JEE Main 2022 (25 July Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let $a, b$ be two non-zero real numbers. If $p$ and $r$ are the roots of the equation $x^{2} - 8 a x + 2 a = 0$ and $q$ and $s$ are the roots of the equation $x^{2} + 12 b x + 6 b = 0$ , such that $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s}$ are in A.P., then $a^{- 1} - b^{- 1}$ is equal to ______.

Q: How do I know the order of the A.P. terms? Could it be $\frac{1}{s}, \frac{1}{r}, \frac{1}{q}, \frac{1}{p}$?

The problem explicitly states that $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s}$ are in A.P. In mathematics, the order of a sequence is defined by the order in which the terms are listed. Therefore, $\frac{1}{p}$ is the first term, $\frac{1}{q}$ is the second, and so on. Always stick to the order given in the problem statement to avoid unnecessary confusion!

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the relationship between the roots and the coefficients of the quadratic equations. The problem provides two equations,

x^{2} - 8 a x + 2 a = 0

and

x^{2} + 12 b x + 6 b = 0

. Notice that the sum of the reciprocals of the roots,

\frac{1}{α} + \frac{1}{β}

, can be expressed simply as

\frac{α + β}{α β}

, which is

- \frac{coefficient of x}{constant term}

Hint 2

Hint 2 (The Arsenal): Use Vieta's formulas to find the sum and product of the roots for both equations. For the first equation, the sum of reciprocals

\frac{1}{p} + \frac{1}{r}

will be a constant. Do the same for the second equation with roots

q

and

s

. This will give you two numerical values that constrain the terms of the Arithmetic Progression.

Hint 3

Hint 3 (The Execution): Let the A.P. be defined as

x, x + d, x + 2 d, x + 3 d

corresponding to

\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s}

. Use the sums of reciprocals you calculated to set up a system of linear equations:

(x) + (x + 2 d) = sum_{1}

and

(x + d) + (x + 3 d) = sum_{2}

. Solve for

x

and

d

to find all four terms.

Visualized Solution

Analyze the First Equation $x^{2} - 8 a x + 2 a = 0$

Given equation: $x^{2} - 8 a x + 2 a = 0$ with roots $p$ and $r$ .
Sum of roots: $p + r = - (- 8 a) = 8 a$ .
Product of roots: $p r = 2 a$ .

Calculate Sum of Reciprocals $\frac{1}{p} + \frac{1}{r}$

We need the sum of reciprocals for the roots $p$ and $r$ .
$\frac{1}{p} + \frac{1}{r} = \frac{p + r}{p r}$
Substituting the values: $\frac{1}{p} + \frac{1}{r} = \frac{8 a}{2 a} = 4$ .

Analyze the Second Equation $x^{2} + 12 b x + 6 b = 0$

Given equation: $x^{2} + 12 b x + 6 b = 0$ with roots $q$ and $s$ .
Sum of roots: $q + s = - 12 b$ .
Product of roots: $q s = 6 b$ .

Calculate Sum of Reciprocals $\frac{1}{q} + \frac{1}{s}$

We need the sum of reciprocals for the roots $q$ and $s$ .
$\frac{1}{q} + \frac{1}{s} = \frac{q + s}{q s}$
Substituting the values: $\frac{1}{q} + \frac{1}{s} = \frac{- 12 b}{6 b} = - 2$ .

Define the Arithmetic Progression (A.P.)

Let the A.P. terms be $x, x + d, x + 2 d, x + 3 d$ .
$\frac{1}{p} = x$
$\frac{1}{q} = x + d$
$\frac{1}{r} = x + 2 d$
$\frac{1}{s} = x + 3 d$

Solve for A.P. Terms: Finding $\frac{1}{q}$

Using $\frac{1}{p} + \frac{1}{r} = 4$ :
$x + (x + 2 d) = 4 ⟹ 2 x + 2 d = 4$
Dividing by $2$ : $x + d = 2$
Since $\frac{1}{q} = x + d$ , we have $\frac{1}{q} = 2$ .

