How many terms of the A.P …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Vivek R Nair 6 years, 6 months ago
- 1 answers
Test
Related Questions
Posted by Sahil Sahil 1 year, 4 months ago
- 2 answers
Posted by Hari Anand 6 months, 1 week ago
- 0 answers
Posted by Parinith Gowda Ms 3 months, 2 weeks ago
- 1 answers
Posted by Kanika . 1 month ago
- 1 answers
Posted by Vanshika Bhatnagar 1 year, 4 months ago
- 2 answers
Posted by Parinith Gowda Ms 3 months, 2 weeks 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
Sia ? 6 years, 6 months ago
According to the question,we have,
a=9. Therefore, common difference d =17-9=8
let the required number of terms be n.
Therefore, Sn=636
{tex}\Rightarrow{/tex}{tex}\frac{n}{2}{/tex}[2a+(n-1)d]=636
{tex}\Rightarrow{/tex}{tex}\frac{n}{2}{/tex}[2(9)+(n-1)8]=636
{tex}\Rightarrow{/tex}n[18+8n-8]=1272
{tex}\Rightarrow{/tex}8n2+10n-1272=0
{tex}\Rightarrow{/tex}4n2+5n-636=0
{tex}\Rightarrow{/tex}4n2+53n-48n-636=0
{tex}\Rightarrow{/tex}n(4n+53)-12(4n+53)=0
{tex}\Rightarrow{/tex}(4n+53)(n-12)=0
{tex}\Rightarrow{/tex}4n+53=0 or n-12=0
{tex}\Rightarrow{/tex}n={tex}\frac{{ - 53}}{4}{/tex} or n=12
Since number of terms cannot neither be negative nor fraction, n=12
hence, the required number of terms is 12.
0Thank You