JEE Main 2023 (11 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let $S = 109 + \frac{108}{5} + \frac{107}{5 ^{2}} + \dots + \frac{2}{5 ^{107}} + \frac{1}{5 ^{108}}$ . Then the value of $(16 S - (25)^{- 54})$ is equal to

Q: The final answer is an integer, but my calculation involves powers of $5$. Did I do something wrong?

Not at all! That is the beauty of this problem. The term $(25)^{-54}$ is actually $(5^2)^{-54} = 5^{-108} = \frac{1}{5^{108}}$. When you simplify $16S$, you will find a term involving $\frac{1}{5^{108}}$ that perfectly cancels out with the term you are subtracting. If you reach a point where the fractional parts don't cancel, re-check your algebraic expansion of the bracketed G.P. sum.

Enter Numerical Value:

Visualized Solution

Analyze the Given Series $S$

Given series: $S = 109 + \frac{108}{5} + \frac{107}{5 ^{2}} + \dots + \frac{2}{5 ^{107}} + \frac{1}{5 ^{108}}$
This is an Arithmetico-Geometric Progression (AGP) where:
Numerators form an A.P.: $109, 108, 107, \dots, 1$
Denominators form a G.P.: $1, 5, 5^{2}, \dots, 5^{108}$

Multiply by the Common Ratio $\frac{1}{5}$

Multiply $S$ by the common ratio $r = \frac{1}{5}$ :
$\frac{S}{5} = \frac{109}{5} + \frac{108}{5 ^{2}} + \frac{107}{5 ^{3}} + \dots + \frac{2}{5 ^{108}} + \frac{1}{5 ^{109}}$

Subtract $\frac{S}{5}$ from $S$

Subtracting the two equations:
$S - \frac{S}{5} = 109 + (\frac{108 - 109}{5}) + (\frac{107 - 108}{5 ^{2}}) + \dots + (\frac{1 - 2}{5 ^{108}}) - \frac{1}{5 ^{109}}$
$\frac{4 S}{5} = 109 - \frac{1}{5} - \frac{1}{5 ^{2}} - \dots - \frac{1}{5 ^{108}} - \frac{1}{5 ^{109}}$

Identify the Geometric Progression

Group the terms to form a G.P.:
$\frac{4 S}{5} = 109 - (\frac{1}{5} + \frac{1}{5 ^{2}} + \dots + \frac{1}{5 ^{108}}) - \frac{1}{5 ^{109}}$
The term in the bracket is a G.P. with $a = \frac{1}{5}$ , $r = \frac{1}{5}$ , and $n = 108$ terms.

Apply the GP Sum Formula

Sum of G.P. $S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}$
Sum $= \frac{\frac{1}{5} ( 1 - ( \frac{1}{5} ) ^{108} )}{1 - \frac{1}{5}} = \frac{\frac{1}{5} ( 1 - \frac{1}{5 ^{108}} )}{\frac{4}{5}}$

Simplify the GP Sum

Simplify the denominator:
$1 - \frac{1}{5} = \frac{4}{5}$
The $\frac{1}{5}$ terms cancel out:
Sum $= \frac{1}{4} (1 - \frac{1}{5 ^{108}})$

Substitute Back into the Equation

Substitute the sum back:
$\frac{4 S}{5} = 109 - \frac{1}{4} (1 - \frac{1}{5 ^{108}}) - \frac{1}{5 ^{109}}$

Multiply by $4$ to Clear Fractions

Multiply by $4$ to clear the fraction:
$\frac{16 S}{5} = 436 - (1 - \frac{1}{5 ^{108}}) - \frac{4}{5 ^{109}}$

Simplify the Expression for $\frac{16 S}{5}$

Simplify the terms:
$\frac{16 S}{5} = 435 + \frac{1}{5 ^{108}} - \frac{4}{5 ^{109}}$
Convert to common denominator $5^{109}$ :
$\frac{16 S}{5} = 435 + \frac{5 - 4}{5 ^{109}} = 435 + \frac{1}{5 ^{109}}$

Calculate the Value of $16 S$

Multiply by $5$ to find $16 S$ :
$16 S = 435 \times 5 + \frac{5}{5 ^{109}}$
$16 S = 2175 + \frac{1}{5 ^{108}}$

