Let n = 1! + 4! + 7! + . . . + 400!. Then ten's digit of n is fr

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321.

Let n = 1! + 4! + 7! + . . . + 400!. Then ten's digit of n is

  • 1

  • 6

  • 2

  • 7


B.

6

Given, n = 1! + 4! + 7! + ... + 400!

1! = 1, 4! = 24, 7! = 5040, 10! = 3628800

and further the terms has last two digits = 00

So, n = 1! + 4! + 7! + ... + 400! = . . . 65

Here ten digit is 6.

Hence, 6 is the answer


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322.

Let a = 10nn! for n = 1, 2, 3 . . . then the greatest  value of n for which a is the greatest is

  • 11

  • 20

  • 10

  • 8


323.

If (1 + 2x + 3x2)10 = a0 + a1x + a2x2 + . . . + a20x20, then a2a1 = ?

  • 10.5

  • 21

  • 10

  • 5.5


324.

The condition that the x3 - bx2 + cx - d = 0 are in progression is

  • c3 = b3d

  • c2 = b2d

  • c = bd3

  • c = bd2


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325.

n = 1 2n2n + 1! = ?

  • 1e

  • e2

  • e

  • 2e


326.

If 12 × 4 + 14 × 6 + 16 × 8 + . . . n terms = knn × 1,then k =?

  • 14

  • 12

  • 1

  • 18


327.

k = 1r = 0k13kCrk = ?

  • 13

  • 23

  • 1

  • 2


328.

1xx + 1x + 2 . . . x + n = A0x + A1x + 1 + Anx + n, 0  i  r  Ar = ?

  •  - 1rr!n - r!

  • - 1rr!n - r!

  •  1r!n - r!

  •  r!n - r!


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329.

1 + 13 . 22 + 15 . 24 + 17 . 26 + ... =?

  • loge2

  • loge3

  • loge4

  • loge5


330.

Given that, 2 + 2 + c  0 and that the system of equations

   + bx + ay + bz = 0;    + cx +by +cz = 0; + by +  +cz = 0has a non-trival solution, then a, b and c lie in

  • Arithmetic Progression

  • Geometric Progression

     

  • Harmonic Progression

  • Arithmetico- geometric Progression


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