The 6th and 17th terms of …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Anubhab Majumder 6 years, 9 months ago
- 1 answers
Related Questions
Posted by M.Yashas Aradhya 2 days ago
- 0 answers
Posted by Uday Gaikwad 17 hours ago
- 0 answers
Posted by Sarthak Kedare 1 day, 14 hours ago
- 0 answers
Posted by Sara Ali 1 day, 14 hours ago
- 1 answers
Posted by M.Yashas Aradhya 2 days ago
- 0 answers
Posted by Aprajita Singh Aprajita Singh 12 hours ago
- 1 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
Rasmi Rv 6 years, 9 months ago
We have nth term of an A.P is {tex}{ T }_{ n }=a+\left( n-1 \right) d{/tex}
Now {tex}{ T }_{ 6 }=19\Rightarrow a+5d=19.............(i)\\ { T }_{ 17 }=41\Rightarrow a+16d=41..........(ii){/tex}
Subtracting equation(i) from(ii), we get
{tex}11d=22\Rightarrow d=\frac { 22 }{ 11 } =2{/tex}
Now from equation (i) we get
{tex}a+10=19\Rightarrow a=9{/tex}
Hence we get {tex}{ T }_{ 40 }=a+39d=9+39\left( 2 \right) =9+78=87{/tex}
So the 40th term is 87.
0Thank You