JEE Main 2022 (29 July Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, a_{3}, \dots$ be an AP. If $\sum_{r = 1}^{\infty} \frac{a _{r}}{2 ^{r}} = 4$ , then $4 a_{2}$ is equal to ________.

Enter Numerical Value:

Progressive Hints

Hint 1

Look closely at the structure of the summation. You have an arithmetic progression in the numerator and a geometric progression in the denominator. This is not a standard series; it is an Arithmetico-Geometric Progression (AGP). Your first task is to visualize the terms and identify this hybrid nature.

Hint 2

When dealing with an infinite AGP, the most powerful tool in your arsenal is the 'Shift-and-Subtract' method. You need to multiply the entire sum by the common ratio of the geometric part—in this case,

1/2

—and then align the terms to simplify the expression.

Hint 3

Start by writing out the sum

S = \frac{a _{1}}{2} + \frac{a _{1} + d}{4} + \frac{a _{1} + 2 d}{8} + \dots

. Now, write

\frac{S}{2} = \frac{a _{1}}{4} + \frac{a _{1} + d}{8} + \dots

directly below it. Subtracting these two equations will cause the arithmetic part to collapse into a simple infinite geometric series.

Visualized Solution

Define the AP Terms

Let the AP be $a_{1}, a_{2}, a_{3}, \dots$
The $r^{t h}$ term is $a_{r} = a_{1} + (r - 1) d$
Terms are: $a_{1}, a_{1} + d, a_{1} + 2 d, \dots$

Expand the Given Series

Given sum: $S = \sum_{r = 1}^{\infty} \frac{a _{r}}{2 ^{r}} = 4$
Expand the summation: $S = \frac{a _{1}}{2} + \frac{a _{2}}{2 ^{2}} + \frac{a _{3}}{2 ^{3}} + \dots$

Identify the AGP Structure

Substitute AP terms into the series:
$S = \frac{a _{1}}{2} + \frac{a _{1} + d}{4} + \frac{a _{1} + 2 d}{8} + \dots$
This is an Arithmetico-Geometric Progression (AGP).

The Shift-and-Subtract Method

Standard AGP method: Multiply $S$ by the common ratio of the GP part.
Here, the common ratio is $r = \frac{1}{2}$ .
Multiply both sides by $\frac{1}{2}$ :
$\frac{S}{2} = \frac{a _{1}}{4} + \frac{a _{1} + d}{8} + \frac{a _{1} + 2 d}{16} + \dots$

Subtracting the Series

Align terms with the same denominators and subtract:
$S = \frac{a _{1}}{2} + \frac{a _{1} + d}{4} + \frac{a _{1} + 2 d}{8} + \dots$
$\frac{S}{2} = 0 + \frac{a _{1}}{4} + \frac{a _{1} + d}{8} + \dots$
$S - \frac{S}{2} = \frac{a _{1}}{2} + (\frac{a _{1} + d - a _{1}}{4}) + (\frac{a _{1} + 2 d - ( a _{1} + d )}{8}) + \dots$

Analyzing the Resulting Series

Result of subtraction:
$\frac{S}{2} = \frac{a _{1}}{2} + \frac{d}{4} + \frac{d}{8} + \frac{d}{16} + \dots$
Notice the terms from the second position onwards.

Summing the Infinite GP

The series $\frac{d}{4} + \frac{d}{8} + \frac{d}{16} + \dots$ is an infinite Geometric Progression.
First term $a^{'} = \frac{d}{4}$
Common ratio $r^{'} = \frac{1}{2}$
Sum of infinite GP: $S_{\infty} = \frac{a ^{'}}{1 - r ^{'}}$

Simplifying the GP Sum

Substitute values into the GP sum formula:
$S_{\infty} = \frac{\frac{d}{4}}{1 - \frac{1}{2}}$
$S_{\infty} = \frac{\frac{d}{4}}{\frac{1}{2}} = \frac{d}{2}$

Simplifying the Expression for $S$

Substitute the GP sum back into the main equation:
$\frac{S}{2} = \frac{a _{1}}{2} + \frac{d}{2}$
Multiply the entire equation by $2$ :
$S = a_{1} + d$

Connecting $S$ to the Second Term

Recall the definition of the AP terms: $a_{2} = a_{1} + d$
Therefore, $S = a_{2}$
We are given that $S = 4$
So, $a_{2} = 4$

Final Calculation

The question asks for the value of $4 a_{2}$
Substitute $a_{2} = 4$ :
$4 a_{2} = 4 \times 4 = 16$
Final Answer: 16

00:00 / 00:00

The Sigma Insight: Arithmetic-Geometric Progression (A.G.P.)