Evaluate the Final Expression

Target expression: $16 S - (25)^{- 54}$
Simplify the second term: $(25)^{- 54} = (5^{2})^{- 54} = 5^{- 108} = \frac{1}{5 ^{108}}$

Final Answer

Substitute $16 S$ :
$(2175 + \frac{1}{5 ^{108}}) - \frac{1}{5 ^{108}}$
The terms $\frac{1}{5 ^{108}}$ cancel out.
Final Result: $2175$

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

We are dealing with an Arithmetico-Geometric Progression (AGP). The sequence is defined as:

S = 109 + \frac{108}{5} + \frac{107}{5 ^{2}} + \dots + \frac{2}{5 ^{107}} + \frac{1}{5 ^{108}}

In this series, the numerators form an arithmetic progression (

109, 108, \dots, 1

) and the denominators form a geometric progression (

1, 5, 5^{2}, \dots, 5^{108}

The Art of the Shift

To solve this, we employ the 'Shift and Subtract' method. We multiply the entire series by the common ratio

r = \frac{1}{5}

\frac{S}{5} = \frac{109}{5} + \frac{108}{5 ^{2}} + \frac{107}{5 ^{3}} + \dots + \frac{2}{5 ^{108}} + \frac{1}{5 ^{109}}

Now, we subtract

\frac{S}{5}

from

S

S - \frac{S}{5} = 109 + (\frac{108 - 109}{5}) + (\frac{107 - 108}{5 ^{2}}) + \dots + (\frac{1 - 2}{5 ^{108}}) - \frac{1}{5 ^{109}}

This simplifies to:

\frac{4 S}{5} = 109 - (\frac{1}{5} + \frac{1}{5 ^{2}} + \dots + \frac{1}{5 ^{108}}) - \frac{1}{5 ^{109}}

The Hidden Geometric Progression

The term inside the parentheses is a finite geometric progression with

a = \frac{1}{5}

r = \frac{1}{5}

, and

n = 108

terms. Using the sum formula

S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}

, we get:

Sum = \frac{\frac{1}{5} ( 1 - \frac{1}{5 ^{108}} )}{1 - \frac{1}{5}} = \frac{\frac{1}{5} ( 1 - \frac{1}{5 ^{108}} )}{\frac{4}{5}} = \frac{1}{4} (1 - \frac{1}{5 ^{108}})

Substituting this back into our equation for

\frac{4 S}{5}

\frac{4 S}{5} = 109 - \frac{1}{4} (1 - \frac{1}{5 ^{108}}) - \frac{1}{5 ^{109}}

The Final Calculation

To isolate

16 S

, we multiply the entire equation by

4

\frac{16 S}{5} = 436 - (1 - \frac{1}{5 ^{108}}) - \frac{4}{5 ^{109}}

\frac{16 S}{5} = 435 + \frac{1}{5 ^{108}} - \frac{4}{5 ^{109}}

Combining the fractional terms with a common denominator of

5^{109}

\frac{16 S}{5} = 435 + \frac{5 - 4}{5 ^{109}} = 435 + \frac{1}{5 ^{109}}

Multiplying by

5

yields:

16 S = 2175 + \frac{1}{5 ^{108}}

Given that

(25)^{- 54} = (5^{2})^{- 54} = 5^{- 108} = \frac{1}{5 ^{108}}

, the expression

16 S - (25)^{- 54}

becomes:

2175 + \frac{1}{5 ^{108}} - \frac{1}{5 ^{108}} = 2175

The final result is 2175.

Common Traps & FAQs

Why do we multiply by

1/5

and then subtract? Can't we just sum it directly?

That is a great question! Summing an AGP directly is extremely difficult because the terms don't share a common factor. By multiplying by the common ratio (

1/5

) and shifting the terms, we create a situation where the difference between consecutive terms in the numerator becomes constant (in this case,

108 - 109 = - 1

107 - 108 = - 1

, etc.). This 'telescoping' effect transforms the complex AGP into a simple G.P., which we know how to sum perfectly.

How do I know how many terms are in the G.P. part after subtraction?

Counting terms is where most students stumble! Look at the series: the original series has terms from

5^{0}

5^{108}

, which is

109

terms. When you subtract

\frac{S}{5}

from

S

, the G.P. part starts from the term with

5^{1}

and goes up to

5^{108}

. That is exactly

108

terms. Always write out the first and last few terms of the subtraction to be absolutely sure of your count.

