Question

Let $b_i>1fori=1,2,...,101.$ Suppose $lo{g}_e{b}_1.lo{g}_e{b}_2,......,lo{g}_e{b}_{101}$ are in Arithmetic Progression (A.P) with the common diffrence $lo{g}_{e}2.$ Suppose $a_1,a_2,.....,a_{101}$ are in A.P. such that $a_1=b_1$ and $a_{51}=b_{51}.$ If $t=b_1+b_2+...+b_{51}$ and $s=a_1+a_2+......+a{5}_3.$ then

• A. $s>tand{a}_{101}>b_{101}$
• B. $s>tand{a}_{101}<b_{101}$
• C. $s<tand{a}_{101}>b_{101}$
• D. $s<tand{a}_{101}<b_{101}$

Answer 1

The correct option is B $s>tand{a}_{101}<b_{101}$

$lo{g}_c,b_1,lo{g}_c{b}_2,......lo{g}_c{b}_{101}$ are in A.P.
$\Rightarrow b_1,b_2,.....b_{101}$ are in G.P.
Also $a_1,a_2,...a_{101}$ are in A.P.
where $a_1=b_1$ are $a_{51}=b_{51}.$
$\therefore b_2,b_3,...b_{50}andG{M}^{\prime }sand{a}_2,a_3,...,a_{50}$ are AM's between $b_1$ and $b_{51}.$
$\therefore GM<AM\Rightarrow b_2<a_2,b_3<a_3,...b_{50}<a_{50}\therefore b_1+b_2+...+b_{51}<a_1+a_2+...+a_{51}\Rightarrow t<sAlso{a}_1,a_{51},a_{101}areinAP{b}_1,b_{51},b_{101}areinGP\because a_1=b_{1}and{a}_{51}=b_{51}\therefore b_{101}>a_{101}$