JEE Main 2020 (3 Sep Evening)

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Animated Solution for Mathematics - Sequence and Series: If $m$ arithmetic means (A. Ms) and three geometric means (G. Ms) are inserted between 3 and 243 such that $4^{t h}$ A.M. is equal to $2^{n d}$ G.M., then $m$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Sequences

Given numbers: $3$ and $243$
A.P. Sequence: $3, A_{1}, A_{2}, \dots, A_{m}, 243$
G.P. Sequence: $3, G_{1}, G_{2}, G_{3}, 243$

Analyzing the A.P. Structure

First term of A.P. ( $a$ ) = $3$
Last term ( $T_{m + 2}$ ) = $243$
Total terms ( $n$ ) = $m + 2$

Setting up the A.P. Equation

Using $T_{n} = a + (n - 1) d$ :
$243 = 3 + (m + 2 - 1) d$
$243 = 3 + (m + 1) d$

Solving for Common Difference $d$

$243 - 3 = (m + 1) d$
$240 = (m + 1) d$
$d = \frac{240}{m + 1}$

Finding the $4^{t h}$ A.M.

$4^{t h}$ A.M. ( $A_{4}$ ) = $5^{t h}$ term of A.P.
$A_{4} = a + 4 d$
Substitute $a = 3$ and $d = \frac{240}{m + 1}$ :
$A_{4} = 3 + 4 (\frac{240}{m + 1})$

Simplifying $A_{4}$

$A_{4} = 3 + \frac{960}{m + 1}$

Analyzing the G.P. Structure

First term of G.P. ( $a^{'}$ ) = $3$
Last term ( $T_{5}^{'}$ ) = $243$
Total terms ( $n^{'}$ ) = $3 + 2 = 5$

Setting up the G.P. Equation

Using $T_{n}^{'} = a^{'} (r^{'})^{n - 1}$ :
$243 = 3 (r^{'})^{5 - 1}$
$243 = 3 (r^{'})^{4}$

Solving for Common Ratio $r^{'}$

$(r^{'})^{4} = \frac{243}{3} = 81$
$(r^{'})^{4} = 3^{4}$
$r^{'} = 3$ (taking positive ratio)

Finding the $2^{n d}$ G.M.

$2^{n d}$ G.M. ( $G_{2}$ ) = $3^{r d}$ term of G.P.
$G_{2} = a^{'} (r^{'})^{2}$
$G_{2} = 3 (3)^{2} = 3 \times 9 = 27$

Equating $A_{4}$ and $G_{2}$

Given: $A_{4} = G_{2}$
$3 + \frac{960}{m + 1} = 27$

Solving for $m$ (Part 1)

$\frac{960}{m + 1} = 27 - 3$
$\frac{960}{m + 1} = 24$

Solving for $m$ (Part 2)

$m + 1 = \frac{960}{24}$
$m + 1 = 40$

Final Value of $m$

$m = 40 - 1$
$m = 39$

00:00 / 00:00

The Sigma Insight: Relation Between A.M., G.M., and H.M.

Solution Diagram

Analyzing the Setup

Imagine you are standing on a vast plain with two markers placed at

3

and

243

. Your task is to build a bridge between them using two different structural methods.

In the first scenario, you construct an Arithmetic Progression (A.P.) with a constant step size

d

. In the second, you construct a Geometric Progression (G.P.) with a constant multiplier

r

The Arithmetic Journey

We start at

a = 3

and end at

T_{m + 2} = 243

. By inserting

m

arithmetic means, the total number of terms in the sequence becomes

n = m + 2

Using the fundamental formula for the

n

-th term of an A.P.,

T_{n} = a + (n - 1) d

, we substitute our known values:

243 = 3 + (m + 2 - 1) d

This simplifies to:

240 = (m + 1) d ⟹ d = \frac{240}{m + 1}

The

4^{t h}

arithmetic mean corresponds to the

5^{t h}

term of the sequence. Therefore, we calculate:

