If {x_1}, {x_2}, {x_3}, dots, {x_n} are in A. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that ${x_1}, {x_2}, \dots, {x_n}$ are in $A.P.$ with common difference $\alpha$,we have ${x_{k+1}} - {x_k} = \alpha$ for all $k = 1, 2, \dot

Vedclass · Accepted Answer

$\frac{\sin (n-1)\alpha}{\cos {x_1} \cos {x_n}}$

Vedclass · Answer

(A) Given that ${x_1}, {x_2}, \dots, {x_n}$ are in $A.P.$ with common difference $\alpha$,we have ${x_{k+1}} - {x_k} = \alpha$ for all $k = 1, 2, \dots, n-1$.We can write the expression as:$S = \sum_{k=1}^{n-1} \sin \alpha \sec {x_k} \sec {x_{k+1}}$Since $\sin \alpha = \sin ({x_{k+1}} - {x_k})$,we have:$S = \sum_{k=1}^{n-1} \frac{\sin ({x_{k+1}} - {x_k})}{\cos {x_k} \cos {x_{k+1}}}$Using the formula $	an A - 	an B = \frac{\sin (A-B)}{\cos A \cos B}$,we get:$S = \sum_{k=1}^{n-1} (	an {x_{k+1}} - 	an {x_k})$This is a telescoping sum:$S = (	an {x_2} - 	an {x_1}) + (	an {x_3} - 	an {x_2}) + \dots + (	an {x_n} - 	an {x_{n-1}})$$S = 	an {x_n} - 	an {x_1} = \frac{\sin ({x_n} - {x_1})}{\cos {x_n} \cos {x_1}}$Since ${x_n} = {x_1} + (n-1)\alpha$,we have ${x_n} - {x_1} = (n-1)\alpha$.Thus,$S = \frac{\sin (n-1)\alpha}{\cos {x_1} \cos {x_n}}$.

If ${x_1}, {x_2}, {x_3}, \dots, {x_n}$ are in $A.P.$ whose common difference is $\alpha$,then the value of $\sin \alpha (\sec {x_1} \sec {x_2} + \sec {x_2} \sec {x_3} + \dots + \sec {x_{n-1}} \sec {x_n}) = $

Similar Questions

The value of $\sum\limits_{n = 1}^\infty {\left( {{{\tan }^{ - 1}}\left( {\frac{n}{{n + 2}}} \right) - {{\tan }^{ - 1}}\left( {\frac{{n - 1}}{{n + 1}}} \right)} \right)} $ is equal to-

Statement-$1$: The sum of the series $1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \dots + (361 + 380 + 400)$ is $8000$.
Statement-$2$: For any natural number $n$,$\sum_{k=1}^n (k^3 - (k-1)^3) = n^3$.

If ${a_1}, {a_2}, \dots, {a_{n+1}}$ are in $A.P.$,then $\frac{1}{{{a_1}{a_2}}} + \frac{1}{{{a_2}{a_3}}} + \dots + \frac{1}{{{a_n}{a_{n+1}}}}$ is

If ${t_n} = \frac{1}{4}(n + 2)(n + 3)$ for $n = 1, 2, 3, \dots$,then $\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + \dots + \frac{1}{t_{2003}} = $

The value of $\mathop {\lim }\limits_{n \to \infty } \left( \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2n - 1)(2n + 1)} \right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module