Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Let a1, a2 , a3 , .........a49 be in A. P such that Summation a 4k+1= 416 ( where lower limit k=0 and upper limit 12 ) and a9 + a43 =66 . if a1^2 +a2 ^2 +... +a17^2 =140m , then m is equal to

Let a1, a2 , a3 , .........a49 be in A. P such that Summation a 4k+1= 416 ( where lower limit k=0 and upper limit 12 ) and a9 + a43 =66 . if a1^2 +a2 ^2 +... +a17^2 =140m , then m is equal to

Answers (1)

best_answer

Solution:

               sum_k=0^12a_4k+1=416

  Rightarrow      frac132[2a_1+48d]=416hspace0.5cmRightarrow a_1+24d=32    .......(1)

Rightarrow       a_9+a_43=66Rightarrow 2a_1+50d=66       .......(2)

        From     (1)  and  (2)       d=1   and    a_1=8

Rightarrow       140m=sum_r=1^17a_r^2=sum_r=1^17[8+(r-1)cdot 1]^2

Rightarrow       140m=sum_r=1^17(r+7)^2=sum_r=1^24r^2-sum_r=1^7r^2

Rightarrow      140m=frac24cdot 25cdot 496-frac7cdot 8cdot 156=frac7cdot 8cdot 56[105-3]=280cdot 17

Rightarrow            m=34.

 

 

 

 

 

Posted by

Deependra Verma

View full answer

Similar Questions

Latest Question