A_{4} = a + 4 d = 3 + 4 (\frac{240}{m + 1})

This yields our first pillar:

A_{4} = 3 + \frac{960}{m + 1}

The Geometric Journey

Now, we pivot to the G.P. bridge. We start at

3

and end at

243

, inserting exactly

3

geometric means. This makes the total number of terms

n^{'} = 3 + 2 = 5

The general term for a G.P. is

T_{n} = a (r)^{n - 1}

. Plugging in our values:

243 = 3 (r)^{5 - 1} ⟹ 243 = 3 (r)^{4}

Dividing by

3

, we find

r^{4} = 81

. Since

81 = 3^{4}

, the common ratio is

r = 3

The

2^{n d}

geometric mean is the

3^{r d}

term of the G.P., given by

G_{2} = a (r)^{2}

. Substituting our values:

G_{2} = 3 (3)^{2} = 3 \times 9 = 27

The Convergence

We have arrived at the climax of our journey. The problem states that the

4^{t h}

A.M. is equal to the

2^{n d}

G.M. We equate our findings:

3 + \frac{960}{m + 1} = 27

Subtracting

3

from both sides, we obtain:

\frac{960}{m + 1} = 24

Rearranging the equation to solve for

m

m + 1 = \frac{960}{24} = 40

Thus, the final value is $m = 39$ .

Common Traps & FAQs

Wait, why is the 4th A.M. the 5th term? Shouldn't it be the 4th term?

This is the most common trap in sequence problems! Think of it this way: the sequence starts with the first number (the pillar). The 1st mean is the 2nd term, the 2nd mean is the 3rd term, and so on. Therefore, the

k

-th mean is always the

(k + 1)

-th term. Always draw out the first few terms if you feel confused!

Why did we take the positive root for the common ratio

r

In this specific problem, we are dealing with a sequence between 3 and 243. While mathematically

r^{4} = 81

allows for

r = - 3

, a negative ratio would cause the terms to oscillate between positive and negative values. In the context of standard mean problems, we look for the principal real root unless otherwise specified.

Can we solve this using properties of means directly without finding

d

and

r

You could, but it often leads to more complex algebraic manipulation. The method of finding the common difference

d

and common ratio

r

first is the most robust and least error-prone approach. It builds a solid foundation for more complex problems where the means might not be integers.

Animated Solution for Mathematics - Sequence and Series: If $m$ arithmetic means (A. Ms) and three geometric means (G. Ms) are inserted between 3 and 243 such that $4^{t h}$ A.M. is equal to $2^{n d}$ G.M., then $m$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Sequences

Analyzing the A.P. Structure

Setting up the A.P. Equation

Solving for Common Difference $d$

Finding the $4^{t h}$ A.M.

Simplifying $A_{4}$

Analyzing the G.P. Structure

Setting up the G.P. Equation

Solving for Common Ratio $r^{'}$

Finding the $2^{n d}$ G.M.

Equating $A_{4}$ and $G_{2}$

Solving for $m$ (Part 1)

Solving for $m$ (Part 2)

Final Value of $m$

The Sigma Insight: Relation Between A.M., G.M., and H.M.

Analyzing the Setup

The Arithmetic Journey

The Geometric Journey

The Convergence

Similar Questions

If $m$ is the A.M. of two distinct real numbers $l$ and $n$ ( $l, n > 1$ ) and $G_{1}, G_{2}$ and $G_{3}$ are three geometric means between $l$ and $n$ , then $G_{1}^{4} + 2 G_{2}^{4} + G_{3}^{4}$ equals:

Let $A_{1}$ and $A_{2}$ be two arithmetic means and $G_{1}, G_{2}$ and $G_{3}$ be three geometric means of two distinct positive numbers. Then $G_{1}^{4} + G_{2}^{4} + G_{3}^{4} + G_{1}^{2} G_{3}^{2}$ is equal to

Let $a, b, c$ be positive integers such that $b / a$ is an integer. If $a, b, c$ are in geometric progression and the arithmetic mean of $a, b, c$ is $b + 2$ , then the value of $\frac{a ^{2} + a - 14}{a + 1}$ is \dots.

