Wednesday, September 7, 2011

Selected And compiled Problems (W.B.J.E.E.) on Geometrical Progression


     Selected And compiled Problems (W.B.J.E.E.) on Geometrical Progression – © Anuran Chowdhury , 8.9.2011                                                

Problem 1|| log10 2, log10 (2x-1) and log10 (2x+3) are in Arithmetic Progression; then what is the value of x?
(a)   log2 5 (b) log 5 2 (c) 1 (d) 0


Solution: (a) is correct
As they are in AP .. . log10 (2x-1)-log10 2 = log10 (2x+3)-log10(2x-1)
                                or; 2log10(2x-1)=log10 2+log10 (2x+3)

Therefore, (2x-1)2 = 2(2x+3)
     Or, 22x -2.2x + 1 = 2.2x +6
     Or, 22x-4.2x -5 =0
     Or, (2x-5)(2x+1)=0
     Or, 2x=5 or 2x=-1 (Not Possible)
     Or, x= log2 5


Problem 2|| Find the sum to n terms of the series 11+102+1003+10004+…Tn

Solution : 11+102+1003+10004+…Tn
               =(10+1)+(100+2)+(1000+3)+(10000+4)+….n terms
              = (10+100+1000+10000+…n terms)+(1+2+3+4+… n terms)
              = 10(10n-1)  + n(n+1)
9                              2
 Problem 3 || Find the sum of 1+4+13+40+ … n terms

Solution : Sn = 1+ 4 + 13 + 40 + ……… Tn  ……………(i)
                         Sn =       1 + 4   + 13 + ………. Tn-1 + Tn …………(ii)

(i)-(ii) => 0= 1+3+9+… n terms - Tn
                 or, Tn = 1(3n-1)
                            2

Put n=1,2,3 and then add respectively-
1(3-1)  +  1(32-1)  +  1(33-1)  +……1(3n-1)
     2              2                 2                      2

=1 {(3+32+33+34+….3n)-(1+1+1….n)}
  2
= 3(3n   -1 ) -n
         4