JEE Advanced 1999

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, \dots, a_{10}$ be in A.P. and $h_{1}, h_{2}, \dots, h_{10}$ be in H.P. If $a_{1} = h_{1} = 2$ and $a_{10} = h_{10} = 3$ , then $a_{4} h_{7}$ is

Q: Why is the common difference $D$ for the H.P. reciprocal sequence negative?

That's a sharp observation! Since $h_1=2$ and $h_{10}=3$, the H.P. is increasing. If the terms are increasing, their reciprocals must be decreasing. A negative common difference $D$ in the reciprocal A.P. is exactly what we need to make the H.P. values grow.

Select Answer:

Visualized Solution

Orient: Defining the Sequences

Given:
A.P. sequence: $a_{1}, a_{2}, \dots, a_{10}$ with $a_{1} = 2$ and $a_{10} = 3$
H.P. sequence: $h_{1}, h_{2}, \dots, h_{10}$ with $h_{1} = 2$ and $h_{10} = 3$
To find: The product of $a_{4}$ and $h_{7}$

Logic Bridge: A.P. General Term Formula

For an A.P., the $n^{th}$ term is given by:
$a_{n} = a_{1} + (n - 1) d$
where $d$ is the common difference.

Raw Setup: Substituting for $a_{10}$

Substituting $n = 10$ , $a_{1} = 2$ , and $a_{10} = 3$ :
$3 = 2 + (10 - 1) d$
$3 = 2 + 9 d$

Atomic Compute: Finding $d$

Subtracting $2$ from both sides:
$1 = 9 d$
Dividing by $9$ :
$d = \frac{1}{9}$

Atomic Compute: Calculating $a_{4}$

To find $a_{4}$ :
$a_{4} = a_{1} + 3 d$
$a_{4} = 2 + 3 (\frac{1}{9})$
$a_{4} = 2 + \frac{1}{3} = \frac{7}{3}$

Logic Bridge: H.P. and Reciprocal A.P.

If $h_{1}, h_{2}, \dots, h_{10}$ are in H.P., then:
$\frac{1}{h _{1}}, \frac{1}{h _{2}}, \dots, \frac{1}{h _{10}}$ are in A.P.
Let the common difference of this reciprocal A.P. be $D$ .

Raw Setup: Substituting for $\frac{1}{h _{10}}$

Using the A.P. formula for the reciprocals:
$\frac{1}{h _{10}} = \frac{1}{h _{1}} + 9 D$
Substituting $h_{1} = 2$ and $h_{10} = 3$ :
$\frac{1}{3} = \frac{1}{2} + 9 D$

Atomic Compute: Finding $D$

Rearranging to solve for $9 D$ :
$9 D = \frac{1}{3} - \frac{1}{2}$
$9 D = \frac{2 - 3}{6} = - \frac{1}{6}$
Dividing by $9$ :
$D = - \frac{1}{54}$

Atomic Compute: Calculating $h_{7}$

To find $h_{7}$ , first find $\frac{1}{h _{7}}$ :
$\frac{1}{h _{7}} = \frac{1}{h _{1}} + 6 D$
$\frac{1}{h _{7}} = \frac{1}{2} + 6 (- \frac{1}{54}) = \frac{1}{2} - \frac{1}{9}$
$\frac{1}{h _{7}} = \frac{9 - 2}{18} = \frac{7}{18} \Rightarrow h_{7} = \frac{18}{7}$

Atomic Compute: Final Product $a_{4} h_{7}$

Calculating the product:
$a_{4} \times h_{7} = \frac{7}{3} \times \frac{18}{7}$
Canceling the $7$ s:
$a_{4} \times h_{7} = \frac{18}{3} = 6$

The Way Forward: Summary and Takeaways

Key Takeaway:
To solve H.P. problems, always convert to the reciprocal A.P.
Next Challenges:
What if the number of terms $n$ was different?
Can you find a general expression for $a_{k} \times h_{n - k + 1}$ ?

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The Sigma Insight: Harmonic Progression (H.P.)

Analyzing the Arithmetic Progression

The general term of an A.P. is defined as

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

. Given that

a_{1} = 2

and

a_{10} = 3

, we set up the following equation:

3 = 2 + (10 - 1) d

This simplifies to

1 = 9 d

, which yields a common difference of

d = \frac{1}{9}

To find the fourth term, we perform the calculation:

a_{4} = a_{1} + 3 d = 2 + 3 (\frac{1}{9}) = 2 + \frac{1}{3} = \frac{7}{3}

We have successfully secured our first piece of the puzzle.

The Harmonic Transformation

The secret to mastering an H.P. is to work with its reciprocal, which forms an A.P. If

h_{1}, h_{2}, \dots, h_{10}

are in H.P., then their reciprocals

\frac{1}{h _{1}}, \frac{1}{h _{2}}, \dots, \frac{1}{h _{10}}

form an A.P.

Let the common difference of this reciprocal sequence be

D

. We know

\frac{1}{h _{1}} = \frac{1}{2}

and

\frac{1}{h _{10}} = \frac{1}{3}

Applying the A.P. formula to the reciprocals, we get:

\frac{1}{h _{10}} = \frac{1}{h _{1}} + 9 D \Rightarrow \frac{1}{3} = \frac{1}{2} + 9 D

Solving for

D

, we find:

9 D = \frac{1}{3} - \frac{1}{2} = - \frac{1}{6} \Rightarrow D = - \frac{1}{54}

The Final Synthesis

With

D

determined, we find the seventh term of the H.P. by first calculating its reciprocal:

\frac{1}{h _{7}} = \frac{1}{h _{1}} + 6 D = \frac{1}{2} + 6 (- \frac{1}{54}) = \frac{1}{2} - \frac{1}{9}

Using a common denominator of

18

, we obtain:

\frac{1}{h _{7}} = \frac{9 - 2}{18} = \frac{7}{18} \Rightarrow h_{7} = \frac{18}{7}

Finally, we calculate the product

a_{4} \times h_{7}

a_{4} \times h_{7} = \frac{7}{3} \times \frac{18}{7} = \frac{18}{3} = 6

The final result of this journey is 6.

