Question No. 40

Let $b_i>1$ for $i=1,2,\dots ,101$. Suppose ${\log }_e{b}_1,{\log }_e{b}_2,\dots ,{\log }_e{b}_{101}$ are in Arithmetic Progression (A.P.) with the common difference ${\log }_{e}2$. Suppose $a_1,a_2,\dots ,a_{101}$ are in A.P. such that $a_1=b_1$ and $a_{51}=b_{51}$. If $t=b_1+b_2+\dots +b_{51}$ and $s=a_1+a_2+\dots +a_{51}$, 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 $a_{101}<b_{101}$

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Solution:

Step 1: Analyze the Sequence $b_i$

Given: ${\log }_e{b}_1,{\log }_e{b}_2,\dots ,{\log }_e{b}_{101}$ are in A.P. with common difference $d={\log }_{e}2$.
So,

$${\log }_e{b}_k={\log }_e{b}_1+(k-1){\log }_{e}2={\log }_e(b_1\cdot 2^{k-1})$$

Thus,

$$b_k=b_1\cdot 2^{k-1}$$

This is a geometric progression (G.P.) with first term $b_1$ and common ratio 2.

Step 2: Analyze the Sequence $a_i$

Given: $a_1,a_2,\dots ,a_{101}$ are in A.P. with $a_1=b_1$ and $a_{51}=b_{51}$.
Let the common difference of the A.P. be $\delta$. Then,

$$a_k=a_1+(k-1)\delta$$

In particular,

$$a_{51}=a_1+50\delta =b_{51}$$

But from the G.P.,

$$b_{51}=b_1\cdot 2^{50}$$

So,

$$a_1+50\delta =a_1\cdot 2^{50}\Rightarrow 50\delta =a_1(2^{50}-1)$$$$\delta =\frac{a_1(2^{50}-1)}{50}$$

Step 3: Compare $s$ and $t$

We have:

• $t={\sum }_{k=1}^{51}{b}_k={\sum }_{k=1}^{51}{a}_1\cdot 2^{k-1}=a_1(2^{51}-1)$

• $s={\sum }_{k=1}^{51}{a}_k={\sum }_{k=1}^{51}[a_1+(k-1)\delta ]=51a_1+\delta {\sum }_{k=1}^{51}(k-1)=51a_1+\delta \cdot \frac{50\cdot 51}{2}=51a_1+1275\delta$

Substitute $\delta$:

$$s=51a_1+1275\cdot \frac{a_1(2^{50}-1)}{50}=51a_1+25.5a_1(2^{50}-1)$$$$=a_1(51+25.5\cdot 2^{50}-25.5)=a_1(25.5+25.5\cdot 2^{50})=25.5a_1(1+2^{50})$$

Now, compare $s$ and $t$:

$$t=a_1(2^{51}-1)=a_1(2\cdot 2^{50}-1)$$$$s=25.5a_1(1+2^{50})=\frac{51}{2}{a}_1(1+2^{50})$$

Since $a_1>1$, we can ignore it for inequality.
Note: $2^{51}-1$ vs $\frac{51}{2}(1+2^{50})$
Multiply both by 2:

• $2t=2a_1(2^{51}-1)=2a_1(2^{51}-1)$

• $2s=51a_1(1+2^{50})$

Now, $2^{51}=2\cdot 2^{50}$, so:

$$2t=2a_1(2\cdot 2^{50}-1)=4a_1\cdot 2^{50}-2a_1$$$$2s=51a_1+51a_1\cdot 2^{50}$$

Clearly, for large $2^{50}$, the term $51a_1\cdot 2^{50}$ in $2s$ is greater than $4a_1\cdot 2^{50}$ in $2t$, since $51>4$. Thus, $s>t$.

Step 4: Compare $a_{101}$ and $b_{101}$

• $a_{101}=a_1+100\delta =a_1+100\cdot \frac{a_1(2^{50}-1)}{50}=a_1+2a_1(2^{50}-1)=a_1(1+2^{51}-2)=a_1(2^{51}-1)$

• $b_{101}=b_1\cdot 2^{100}=a_1\cdot 2^{100}$

Clearly, $2^{100}>2^{51}-1$, so $b_{101}>a_{101}$.

Step 5: Conclusion

• $s>t$

• $a_{101}<b_{101}$

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Final Answer:

$$\boxed{{\displaystyle B}}$$