Home
Class 12
MATHS
If n is a positive integer satisfying th...

If n is a positive integer satisfying the equation `2+(6*2^(2)-4*2)+(6*3^(2)-4*3)+"......."+(6*n^(2)-4*n)=140` then the value of n is

Text Solution

Verified by Experts

`:.2+(6*2^(2)-4*2)+(6*3^(2)-4*3)+"......."+(6*n^(2)-4*n)=140`
`2+6(2^(2)+3^(2)+"......."+n^(2))-4*(2+3+"....."+n)=140`
`implies 2+6((n(n+1)(2n-1))/(6)-1)-4((n(n+1))/(2)-1)=140`
`implies 2+n(n+1)(2n+1)-6-2n(n+1)+4=140`
`implies n(n+1)(2n+1)-2n(n+1)-140=0`
`implies 2n^(3)3n^(2)+n-2n^(2)-2n-140=0`
`implies 2n^(3)+n^(2)-n-140=0`
`implies (n-4)+(2n^(2)+9n+35)=0`
`implies n=4 " or " 2n^(2)+9n+35=0`
`implies 2n^(2)+9n+35=0`
`implies n=(-9pmsqrt(81-280))/(4)`
`:. n=(9pmsqrt(-199))/(4) " " [" complex values "]`
Only positive integer value of n is 4.
Promotional Banner

Topper's Solved these Questions

  • SEQUENCES AND SERIES

    ARIHANT MATHS|Exercise Exercise (Matching Type Questions)|3 Videos
  • SEQUENCES AND SERIES

    ARIHANT MATHS|Exercise Matching Type Questions|1 Videos
  • SEQUENCES AND SERIES

    ARIHANT MATHS|Exercise Exercise (Passage Based Questions)|24 Videos
  • PROPERTIES AND SOLUTION OF TRIANGLES

    ARIHANT MATHS|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|21 Videos
  • SETS, RELATIONS AND FUNCTIONS

    ARIHANT MATHS|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|12 Videos

Similar Questions

Explore conceptually related problems

If n is a positive integer, then (sqrt(3)+1)^(2n)-(sqrt(3)-1)^(2n) is

The number of positive integers satisfying the inequality C(n+1,n-2) - C(n+1,n-1)<=100 is

The positive integer value of n >3 satisfying the equation 1/(sin(pi/n))=1/(sin((2pi)/n))+1/(sin((3pi)/n))i s

If n is any positive integer , show that 2^(3n +3) -7n - 8 is divisible by 49 .

Solve the following linear equations : n/2-(3n)/4+(5n)/6=21

If m and n are the smallest positive integers satisfying the relation (2CiS pi/6)^m=(4CiSpi/4)^n , where i = sqrt(-1), (m+ n) equals to

Let n be a positive integer such that sin (pi/(2n))+cos(pi/(2n))=(sqrt(n))/2dot Then

If a root of the equation n^(2)sin^(2)x-2sinx-(2n+1)=0 lies in [0,pi//2] the minimum positive integer value of n is