Solve for A.P. Terms: Finding $\frac{1}{r}$

Using $\frac{1}{q} + \frac{1}{s} = - 2$ :
$(x + d) + (x + 3 d) = - 2 ⟹ 2 x + 4 d = - 2$
Dividing by $2$ : $x + 2 d = - 1$
Since $\frac{1}{r} = x + 2 d$ , we have $\frac{1}{r} = - 1$ .

Find Common Difference $d$ and Remaining Terms

Common difference $d = \frac{1}{r} - \frac{1}{q} = - 1 - 2 = - 3$ .
$\frac{1}{p} = \frac{1}{q} - d = 2 - (- 3) = 5$ .
$\frac{1}{s} = \frac{1}{r} + d = - 1 + (- 3) = - 4$ .

Calculate $a^{- 1}$

From the first equation, $p r = 2 a ⟹ a = \frac{p r}{2}$ .
Therefore, $a^{- 1} = \frac{1}{a} = \frac{2}{p r} = 2 \cdot \frac{1}{p} \cdot \frac{1}{r}$ .
Substituting values: $a^{- 1} = 2 (5) (- 1) = - 10$ .

Calculate $b^{- 1}$

From the second equation, $q s = 6 b ⟹ b = \frac{q s}{6}$ .
Therefore, $b^{- 1} = \frac{1}{b} = \frac{6}{q s} = 6 \cdot \frac{1}{q} \cdot \frac{1}{s}$ .
Substituting values: $b^{- 1} = 6 (2) (- 4) = - 48$ .

Final Calculation: $a^{- 1} - b^{- 1}$

We need to find $a^{- 1} - b^{- 1}$ .
$a^{- 1} - b^{- 1} = - 10 - (- 48)$
$a^{- 1} - b^{- 1} = - 10 + 48 = 38$ .
The final answer is 38.

00:00 / 00:00

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Quadratic Foundation

We begin with our first equation:

x^{2} - 8 a x + 2 a = 0

. Its roots are

p

and

r

Your first instinct should always be Vieta's formulas, as they are the heartbeat of quadratic problems. We know the sum of the roots is

p + r = 8 a

and the product is

p r = 2 a

Now, look at the second equation:

x^{2} + 12 b x + 6 b = 0

, with roots

q

and

s

. Similarly,

q + s = - 12 b

and

q s = 6 b

These are our building blocks. Do not rush to solve for

p, r, q,

s

; instead, focus on the relationship between their reciprocals.

The Reciprocal Shortcut

The problem states that

\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s}

are in an Arithmetic Progression (A.P.). This is our bridge.

Let us calculate the sum of the reciprocals for the first pair:

\frac{1}{p} + \frac{1}{r} = \frac{p + r}{p r} = \frac{8 a}{2 a} = 4

The variable

a

vanishes! It is the beauty of algebra at work. Now, for the second pair:

\frac{1}{q} + \frac{1}{s} = \frac{q + s}{q s} = \frac{- 12 b}{6 b} = - 2

We have reduced the entire problem to two simple constants:

4

and

- 2

The A.P

Bridge

Let the terms of the A.P. be

x, x + d, x + 2 d, x + 3 d

. Mapping these to our reciprocals, we set:

\frac{1}{p} = x

\frac{1}{q} = x + d

\frac{1}{r} = x + 2 d

, and

\frac{1}{s} = x + 3 d

We derive two equations from our reciprocal sums:

\frac{1}{p} + \frac{1}{r} = x + (x + 2 d) = 2 x + 2 d = 4

, which simplifies to

x + d = 2

. Since

x + d

\frac{1}{q}

, we have found

\frac{1}{q} = 2

\frac{1}{q} + \frac{1}{s} = (x + d) + (x + 3 d) = 2 x + 4 d = - 2

, which simplifies to

x + 2 d = - 1

. Since

x + 2 d

\frac{1}{r}

, we have found

\frac{1}{r} = - 1

Final Calculation

We have

\frac{1}{q} = 2

and

\frac{1}{r} = - 1

. The common difference

d

is:

d = \frac{1}{r} - \frac{1}{q} = - 1 - 2 = - 3

Now we find the remaining terms:

