Get Answers to all your Questions

header-bg qa

Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of theseries
1^{2}+2\cdot 2^{2}+3^{2}+2\cdot 4^{2}+5^{2}+2\cdot 6^{2}+.....
If B−2A=100λ, then λ is equal to :

  • Option 1)

    496

  • Option 2)

    232

  • Option 3)

    248

  • Option 4)

    464

 

Answers (2)

B-2A= \sum_{r=1}^{40} T_{r}-2 \sum_{r=1}^{20}T_{r}

= \sum_{r=21}^{40} T_{r}- \sum_{r=1}^{20}T_{r}

\left \{ 21^{2}+22^{2} +23^{2}+....40^{2}\right \}- \left \{ 1^{2} +2^{2}+3^{2+.....}\right \}

\left ( 21^{2} -1^{2}\right )+2(22^{2}-2^{2}).........

20\left ( 22+24+26+28+..... \right )

20\left ( 22+24..... 60\right )+20(24+28+.....60)= 248

 

Sum of n terms of an AP -

S_{n}= \frac{n}{2}\left [ 2a +\left ( n-1 \right )d\right ]

 

and

Sum of n terms of an AP

 

S_{n}= \frac{n}{2}\left [ a+l\right ]

- wherein

a\rightarrow first term

d\rightarrow common difference

n\rightarrow number of terms

 

 


Option 1)

496

Option 2)

232

Option 3)

248

Option 4)

464

Posted by

Vakul

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE