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 (2^x-1) and log₁₀ (2^x+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 (2^x-1)-log₁₀ 2 = log₁₀ (2^x+3)-log₁₀(2^x-1)

                                or; 2log₁₀(2^x-1)=log₁₀ 2+log₁₀ (2^x+3)

Therefore, (2^x-1)2 = 2(2^x+3)

     Or, 2^2x -2.2^x + 1 = 2.2^x +6

     Or, 2^2x-4.2^x -5 =0

     Or, (2^x-5)(2^x+1)=0

     Or, 2^x=5 or 2^x=-1 (Not Possible)

     Or, x= log₂ 5

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

Solution : 11+102+1003+10004+…T_n

               =(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 : S_n = 1+ 4 + 13 + 40 + ……… T_{n  ……………(i)}

                         S_n =       1 + 4   + 13 + ………. T_{n-1 + Tn …………(ii)}

(i)-(ii) => 0= 1+3+9+… n terms - T_n

                 _or, T_n = 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          2