Analyzing the Setup

We are given the infinite sum:

r = 1 \sum \infty \frac{a _{r}}{2 ^{r}} = 4

Here,

a_{r}

represents the

r^{t h}

term of an arithmetic progression (AP), defined as

a_{r} = a_{1} + (r - 1) d

. This structure, where a linear term is multiplied by a geometric term, is known as an Arithmetico-Geometric Progression (AGP).

The Shift-and-Subtract Method

To evaluate the sum

S

, we write out the terms explicitly:

S = \frac{a _{1}}{2} + \frac{a _{1} + d}{4} + \frac{a _{1} + 2 d}{8} + \frac{a _{1} + 3 d}{16} + \dots

We multiply the entire equation by the common ratio of the geometric part, which is

\frac{1}{2}

\frac{S}{2} = \frac{a _{1}}{4} + \frac{a _{1} + d}{8} + \frac{a _{1} + 2 d}{16} + \dots

Simplifying the Expression

Now, we subtract the second equation from the first. By aligning terms with identical denominators, we observe a systematic cancellation:

S - \frac{S}{2} = \frac{a _{1}}{2} + (\frac{a _{1} + d}{4} - \frac{a _{1}}{4}) + (\frac{a _{1} + 2 d}{8} - \frac{a _{1} + d}{8}) + \dots

This simplifies to:

\frac{S}{2} = \frac{a _{1}}{2} + \frac{d}{4} + \frac{d}{8} + \frac{d}{16} + \dots

Final Calculation

The tail end of the equation is an infinite geometric series with first term

A = \frac{d}{4}

and common ratio

r = \frac{1}{2}

. Using the sum formula

S_{\infty} = \frac{A}{1 - r}

, the tail sums to:

\frac{\frac{d}{4}}{1 - \frac{1}{2}} = \frac{d}{2}

Substituting this back into our equation, we obtain:

\frac{S}{2} = \frac{a _{1}}{2} + \frac{d}{2}

Multiplying by

2

, we find

S = a_{1} + d

. Since

a_{1} + d

is the definition of the second term

a_{2}

, we conclude that

a_{2} = 4

. Therefore, the final result is:

$4 a_{2} = 4 \times 4 = 16$

Common Traps & FAQs

Wait, why can't I just use the sum formula for an AP? This is an AP, right?

That is a very common trap! While the numerators form an AP, the presence of the

2^{r}

in the denominator changes everything. It is an Arithmetico-Geometric Progression (AGP). You cannot use the standard AP sum formula because the geometric decay in the denominator dominates the behavior of the series.

Why do we multiply by

1/2

and shift? Is there a reason for this specific step?

Think of it as a 'cleaning' process. By multiplying by the common ratio and shifting, you align terms with identical denominators. This alignment allows the arithmetic differences to emerge, effectively 'peeling away' the complexity of the AP and leaving you with a clean, manageable infinite geometric series.

What if the common ratio was something else, like

1/3

? Would the method still work?

Absolutely! The 'Shift-and-Subtract' method is universal for AGPs. If the ratio were

1/3

, you would multiply by

1/3

and shift. The core principle is to create a structure where the arithmetic part cancels out, leaving you with a geometric series that you can easily sum using the formula

S_{\infty} = \frac{a}{1 - r}

Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, a_{3}, \dots$ be an AP. If $\sum_{r = 1}^{\infty} \frac{a _{r}}{2 ^{r}} = 4$ , then $4 a_{2}$ is equal to ________.

Enter Numerical Value:

Visualized Solution

Define the AP Terms

Expand the Given Series

Identify the AGP Structure

The Shift-and-Subtract Method

Subtracting the Series

Analyzing the Resulting Series

Summing the Infinite GP

Simplifying the GP Sum

Simplifying the Expression for $S$

Connecting $S$ to the Second Term

Final Calculation

The Sigma Insight: Arithmetic-Geometric Progression (A.G.P.)

