Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Tanisha Thakur 7 years, 9 months ago
- 1 answers
Posted by Avika Sharma 7 years, 9 months ago
- 0 answers
Posted by Abir Chakroborty 7 years, 9 months ago
- 4 answers
Posted by Shakshi Sharma 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
According to question we are given that, a = 4, l = 49 and Sn = 265.we know that Sn = {tex}\frac n2{/tex}(a+l)
{tex}\therefore{/tex} 265 = {tex}\frac{n}{2}{/tex}(4 + 49) {tex}\Rightarrow{/tex} 530 = 53n {tex}\Rightarrow{/tex} n = 10
{tex}\therefore l{/tex} = a10 = a + 9d
{tex}\Rightarrow{/tex} 49 = 4 + 9d {tex}\Rightarrow{/tex} 9d = 45 {tex}\Rightarrow{/tex} d = 5
{tex}\therefore{/tex} Common difference of = 5.
Posted by Aarav Balan 7 years, 9 months ago
- 1 answers
Megha A 7 years, 9 months ago
Posted by . . 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Given: {tex}\frac { x } { a } \cos \theta + \frac { y } { b } \sin \theta = 1 \text { and } \frac { x } { a } \sin \theta - \frac { y } { b } \cos \theta = 1 , {/tex}
To prove: {tex}\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 2{/tex}
Now, {tex}{\left( {\frac{x}{a}\cos \theta + \frac{y}{b}\sin \theta } \right)^2} + {\left( {\frac{x}{a}\sin \theta - \frac{y}{b}\cos \theta } \right)^2} = {(1)^2} + {(1)^2}{/tex}
{tex} = \frac{{{x^2}}}{{{a^2}}}{\cos ^2}\theta + \frac{{{y^2}}}{{{b^2}}}{\sin ^2}\theta + 2\frac{x}{a}\cos \theta \frac{y}{b}\sin \theta + \frac{{{x^2}}}{{{a^2}}}{\sin ^2}\theta + \frac{{{y^2}}}{{{b^2}}}{\cos ^2}\theta - 2\frac{x}{a}\sin \theta \frac{y}{b}\cos \theta = 1 + 1{/tex}
{tex} = \frac{{{x^2}}}{{{a^2}}}{\cos ^2}\theta + \frac{{{x^2}}}{{{a^2}}}{\sin ^2}\theta + \frac{{{y^2}}}{{{b^2}}}{\sin ^2}\theta + \frac{{{y^2}}}{{{b^2}}}{\cos ^2}\theta = 2{/tex}
{tex} = \frac{{{x^2}}}{{{a^2}}}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) + \frac{{{y^2}}}{{{b^2}}}\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = 2{/tex}
{tex} = \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 2{/tex} {tex}\left[ {\because {{\cos }^2}\theta + {{\sin }^2}\theta = 1} \right]{/tex}
Hence proved.
Posted by Priyanka Sharma 7 years, 9 months ago
- 5 answers
Posted by Dilip Thakkar 7 years, 9 months ago
- 1 answers
Posted by Amit Singh 7 years, 9 months ago
- 4 answers
Posted by Tarun Sharma 7 years, 9 months ago
- 1 answers
Ayushi Shrivastava 7 years, 9 months ago
Posted by Avni Neil Khanna 7 years, 9 months ago
- 1 answers
Posted by Avni Neil Khanna 7 years, 9 months ago
- 0 answers
Posted by Hrithik Khoker 7 years, 9 months ago
- 1 answers
Aisha Massey 7 years, 9 months ago
Posted by Yashika Dalal 7 years, 9 months ago
- 2 answers
Posted by Komal Khipal 7 years, 9 months ago
- 0 answers
Posted by Tek Chand Sameq 7 years, 9 months ago
- 2 answers
Posted by Abhishek Rajpoot 7 years, 9 months ago
- 4 answers
Posted by Sundarresh Vellala 7 years, 9 months ago
- 0 answers
Posted by Sakshi Chauhan 7 years, 9 months ago
- 1 answers
Shivansh Srivastav 7 years, 9 months ago
Posted by Komal Khipal 7 years, 9 months ago
- 2 answers
Posted by Ravi Jonwar 7 years, 9 months ago
- 2 answers
Posted by Alka Rathi 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Let a be the first term and d be the common difference of the given AP. Then,
Sm = sum of first m terms of the given AP;
Sn = sum of first n terms of the given AP.
{tex}\frac { S _ { m } } { S _ { n } } = \frac { m ^ { 2 } } { n ^ { 2 } } \Rightarrow \frac { \frac { m } { 2 } [ 2 a + ( m - 1 ) d ] } { \frac { n } { 2 } [ 2 a + ( n - 1 ) d ] } = \frac { m ^ { 2 } } { n ^ { 2 } }{/tex}
{tex} \Rightarrow \frac { 2 a + ( m - 1 ) d } { 2 a + ( n - 1 ) d } = \frac { m } { n }{/tex}
{tex}\Rightarrow{/tex} 2an+mnd-nd= 2am+mnd-md
{tex}\Rightarrow{/tex} 2an-2am=nd-md
{tex}\Rightarrow{/tex} 2a(n - m) = d(n - m) {tex}\Rightarrow{/tex}2a=d...(i)
{tex}\therefore \quad \frac { T _ { m } } { T _ { n } } = \frac { a + ( m - 1 ) d } { a + ( n - 1 ) d } = \frac { a + ( m - 1 ) \cdot 2 a } { a + ( n - 1 ) \cdot 2 a }{/tex} [from (i)]
{tex}= \frac { a + 2 a m - 2 a } { a + 2 a n - 2 a } = \frac { 2 a m - a } { 2 a n - a } = \frac { a ( 2 m - 1 ) } { a ( 2 n - 1 ) } = \frac { 2 m - 1 } { 2 n - 1 }{/tex}.
Posted by Aasmaan Gupta 7 years, 9 months ago
- 0 answers
Posted by A B 7 years, 9 months ago
- 3 answers
Posted by Abhi Bhai 7 years, 9 months ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