Common Traps & FAQs

Wait, why do we take the reciprocal of the H.P. terms? Can't we just solve it directly?

Great question! H.P. is defined by the property that its reciprocals form an A.P. There is no simple 'linear' formula for H.P. terms directly. By taking the reciprocal, we map a complex harmonic problem into the familiar, linear world of A.P., where we have powerful tools like the general term formula.

Why is the common difference

D

for the H.P. reciprocal sequence negative?

That's a sharp observation! Since

h_{1} = 2

and

h_{10} = 3

, the H.P. is increasing. If the terms are increasing, their reciprocals must be decreasing. A negative common difference

D

in the reciprocal A.P. is exactly what we need to make the H.P. values grow.

Is there a shortcut for

a_{4} \times h_{7}

While there are properties relating terms of A.P. and H.P., the most robust way to solve this in an exam is to calculate the terms individually. Don't look for shortcuts that might lead to errors; trust the process of finding the common difference first!

Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, \dots, a_{10}$ be in A.P. and $h_{1}, h_{2}, \dots, h_{10}$ be in H.P. If $a_{1} = h_{1} = 2$ and $a_{10} = h_{10} = 3$ , then $a_{4} h_{7}$ is

Select Answer:

Visualized Solution

Orient: Defining the Sequences

Logic Bridge: A.P. General Term Formula

Raw Setup: Substituting for $a_{10}$

Atomic Compute: Finding $d$

Atomic Compute: Calculating $a_{4}$

Logic Bridge: H.P. and Reciprocal A.P.

Raw Setup: Substituting for $\frac{1}{h _{10}}$

Atomic Compute: Finding $D$

Atomic Compute: Calculating $h_{7}$

Atomic Compute: Final Product $a_{4} h_{7}$

The Way Forward: Summary and Takeaways

The Sigma Insight: Harmonic Progression (H.P.)

Analyzing the Arithmetic Progression

The Harmonic Transformation

The Final Synthesis

Similar Questions

If $a_{1}, a_{2}, \dots, a_{n}$ are in H.P., then the expression $a_{1} a_{2} + a_{2} a_{3} + \dots + a_{n - 1} a_{n}$ is equal to

Let the positive numbers $a, b, c, d$ be in A.P. Then $ab c, ab d, a c d, b c d$ are

Let $\frac{1}{x _{1}}, \frac{1}{x _{2}}, \dots, \frac{1}{x _{n}}$ ( $x_{i} \neq = 0$ for $i = 1, 2, \dots, n$ ) be in A.P. such that $x_{1} = 4$ and $x_{21} = 20$ . If n is the least positive integer for which $x_{n} > 50$ , then $\sum_{i = 1}^{n} (\frac{1}{x _{i}})$ is equal to :

If $x = \sum_{n = 0}^{\infty} a^{n}, y = \sum_{n = 0}^{\infty} b^{n}, z = \sum_{n = 0}^{\infty} c^{n}$ where $a, b, c$ are in A.P and $∣ a ∣ < 1, ∣ b ∣ < 1, ∣ c ∣ < 1$ then $x, y, z$ are in

Let $a_{1}, a_{2}, a_{3} \dots$ be in harmonic progression with $a_{1} = 5$ and $a_{20} = 25$ . The least positive integer $n$ for which $a_{n} < 0$ is

If $x > 1, y > 1, z > 1$ are in G.P., then $\frac{1}{1 + l n x}, \frac{1}{1 + l n y}, \frac{1}{1 + l n z}$ are in

Let $R_{1}$ and $R_{2}$ respectively be the maximum ranges up and down an inclined plane and $R$ be the maximum range on the horizontal plane. Then $R_{1}, R, R_{2}$ are in

Select Answer:

Visualized Solution

Orient: Defining the Sequences

Logic Bridge: A.P. General Term Formula

Raw Setup: Substituting for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a10​

Atomic Compute: Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Atomic Compute: Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a4​

Logic Bridge: H.P. and Reciprocal A.P.

Raw Setup: Substituting for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h10​1​

Atomic Compute: Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Atomic Compute: Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h7​

Atomic Compute: Final Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a4​h7​

The Way Forward: Summary and Takeaways

The Sigma Insight: Harmonic Progression (H.P.)

Analyzing the Arithmetic Progression

The Harmonic Transformation

The Final Synthesis

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the positive numbers a,b,c,d be in A.P. Then abc,abd,acd,bcd are

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​,a2​,a3​… be in harmonic progression with a1​=5 and a20​=25. The least positive integer n for which an​<0 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x>1,y>1,z>1 are in G.P., then 1+lnx1​,1+lny1​,1+lnz1​ are in

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Raw Setup: Substituting for $a_{10}$

Atomic Compute: Finding $d$

Atomic Compute: Calculating $a_{4}$

Raw Setup: Substituting for $\frac{1}{h _{10}}$

Atomic Compute: Finding $D$

Atomic Compute: Calculating $h_{7}$

Atomic Compute: Final Product $a_{4} h_{7}$

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Let $a_{1}, a_{2}, a_{3} \dots$ be in harmonic progression with $a_{1} = 5$ and $a_{20} = 25$ . The least positive integer $n$ for which $a_{n} < 0$ is

If $x > 1, y > 1, z > 1$ are in G.P., then $\frac{1}{1 + l n x}, \frac{1}{1 + l n y}, \frac{1}{1 + l n z}$ are in

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