Analyzing the Setup

The Shift-and-Subtract Method

Simplifying the Expression

Final Calculation

Similar Questions

Suppose $a_{1}, a_{2}, 2, a_{3}, a_{4}$ be in an arithmetico-geometric progression. If the common ratio of the corresponding geometric progression is 2 and the sum of all 5 terms of the arithmetico-geometric progression is $\frac{49}{2}$ , then $a_{4}$ is equal to ______________

If $8 = 3 + \frac{1}{4} (3 + p) + \frac{1}{4 ^{2}} (3 + 2 p) + \frac{1}{4 ^{3}} (3 + 3 p) + \dots \infty$ , then the value of $p$ is_____

The sum to infinite term of the series $1 + \frac{2}{3} + \frac{6}{3 ^{2}} + \frac{10}{3 ^{3}} + \frac{14}{3 ^{4}} + \dots$ is

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The sum of the infinite series $1 + \frac{2}{3} + \frac{7}{3 ^{2}} + \frac{12}{3 ^{3}} + \frac{17}{3 ^{4}} + \frac{22}{3 ^{5}} + \dots$ is equal to

The sum $\sum_{k = 1}^{20} \frac{k}{2 ^{k}}$ is equal to-

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If $7 = 5 + \frac{1}{7} (5 + α) + \frac{1}{7 ^{2}} (5 + 2 α) + \frac{1}{7 ^{3}} (5 + 3 α) + \dots \infty$ , then the value of $α$ is :

If $(20)^{19} + 2 (21) (20)^{18} + 3 (21)^{2} (20)^{17} + \dots + 20 (21)^{19} = k (20)^{19}$ , then $k$ is equal to ____.

Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​,a2​,a3​,… be an AP. If ∑r=1∞​2rar​​=4, then 4a2​ is equal to ________.

Enter Numerical Value:

Visualized Solution

Define the AP Terms

Expand the Given Series

Identify the AGP Structure

The Shift-and-Subtract Method

Subtracting the Series

Analyzing the Resulting Series

Summing the Infinite GP

Simplifying the GP Sum

Simplifying the Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Connecting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S to the Second Term

Final Calculation

The Sigma Insight: Arithmetic-Geometric Progression (A.G.P.)

Analyzing the Setup

The Shift-and-Subtract Method

Simplifying the Expression

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, a_{3}, \dots$ be an AP. If $\sum_{r = 1}^{\infty} \frac{a _{r}}{2 ^{r}} = 4$ , then $4 a_{2}$ is equal to ________.

Simplifying the Expression for $S$

Connecting $S$ to the Second Term

If $8 = 3 + \frac{1}{4} (3 + p) + \frac{1}{4 ^{2}} (3 + 2 p) + \frac{1}{4 ^{3}} (3 + 3 p) + \dots \infty$ , then the value of $p$ is_____

The sum to infinite term of the series $1 + \frac{2}{3} + \frac{6}{3 ^{2}} + \frac{10}{3 ^{3}} + \frac{14}{3 ^{4}} + \dots$ is

The value of $2^{1/4} \cdot 4^{1/8} \cdot 8^{1/16} \dots \infty$ is

The sum of the infinite series $1 + \frac{2}{3} + \frac{7}{3 ^{2}} + \frac{12}{3 ^{3}} + \frac{17}{3 ^{4}} + \frac{22}{3 ^{5}} + \dots$ is equal to

The sum $\sum_{k = 1}^{20} \frac{k}{2 ^{k}}$ is equal to-

If the value of $(1 + \frac{2}{3} + \frac{6}{3 ^{2}} + \frac{10}{3 ^{3}} + \dots upto \infty)^{l o g_{(0.25)} (\frac{1}{3} + \frac{1}{3 ^{2}} + \frac{1}{3 ^{3}} + \dots upto \infty)}$ is $l$ , then $l^{2}$ is equal to

Let $S = 2 + \frac{6}{7} + \frac{12}{7 ^{2}} + \frac{20}{7 ^{3}} + \frac{30}{7 ^{4}} + \dots$ . then $4 S$ is equal to

If $7 = 5 + \frac{1}{7} (5 + α) + \frac{1}{7 ^{2}} (5 + 2 α) + \frac{1}{7 ^{3}} (5 + 3 α) + \dots \infty$ , then the value of $α$ is :

If $(20)^{19} + 2 (21) (20)^{18} + 3 (21)^{2} (20)^{17} + \dots + 20 (21)^{19} = k (20)^{19}$ , then $k$ is equal to ____.