Question

Let ${{b}_{i}}>1$ for i = 1, 2,.…., 101. Suppose ${{\log }_{e}}{{b}_{1}},{{\log }_{e}}{{b}_{2}},.....,{{\log }_{e}}{{b}_{101}}$ are in Arithmetic progression (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,
(A). s > t and ${{a}_{101}}>{{b}_{101}}$.
(B). s > t and ${{a}_{101}}<{{b}_{101}}$.
(C). s < t and ${{a}_{101}}>{{b}_{101}}$.
(D). s < t and ${{a}_{101}}<{{b}_{101}}$.

Answer 1

Hint: Find the relation between ${{b}_{1}},{{b}_{2}},{{b}_{3}}$ by using the AP. Thus, ${{b}_{1}},{{b}_{2}},{{b}_{3}}...,{{b}_{101}}$ forms Geometric progression. Find the relation which connects the ${{n}^{th}}$ term of AP and GP. Then take $t={{b}_{1}}+{{b}_{2}}+....{{b}_{51}}$ which is a GP and find its sum. Take $s={{a}_{1}}+{{a}_{2}}+....{{a}_{51}}$ and find its sum. Compare them.

Complete step-by-step solution -
Given to us is an arithmetic progression, ${{\log }_{e}}{{b}_{1}},{{\log }_{e}}{{b}_{2}},{{\log }_{e}}{{b}_{3}}.....,{{\log }_{e}}{{b}_{101}}$.
They have a common difference, which can be marked as ‘d’. It is the difference between the ${{2}^{nd}}$ term and ${{1}^{st}}$term.
Arithmetic progression is the sequence of numbers such that the difference between the consecutive terms is constant.
Here the common difference of AP is given as ${{\log }_{e}}2$.
Common difference $={{2}^{nd}}$ term - ${{1}^{st}}$ term