If the arithmetic mean and geometric mean of the $p^{t h}$ and $q^{t h}$ terms of the sequence $- 16, 8, - 4, 2, \dots$ satisfy the equation $4 x^{2} - 9 x + 5 = 0$ , then $p + q$ is equal to

Comprehension Passage

Which one of the following statements is correct ?

Which one of the following statements is correct ?

Which one of the following statements is correct?

The harmonic mean of two numbers is $4$ . Their arithmetic mean $A$ and the geometric mean $G$ satisfy the relation $2 A + G^{2} = 27$ . Find the two numbers.

Let $p$ be the first of the $n$ arithmetic means between two numbers and $q$ the first of $n$ harmonic means between the same numbers. Show that $q$ does not lie between $p$ and $(\frac{n + 1}{n - 1})^{2} p$ .

Let $0 < z < y < x$ be three real numbers such that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in an arithmetic progression and $x, 2 y, z$ are in a geometric progression. If $x y + y z + z x = \frac{3}{2} x y z$ , then $3 (x + y + z)^{2}$ is equal to

If the first and the $(2 n - 1)$ st terms of an A.P., a G.P. and an H.P. are equal and their $n$ th terms are $a, b$ and $c$ respectively, then

Enter Numerical Value:

Visualized Solution

Visualizing the Sequences

Analyzing the A.P. Structure

Setting up the A.P. Equation

Solving for Common Difference .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Finding the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4th A.M.

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A4​

Analyzing the G.P. Structure

Setting up the G.P. Equation

Solving for Common Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r′

Finding the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2nd G.M.

Equating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A4​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G2​

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m (Part 1)

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m (Part 2)

Final Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

The Sigma Insight: Relation Between A.M., G.M., and H.M.

Analyzing the Setup

The Arithmetic Journey

The Geometric Journey

The Convergence

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If m is the A.M. of two distinct real numbers l and n (l,n>1) and G1​,G2​ and G3​ are three geometric means between l and n, then G14​+2G24​+G34​ equals:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the arithmetic mean and geometric mean of the pth and qth terms of the sequence −16,8,−4,2,… satisfy the equation 4x2−9x+5=0, then p+q is equal to

Comprehension Passage

Which one of the following statements is correct ?

Which one of the following statements is correct ?

Which one of the following statements is correct?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The harmonic mean of two numbers is 4. Their arithmetic mean A and the geometric mean G satisfy the relation 2A+G2=27. Find the two numbers.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the first and the (2n−1)st terms of an A.P., a G.P. and an H.P. are equal and their nth terms are a,b and c respectively, then

Solving for Common Difference $d$

Finding the $4^{t h}$ A.M.

Simplifying $A_{4}$

Solving for Common Ratio $r^{'}$

Finding the $2^{n d}$ G.M.

Equating $A_{4}$ and $G_{2}$

Solving for $m$ (Part 1)

Solving for $m$ (Part 2)

Final Value of $m$

If $m$ is the A.M. of two distinct real numbers $l$ and $n$ ( $l, n > 1$ ) and $G_{1}, G_{2}$ and $G_{3}$ are three geometric means between $l$ and $n$ , then $G_{1}^{4} + 2 G_{2}^{4} + G_{3}^{4}$ equals:

If the arithmetic mean and geometric mean of the $p^{t h}$ and $q^{t h}$ terms of the sequence $- 16, 8, - 4, 2, \dots$ satisfy the equation $4 x^{2} - 9 x + 5 = 0$ , then $p + q$ is equal to

The harmonic mean of two numbers is $4$ . Their arithmetic mean $A$ and the geometric mean $G$ satisfy the relation $2 A + G^{2} = 27$ . Find the two numbers.

If the first and the $(2 n - 1)$ st terms of an A.P., a G.P. and an H.P. are equal and their $n$ th terms are $a, b$ and $c$ respectively, then