The final answer is an integer, but my calculation involves powers of

5

. Did I do something wrong?

Not at all! That is the beauty of this problem. The term

(25)^{- 54}

is actually

(5^{2})^{- 54} = 5^{- 108} = \frac{1}{5 ^{108}}

. When you simplify

16 S

, you will find a term involving

\frac{1}{5 ^{108}}

that perfectly cancels out with the term you are subtracting. If you reach a point where the fractional parts don't cancel, re-check your algebraic expansion of the bracketed G.P. sum.

Animated Solution for Mathematics - Sequence and Series: Let $S = 109 + \frac{108}{5} + \frac{107}{5 ^{2}} + \dots + \frac{2}{5 ^{107}} + \frac{1}{5 ^{108}}$ . Then the value of $(16 S - (25)^{- 54})$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Given Series $S$

Multiply by the Common Ratio $\frac{1}{5}$

Subtract $\frac{S}{5}$ from $S$

Identify the Geometric Progression

Apply the GP Sum Formula

Simplify the GP Sum

Substitute Back into the Equation

Multiply by $4$ to Clear Fractions

Simplify the Expression for $\frac{16 S}{5}$

Calculate the Value of $16 S$

Evaluate the Final Expression

Final Answer

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

Analyzing the Setup

The Art of the Shift

The Hidden Geometric Progression

The Final Calculation

Similar Questions

If $S = \frac{7}{5} + \frac{9}{5 ^{2}} + \frac{13}{5 ^{3}} + \frac{19}{5 ^{4}} + \dots$ , then $160 S$ is equal to .

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

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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

If $S (x) = (1 + x) + 2 (1 + x)^{2} + 3 (1 + x)^{3} + \dots + 60 (1 + x)^{60}, x \neq = 0$ , and $(60)^{2} S (60) = a (b)^{b} + b$ , where $a, b \in N$ , then $(a + b)$ 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 S=109+5108​+52107​+⋯+51072​+51081​. Then the value of (16S−(25)−54) is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Given Series .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Multiply by the Common Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 51​

Subtract .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5S​ from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Identify the Geometric Progression

Apply the GP Sum Formula

Simplify the GP Sum

Substitute Back into the Equation

Multiply by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4 to Clear Fractions

Simplify the Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 516S​

Calculate the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 16S

Evaluate the Final Expression

Final Answer

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

Analyzing the Setup

The Art of the Shift

The Hidden Geometric Progression

The Final Calculation

Similar Questions

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Animated Solution for Mathematics - Sequence and Series: Let $S = 109 + \frac{108}{5} + \frac{107}{5 ^{2}} + \dots + \frac{2}{5 ^{107}} + \frac{1}{5 ^{108}}$ . Then the value of $(16 S - (25)^{- 54})$ is equal to

Analyze the Given Series $S$

Multiply by the Common Ratio $\frac{1}{5}$

Subtract $\frac{S}{5}$ from $S$

Multiply by $4$ to Clear Fractions

Simplify the Expression for $\frac{16 S}{5}$

Calculate the Value of $16 S$

If $S = \frac{7}{5} + \frac{9}{5 ^{2}} + \frac{13}{5 ^{3}} + \frac{19}{5 ^{4}} + \dots$ , then $160 S$ 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

The sum $\sum_{k = 1}^{20} \frac{k}{2 ^{k}}$ 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 :

The sum of the infinite series $1 + \frac{5}{6} + \frac{12}{6 ^{2}} + \frac{22}{6 ^{3}} + \frac{35}{6 ^{4}} + \frac{51}{6 ^{5}} + \frac{70}{6 ^{6}} + \dots$ is equal to:

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

The sum $1 + 2 \cdot 3 + 3 \cdot 3^{2} + \dots + 10 \cdot 3^{9}$ is equal to

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

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

If $S (x) = (1 + x) + 2 (1 + x)^{2} + 3 (1 + x)^{3} + \dots + 60 (1 + x)^{60}, x \neq = 0$ , and $(60)^{2} S (60) = a (b)^{b} + b$ , where $a, b \in N$ , then $(a + b)$ equal to