Discussion Forum

Let $b_{i}>1$ for i=1,2,.....101.  Suppose  $\log_{e} b_{1},\log_{e} b_{2},.......\log_{e} b_{101}$ are  in AP with the common difference  $\log_{e} 2$ . Suppose  $a_{1} ,a_{2},........a_{101}$  are in AP, such that  $a_{1} =b_{1}$ and $a_{51} =b_{51}$. If  $t= b_{1} +b_{2}+.....+b_{51}$ and  $s= a_{1} +a_{2}+.....+a_{51}$ , then

Answer:

Explanation:

If  $\log b_{1}, \log b_{2}......... \log b_{101}$ are in AP, with common difference  $\log_{e} 2$. then  $b_{1}, b_{2}......... b_{101}$ are in GP , with common ratio 2

                $b_{1}=2^{0}b_{1}$

                 $b_{2}=2^{1}b_{1}$

                 $b_{3}=2^{2}b_{1}$

                    :              :            :

                  $b_{101}=2^{100}b_{1}$             ..........(i)

  Also  $a_{1} ,a_{2},........a_{101}$  are in AP

     Given, $a_{1} =b_{1}$  and $a_{51} =b_{51}$

$\Rightarrow  a_{1}+50D=2^{50}b_{1}$

$\Rightarrow  a_{1}+50D=2^{50}a_{1}[\because a_{1}=b_{1}]$     ........(ii)

Now,   $t= b_{1} +b_{2}+.....+b_{51}$

$\Rightarrow t=b_{1}\frac{(2^{51}-1)}{2-1}$........(iii)

and $s= a_{1} +a_{2}+.....+a_{51}$ 

$=\frac{51}{2}(2a_{1}+50D)$   .....(iv)

$\therefore$     $t= a_1(2^{51}-1)[\because a_{1}=b_{1}]$

or   $t= 2^{51}a_{1}-a_{1}<2^{51}a_{1}$   ......(v)

and  $s= \frac{51}{2}[a_{1}+(a_{1}+50D]$   

                                                          [from Eq.(iii)]

   =  $\frac{51}{2}[a_{1}+2^{50}a_{1}]=\frac{51}{2}a_{1}+\frac{51}{2}2^{50}a_{1}$

$\therefore$  $s>2^{51}a_{1}$           ............(vi)

 From Eqs(v) and (vi)

  we get s>t

 Also              $a_{101}=a_{1}+100D$

and    $b_{101}=2^{100}b_{1}$

$\therefore$   $a_{101}=a_{1}+100(\frac{2^{50}a_{1}-a_{1}}{50})$

  and     $b_{101}=2^{100}a_{1}$

$\Rightarrow a_{101}=a_{1}+2^{51}a_{1}-2a_{1}=2^{51}a_{1}-a_{1}$

$\Rightarrow a_{101}<2^{51}a_{1}$

and  $b_{101}>2^{51}a_{1}\Rightarrow b_{101}>a_{101}$