Home
Class 12
MATHS
If n(1), n(2), n(3),…..,n(100) are posit...

If `n_(1)`, `n_(2)`, `n_(3)`,…..,`n_(100)` are positive real numbers such that `n_(1)+n_(2)+n_(3)+…+n_(100)=20` and `k=n_(1)(n_(2)+n_(3)+n_(4))(n_(5)+n_(6)+…+n_(9))(n_(10)+….+n_(16))…(…+n_(100))`, then `k` belongs to

A

A. `(o,100]`

B

B. `(o,128]`

C

C. `(o,144]`

D

D. `(o,1024]`

Text Solution

Verified by Experts

The correct Answer is:
D

`(d)` Using `A.M ge G.M.`
`([n_(1)+(n_(2)+n_(3)+n_(4))+(n_(5)+n_(6)+...+n_(9))+...+(...+n_(100))])/(10) ge 10sqrt(k)(AM ge GM)`
`implies 2 ge 10 sqrt(k)`
`implies 0 lt k le 1024`
Promotional Banner

Topper's Solved these Questions

  • INEQUALITIES INVOLVING MEANS

    CENGAGE PUBLICATION|Exercise Comprehension|2 Videos
  • INEQUALITIES INVOLVING MEANS

    CENGAGE PUBLICATION|Exercise Illustration|29 Videos
  • INEQUALITIES AND MODULUS

    CENGAGE PUBLICATION|Exercise Single correct Answer|21 Videos
  • INTEGRALS

    CENGAGE PUBLICATION|Exercise All Questions|762 Videos

Similar Questions

Explore conceptually related problems

Let n=3^(100) , then for n

Let n_(1)ltn_(2)ltn_(3)ltn_(4)ltn_(5) be positive integers such that n_(1)+n_(2)+n_(3)+n_(4)+n_(5)=20 . Then the number of such distinct arrangements (n_(1),n_(2),n_(3),n_(4),n_(5)) is

If n is a positive integer, show that, (n+1)^(2) + (n+2)^(2) + …+ 4n^(2) = (n)/(6)(2n+1)(7n+1)

If .^(n)C_(3)=k.n(n-1)(n-2) then k=

Let n_1ltn_2ltn_3ltn_4ltn_5 be positive integers such that n_1+n_2+n_3+n_4+n_5="20. then the number of distinct arrangements n_1, n_2, n_3, n_4, n_5 is

Prove that (.^(n)C_(1))/(2) + (.^(n)C_(3))/(4) + (.^(n)C_(5))/(6) + "…." = (2^(n) - 1)/(n+1) .

Prove that .^(n)C_(0) + (.^(n)C_(1))/(2) + (.^(n)C_(2))/(3) + "……" +(. ^(n)C_(n))/(n+1) = (2^(n+1)-1)/(n+1) .

Evaluate .^(n)C_(0).^(n)C_(2)+.^(n)C_(1).^(n)C_(3)+.^(n)C_(2).^(n)C_(4)+"...."+.^(n)C_(n-2).^(n)C_(n) .

For the Paschen series the value of n_(1) and n_(2) in the expression Delta E= Rhc ((1)/(n_(1)^(2))-(1)/(n_(2)^(2))) are

Prove that .^(n)C_(0) +5 xx .^(n)C_(1) + 9 xx .^(n)C_(2) + "…." + (4n+1) xx .^(n)C_(n) = (2n+1) 2^(n) .