Let an-0-4tannx dx- Then a2-a4- a3-a5- a4-a6 are in

The correct option is D HP #160;a_ n =∫ _ 0 ^ #10;π #160; 4 #160;tan ^ n xdx #160; ...(1) #10; #160;⇒ a _ n 2 #10;=∫ _ 0 ^π #160; 4 ...

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Byju's Answer

Standard XII

Mathematics

Functions

Let an=∫0π4...

Question

Let $a_{n} = \int_{0}^{\frac{π}{4}} {tan}^{n} x d x .$ Then $a_{2} + a_{4}, a_{3} + a_{5}, a_{4} + a_{6}$ are in

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Solution

The correct option is D HP
$a_{n} = \int_{0}^{\frac{π}{4}} {tan}^{n} x d x$ ...(1)
$\Rightarrow a_{n - 2} = \int_{0}^{\frac{π}{4}} {tan}^{n - 2} x d x$ ...(2)
Adding (1) and (2), we get
$a_{n} + a_{n - 2} = \int_{0}^{\frac{π}{4}} {tan}^{n} x d x + \int_{0}^{\frac{π}{4}} {tan}^{n - 2} x d x$
$= \int_{0}^{\frac{π}{4}} {tan}^{n - 2} x d x \times ({sec}^{2} x - 1) d x + \int_{0}^{\frac{π}{2}} {tan}^{n - 2} x d x = \int_{0}^{\frac{π}{4}} {tan}^{n} x {sec}^{2} x d x$
$\Rightarrow a_{n} + a_{n - 2} = \frac{1}{n - 2}$
Therefore $a_{2} + a_{4} = \frac{1}{2}, a_{3} + a_{5} = \frac{1}{3}, a_{4} + a_{6} = \frac{1}{4}$ are in H.P

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Similar questions

Q. Let

a_{n} = \int_{0}^{π / 4} t a n^{n} x d x

. Then

a_{2} + a_{4}, a_{3} + a_{5}, a_{4} + a_{6}

are in

Q. Suppose

a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}

are integers such that

\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}

where

0 \leq a < j

for

j = 2, 4, 5, 6, 7.

The sum

a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}

Q. Suppose

a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}

are intagers such that

\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4!} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}

where

0 \leq a_{j} < j

for

j = 2, 3, 4, 5, 6, 7

. The sum

a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}

is-

Suppose $a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$ are integers such that $\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4!} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}$ where 0 $\leq$ a< j for j= 2,4,5,6,7.

The sum $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$ is

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Let an=∫π40tannxdx. Then a2+a4,a3+a5,a4+a6 are in

The correct option is D HPan=∫π40tannxdx ...(1) ⇒an−2=∫π40tann−2xdx ...(2) Adding (1) and (2), we get an+an−2=∫π40tannxdx+∫π40tann−2xdx =∫π40tann−2xdx×(sec2x−1)dx+∫π20tann−2xdx=∫π40tannxsec2xdx ⇒an+an−2=1n−2 Therefore a2+a4=12,a3+a5=13,a4+a6=14 are in H.P

Let $a_{n} = \int_{0}^{\frac{π}{4}} {tan}^{n} x d x .$ Then $a_{2} + a_{4}, a_{3} + a_{5}, a_{4} + a_{6}$ are in