Let bi > 1 for I = 1, 2, ......, 101. Suppose logeb1, logeb2, ......., logeb101 are in Arithmetic Progression (A.P.) with the common difference loge2. Suppose a1, a2, ......, a101 are in A.P. such that a1 = b1 and a51 = b51. If t = b1 + b2 + .... + b51 and s = a1 + a2 + ..... + a51, then

Sequences and Series PYQ — JEE Advanced 2016 · Shift 2 · Q33 with Solution

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Correct answer: B

Interpret the condition on $b_i$ .

Given that $\ln b_1, \ln b_2, \dots, \ln b_{101}$ are in A.P. with common difference $\ln 2$ , we get

\ln b_i = \ln b_1 + (i-1)\ln 2.

Hence,

b_i = b_1\cdot 2^{i-1}.

So $\{b_i\}$ is a geometric progression with common ratio $2$ .

In particular,

b_{51}=b_1\cdot 2^{50}, \qquad b_{101}=b_1\cdot 2^{100}.

Form the arithmetic progression $a_1,a_2,\dots,a_{101}$ .

We are given that $a_1=b_1$ and $a_{51}=b_{51}=b_1 2^{50}$ .

Let the common difference of the A.P. $\{a_i\}$ be $d$ . Then

a_{51}=a_1+50d.

So,

b_1+50d=b_1 2^{50}

which gives

50d=b_1(2^{50}-1)

d=\frac{b_1(2^{50}-1)}{50}.

Thus,

a_{101}=a_1+100d=b_1+100\cdot \frac{b_1(2^{50}-1)}{50}.

So,

a_{101}=b_1+2b_1(2^{50}-1)=b_1(2^{51}-1).

Compare with

b_{101}=b_1 2^{100}.

Since

2^{51}-1 < 2^{100},

we have

a_{101}<b_{101}.

So any option claiming $a_{101}>b_{101}$ is false.

Compute $t=b_1+b_2+\cdots+b_{51}$ .

Since $b_i=b_1 2^{i-1}$ ,

t=b_1(1+2+2^2+\cdots+2^{50}).

This is a GP sum:

t=b_1\frac{2^{51}-1}{2-1}=b_1(2^{51}-1).

Compute $s=a_1+a_2+\cdots+a_{51}$ .

Since $a_1$ to $a_{51}$ are in A.P.,

s=\frac{51}{2}(a_1+a_{51}).

Using $a_1=b_1$ and $a_{51}=b_1 2^{50}$ ,

s=\frac{51}{2}(b_1+b_1 2^{50})

s=\frac{51b_1}{2}(1+2^{50}).

Compare $s$ and $t$ .

We compare

s=\frac{51b_1}{2}(1+2^{50}), \qquad t=b_1(2^{51}-1).

Since $b_1>0$ , divide by $b_1$ :

\text{compare } \frac{51}{2}(1+2^{50}) \text{ and } 2^{51}-1.

Now,

\frac{51}{2}(1+2^{50})-(2^{51}-1) =\frac{51}{2}+\frac{51}{2}2^{50}-2\cdot 2^{50}+1.

So,

=\frac{53}{2}+\left(\frac{51}{2}-2\right)2^{50} =\frac{53}{2}+\frac{47}{2}2^{50}>0.

Hence,

s>t.

Conclusion from both comparisons

We found:

s>t \quad \text{and} \quad a_{101}<b_{101}.

Therefore the correct option is B.

Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So, the derived answer agrees with the stored answer.

More from Sequences and Series

View written solutionFree

Correct answer: B

Interpret the condition on $b_i$ .

Given that $\ln b_1, \ln b_2, \dots, \ln b_{101}$ are in A.P. with common difference $\ln 2$ , we get

\ln b_i = \ln b_1 + (i-1)\ln 2.

Hence,

b_i = b_1\cdot 2^{i-1}.

So $\{b_i\}$ is a geometric progression with common ratio $2$ .

In particular,

b_{51}=b_1\cdot 2^{50}, \qquad b_{101}=b_1\cdot 2^{100}.

Form the arithmetic progression $a_1,a_2,\dots,a_{101}$ .

We are given that $a_1=b_1$ and $a_{51}=b_{51}=b_1 2^{50}$ .

Let the common difference of the A.P. $\{a_i\}$ be $d$ . Then

a_{51}=a_1+50d.

So,

b_1+50d=b_1 2^{50}

which gives

50d=b_1(2^{50}-1)

d=\frac{b_1(2^{50}-1)}{50}.

Thus,

a_{101}=a_1+100d=b_1+100\cdot \frac{b_1(2^{50}-1)}{50}.

So,

a_{101}=b_1+2b_1(2^{50}-1)=b_1(2^{51}-1).

Compare with

b_{101}=b_1 2^{100}.

Since

2^{51}-1 < 2^{100},

we have

a_{101}<b_{101}.

So any option claiming $a_{101}>b_{101}$ is false.

Compute $t=b_1+b_2+\cdots+b_{51}$ .

Since $b_i=b_1 2^{i-1}$ ,

t=b_1(1+2+2^2+\cdots+2^{50}).

This is a GP sum:

t=b_1\frac{2^{51}-1}{2-1}=b_1(2^{51}-1).

Compute $s=a_1+a_2+\cdots+a_{51}$ .

Since $a_1$ to $a_{51}$ are in A.P.,

s=\frac{51}{2}(a_1+a_{51}).

Using $a_1=b_1$ and $a_{51}=b_1 2^{50}$ ,

s=\frac{51}{2}(b_1+b_1 2^{50})

s=\frac{51b_1}{2}(1+2^{50}).

Compare $s$ and $t$ .

We compare

s=\frac{51b_1}{2}(1+2^{50}), \qquad t=b_1(2^{51}-1).

Since $b_1>0$ , divide by $b_1$ :

\text{compare } \frac{51}{2}(1+2^{50}) \text{ and } 2^{51}-1.

Now,

\frac{51}{2}(1+2^{50})-(2^{51}-1) =\frac{51}{2}+\frac{51}{2}2^{50}-2\cdot 2^{50}+1.

So,

=\frac{53}{2}+\left(\frac{51}{2}-2\right)2^{50} =\frac{53}{2}+\frac{47}{2}2^{50}>0.

Hence,

s>t.

Conclusion from both comparisons

We found:

s>t \quad \text{and} \quad a_{101}<b_{101}.

Therefore the correct option is B.

Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So, the derived answer agrees with the stored answer.

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