \[1\].
So, the correct answer is “Option C”.

Note: Students should note that in these types of questions, always calculate the expression carefully, because if you make the mistake in evaluating that, you will end up getting your answer wrong. So, here, getting this $ \dfrac{1}{{{I_n} + {I_{n - 2}}}} = (n - 1), $ term correct was necessary.