Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Suryansh Singh 5 years, 5 months ago
- 2 answers
Posted by Kaushik Herle 5 years, 5 months ago
- 1 answers
Posted by Kaushik Herle 5 years, 5 months ago
- 1 answers
Posted by Jasleen Kaur 5 years, 5 months ago
- 1 answers
Yogita Ingle 5 years, 5 months ago
Let the number of terms required to make the sum of 636 be n and common difference be d.
Given Arithmetic Progression : 9 , 17 , 25 ....
First term = a = 9
Second term = a + d = 17
Common difference = d = a + d - a = 17 - 9 = 8
From the indentities of arithmetic progressions, we know : -
{tex}S_{n}=\dfrac{n}{2}\{2a+(n-1)d\}, where S_{n} {/tex}is the sum of first term to nth term of the AP, a is the first term, d is the common difference and n is the number of terms of AP.
In the given Question, sum of APs is 636.
Therefore,
⇒ {tex}636 = \dfrac{n}{2} \{2(9) + (n - 1)8 \} \\ \\ \\ = > 1272 = n(18 + 8n - 8) \\ \\ = > 1272 = 10n + 8n {}^{2} \\ \\ = > 636 = 5n + 4n {}^{2}{/tex}
⇒ 4n² + 5n - 636 = 0
⇒ 4n² + ( 53 - 48 )n - 636 = 0
⇒ 4n² + 53n - 48n - 636 = 0
⇒ 4n² - 48n + 53n - 636 = 0
⇒ 4n( n - 12 ) + 53( n - 12 ) = 0
⇒ ( n - 12 )( 4n + 53 ) = 0
By Zero Product Rule,
⇒ n - 12 = 0
⇒ n = 12
Hence,
Number of terms of the AP [ 9 , 17 , 25 ] which are required to make the sum of 636 is 12.
Posted by Sarala Gandhi 5 years, 5 months ago
- 0 answers
Posted by Suraj Mishra 5 years, 5 months ago
- 0 answers
Posted by Meenu Singh 5 years, 5 months ago
- 0 answers
Posted by Vedant K 5 years, 5 months ago
- 1 answers
Posted by Het Patel 5 years, 5 months ago
- 0 answers
Posted by Roushan Kumar 5 years, 5 months ago
- 0 answers
Posted by Arshjot Singh 5 years, 5 months ago
- 1 answers
Posted by Jassi Panjeta21 5 years, 5 months ago
- 5 answers
Posted by Belamkar Prakashrao 5 years, 5 months ago
- 0 answers
Posted by Krati Aggarwal 5 years, 5 months ago
- 1 answers
Posted by Joel Joji 5 years, 5 months ago
- 4 answers
Posted by Vimlesh Devi 5 years, 5 months ago
- 4 answers
Navoday Educator 5 years, 5 months ago
Angel Niki??? 5 years, 5 months ago
Kirti Goswami 5 years, 5 months ago
Vaishnavi Belapurkar 5 years, 5 months ago
Posted by Manjeet Singh 5 years, 5 months ago
- 0 answers
Posted by Bhola Nath 5 years, 5 months ago
- 1 answers
Posted by Pragyat Nikunj 5 years, 5 months ago
- 0 answers
Posted by Dharmesh Dubey 5 years, 5 months ago
- 0 answers
Posted by Ravi Prakash Chaudhary 5 years, 5 months ago
- 1 answers
Posted by Pranjal Lobo 5 years, 5 months ago
- 1 answers
Yogita Ingle 5 years, 5 months ago
P(x) = {tex}X^2 +5x+6+k=0{/tex}_(1)
Putting the value 1 in (1)equation.
P(1) = 1^2 +5*1+6+k=0
= 1+5+6+k =0
=6+6+k=0
12+k=0
K= - 12
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
Suryansh Singh 5 years, 5 months ago
0Thank You