& \Rightarrow {{\log }_{e}}{{b}_{2}}-{{\log }_{e}}{{b}_{1}}={{\log }_{e}}2 \\\ & \log \left( \dfrac{{{b}_{2}}}{{{b}_{1}}} \right)=\log 2 \\\ \end{aligned}$$ $$\because $$ We know that, $$\log A-\log B=\log \left( \dfrac{A}{B} \right)$$. Thus, we get $$\dfrac{{{b}_{2}}}{{{b}_{1}}}=2-(1)$$ Now let us find the relation between $${{3}^{rd}}$$ and $${{2}^{nd}}$$. $$\begin{aligned} & {{\log }_{e}}{{b}_{3}}-{{\log }_{e}}{{b}_{3}}={{\log }_{e}}2 \\\ & \log \left( \dfrac{{{b}_{3}}}{{{b}_{2}}} \right)=\log 2 \\\ & \Rightarrow \dfrac{{{b}_{3}}}{{{b}_{2}}}=2-(2) \\\ \end{aligned}$$ From equation (1), we got $$\dfrac{{{b}_{2}}}{{{b}_{1}}}=2$$. $$\therefore {{b}_{2}}=2{{b}_{1}}$$, apply this is in equation (2). In equation (2), $${{b}_{3}}=2{{b}_{2}}$$. $$\begin{aligned} & \therefore {{b}_{3}}=2\times {{b}_{1}} \\\ & {{b}_{3}}=4{{b}_{1}} \\\ \end{aligned}$$ Thus, $${{b}_{1}},{{b}_{2}},{{b}_{3}},....{{b}_{101}}$$ is a Geometric progression, with common ratio 2. Geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called common ratio. Thus, $${{b}_{51}}={{2}^{50}}{{b}_{1}}-(3)$$ Now, given that $${{a}_{1}},{{a}_{2}},{{a}_{3}}....$$ are in arithmetic progression and given that $${{a}_{1}}={{b}_{1}}$$ and $${{a}_{51}}={{b}_{51}}$$, $${{a}_{51}}={{b}_{51}}$$ $${{a}_{51}}={{2}^{50}}{{b}_{1}}$$[From equation (3)] The equation of AP to find $${{n}^{th}}$$ term can be given as, $${{a}_{n}}={{a}_{1}}+\left( n-a \right)d$$ So, to find the value of $${{a}_{51}}$$, $$\begin{aligned} & {{a}_{51}}={{a}_{1}}+\left( 51-1 \right)d \\\ & \Rightarrow {{a}_{51}}={{a}_{1}}+50d \\\ \end{aligned}$$ We found that $${{a}_{51}}={{2}^{50}}{{b}_{1}}$$ and $${{a}_{1}}={{b}_{1}}$$. $$\Rightarrow {{b}_{1}}+50d={{2}^{50}}{{b}_{1}}$$ $$\Rightarrow 50d={{2}^{50}}{{b}_{1}}-{{b}_{1}}$$ $$\therefore d=\dfrac{{{2}^{50}}{{b}_{1}}-{{b}_{1}}}{50}$$. Similarly, $${{a}_{101}}={{a}_{1}}+\left( 101-1 \right)d$$. $$\begin{aligned} & {{a}_{101}}={{a}_{1}}+100d \\\ & {{a}_{101}}={{b}_{1}}+100\times \dfrac{\left( {{2}^{50}}{{b}_{1}}-{{b}_{1}} \right)}{50} \\\ & {{a}_{101}}={{b}_{1}}+2\left( {{2}^{50}}{{b}_{1}}-{{b}_{1}} \right) \\\ & {{a}_{101}}={{b}_{1}}+{{2}^{1}}\times {{2}^{50}}{{b}_{1}}-2{{b}_{1}}=-{{b}_{1}}+{{2}^{51}}{{b}_{1}} \\\ & \therefore {{a}_{101}}={{2}^{51}}{{b}_{1}}-{{b}_{1}} \\\ \end{aligned}$$ We know, $${{b}_{51}}={{2}^{50}}{{b}_{1}}$$. Similarly, $${{b}_{101}}={{2}^{100}}{{b}_{1}}$$. So we can say $${{a}_{101}}$$ will be less than $${{b}_{101}}$$, $$\Rightarrow {{a}_{101}}<{{b}_{101}}-(4)$$ Now, it is given that, $$t={{b}_{1}}+{{b}_{2}}+...+{{b}_{51}}$$. We know, $${{b}_{2}}=2{{b}_{1}},{{b}_{3}}=4{{b}_{1}}$$ etc. $$\begin{aligned} & \therefore t={{b}_{1}}+2{{b}_{2}}+4{{b}_{1}}+... \\\ & t={{b}_{1}}\left[ 1+2+4+....+{{2}^{50}} \right] \\\ \end{aligned}$$ The sum of the series formed can be written as the sum of Geometric Progression. $${{S}_{n}}=\dfrac{{{a}_{1}}\left( 1-{{r}^{n}} \right)}{1-r}$$where, $${{a}_{1}}=2={{2}^{0}},r=2,n=51$$. $$\begin{aligned} & \therefore t={{b}_{1}}\left[ \dfrac{{{2}^{0}}\left( 1-{{2}^{51}} \right)}{1-2} \right]={{b}_{1}}\left[ \dfrac{{{2}^{0}}\left( 1-{{2}^{51}} \right)}{-1} \right] \\\ & t={{b}_{1}}\left[ \dfrac{{{2}^{0}}\left( {{2}^{51}}-1 \right)}{1} \right] \\\ & \therefore t={{b}_{1}}\left( {{2}^{51}}-1 \right)-(5) \\\ \end{aligned}$$ Now, it is given in question that, $$s={{a}_{1}}+{{a}_{2}}+{{a}_{3}}+....+{{a}_{51}}$$ The sum of n terms of AP can be written as, $${{S}_{n}}=\dfrac{n}{2}$$[last term + last term] Here, n = 51, $$a={{a}_{1}}$$ and n = 51. $$\therefore {{S}_{51}}=\dfrac{51}{2}\left[ {{a}_{1}}+{{a}_{51}} \right]$$ We know, $${{a}_{1}}={{b}_{1}}$$ and $${{a}_{51}}={{2}^{50}}{{b}_{1}}$$. $$\begin{aligned} & \therefore {{S}_{51}}=\dfrac{51}{2}\left[ {{b}_{1}}+{{2}^{50}}{{b}_{1}} \right] \\\ & S=\dfrac{51{{b}_{1}}}{2}\left[ {{2}^{50}}+1 \right]-(6) \\\ \end{aligned}$$ Now, let us compare equation (5) and (6) we will get that t is less than s, i.e. t < s i.e. s > t, s is greater than t, and we got that $${{a}_{101}}<{{b}_{101}}$$. $$\therefore $$We got that s > t and $${{a}_{101}}<{{b}_{101}}$$. _Hence option (b) is the correct answer._ Note: Questions like these are a bit tough to understand. You should know how AP and GP series are formed, how their common differences are taken etc. We have used the formula to find the sum of GP and AP here to solve and prove that s > t. Thus the formulae of the series should be thoroughly learned.