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Arithmetic and Progression .
by dhwani bhatt - Monday, 28 May 2012, 10:24 PM
  1) A hexagon is formed by joining the mid-points of sides of a regular hexagon. Again , the mid-points of this smaller hexagon are joined to form a new hexagon. This is done 100 times. The perimeter of original hexagon is [1/(2+root3)]. Then sum of perimeter of all inner hexagon is ?

2) 2+12+36+80+150+..........30 terms ?

3)If a, b , c are in G.P ,, then logaX , logbX , logcX are in ?

4)30th term of the series 3+5+9+15+23+.... is ?

5)Find the sum of first n terms of the series 1.2.3 + 2.3.4 + 3.4.5 +..... ?

These are my doubts. Please assist me with the aforementioned questions ?
Re: Arithmetic and Progression .
by arsh arora - Tuesday, 29 May 2012, 02:41 AM
 

hi dhwani,

1) rt 3 a/2 will be the frst term where a is side of orgnal hexagon,

and cmmn term is rt3/2 so solve now: a=1/6( 1/(2+rt3))

3) HP old cat qstn

4) 815 solve by shifting terms by one, u will get T30=3+2+4+6+8....till 29th term(56)= 3+ 29*28=815

5) take Tn=n* n+1*n+2 then use square and cubic formula, better use options for this rather then goin by this approach.

Re: Arithmetic and Progression .
by dhwani bhatt - Tuesday, 29 May 2012, 05:35 PM
  Hi Arsh ,

Can you please provide me with the detailed solution.
Re: Arithmetic and Progression .
by TG Team - Thursday, 31 May 2012, 02:36 PM
 

Hi Dhwani smile

5) See the pattern:

1×2×3 = (1/4) × [1×2×3×4 - 0×1×2×3]

2×3×4 = (1/4) × [2×3×4×5 -1×2×3×4]

:::::::::::::::::::::::::::::::::::::::::::

n(n + 1)(n + 2) = (1/4) × [n(n + 1)(n + 2)(n + 3) - (n - 1)n(n + 1)(n + 2)]

=> The required sum is = (1/4) × [n(n + 1)(n + 2)(n + 3)] = [(n² + 3n + 1)² - 1]/4.

 

3) If a, b, c are in GP i.e. b² = ac

Taking logx both sides we get,

2logxb = logxa + logxc

=> 2/logbx = 1/logax + 1/logcx.

i.e. logax, logbx and logcx are in HP.

Kamal Lohia  

 

Re: Arithmetic and Progression .
by dhwani bhatt - Saturday, 2 June 2012, 04:39 PM
  Thank you sir ,

5)In last second step you wrote (1/4)x[n(n+1)(n+2)(n+3)-(n-1) n(n+1)(n+2)].
==> This is how I proceeded by taking out n , n+1 , n+2 as common.
==>(1/4)n(n+1)(n+2)[n+3-n+1]
==>n(n+1)(n+2).
Re: Arithmetic and Progression .
by Mohit Sharma - Monday, 4 June 2012, 01:43 AM
  Hi kamal sir,
Please explain question no 2. It forms an ap but after subtracting the same series from itself for 3 times. Please explain how to solve such question.
Regards