\frac{1}{p} = \frac{1}{q} - d = 2 - (- 3) = 5

\frac{1}{s} = \frac{1}{r} + d = - 1 + (- 3) = - 4

Finally, we calculate

a^{- 1} - b^{- 1}

. Since

a = \frac{p r}{2}

, then

a^{- 1} = \frac{2}{p r} = 2 \cdot (\frac{1}{p} \cdot \frac{1}{r}) = 2 (5) (- 1) = - 10

Similarly, since

b = \frac{q s}{6}

, then

b^{- 1} = \frac{6}{q s} = 6 \cdot (\frac{1}{q} \cdot \frac{1}{s}) = 6 (2) (- 4) = - 48

The final result is:

a^{- 1} - b^{- 1} = - 10 - (- 48) = 38

Common Traps & FAQs

Wait, why do we reject the idea of solving for

p, q, r, s

directly using the quadratic formula?

That is a fantastic instinct to question! While you *could* use the quadratic formula to find

p, r, q,

and

s

, you would end up with expressions involving square roots of

a

and

b

. Trying to force those into an A.P. would lead to an algebraic nightmare. In JEE Advanced, the goal is to find the most elegant path. By using Vieta's formulas, we bypass the need to know the roots individually and focus on their symmetric properties, which is exactly what the problem designer intended.

How do I know the order of the A.P. terms? Could it be

\frac{1}{s}, \frac{1}{r}, \frac{1}{q}, \frac{1}{p}

The problem explicitly states that

\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s}

are in A.P. In mathematics, the order of a sequence is defined by the order in which the terms are listed. Therefore,

\frac{1}{p}

is the first term,

\frac{1}{q}

is the second, and so on. Always stick to the order given in the problem statement to avoid unnecessary confusion!

Why do the variables

a

and

b

disappear when we calculate the sum of reciprocals?

This is the 'Aha!' moment of the problem. When you calculate

\frac{1}{p} + \frac{1}{r} = \frac{p + r}{p r}

, you are dividing the sum of roots by the product of roots. In both equations, the sum of roots is a multiple of the variable (

8 a

12 b

) and the product of roots is the same multiple of the variable (

2 a

6 b

). When you divide them, the variable cancels out, leaving you with a pure constant. This is a common pattern in competitive exams—look for these cancellations to simplify your life!

Enter Numerical Value:

Visualized Solution

Analyze the First Equation $x^{2} - 8 a x + 2 a = 0$

Calculate Sum of Reciprocals $\frac{1}{p} + \frac{1}{r}$

Analyze the Second Equation $x^{2} + 12 b x + 6 b = 0$

Calculate Sum of Reciprocals $\frac{1}{q} + \frac{1}{s}$

Define the Arithmetic Progression (A.P.)

Solve for A.P. Terms: Finding $\frac{1}{q}$

Solve for A.P. Terms: Finding $\frac{1}{r}$

Find Common Difference $d$ and Remaining Terms

Calculate $a^{- 1}$

Calculate $b^{- 1}$

Final Calculation: $a^{- 1} - b^{- 1}$

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Quadratic Foundation

The Reciprocal Shortcut

The A.P

Final Calculation

Similar Questions

Let the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$ be $\frac{5}{16}, a > 2$ . If $α$ is such that $a, 4, α, b$ are in A.P., then the equation $α x^{2} - a x + 2 (α - 2 b) = 0$ has:

Let $a_{1}, a_{2}, \dots, a_{n}$ be in A.P. If $a_{5} = 2 a_{7}$ and $a_{11} = 18$ , then $12 (\frac{1}{a _{10} + a _{11}} + \frac{1}{a _{11} + a _{12}} + \dots + \frac{1}{a _{17} + a _{18}})$ is equal to

If $a_{1}, a_{2}, a_{3}, \dots$ and $b_{1}, b_{2}, b_{3}, \dots$ are A.P. and $a_{1} = 2, a_{10} = 3, a_{1} b_{1} = 1 = a_{10} b_{10}$ then $a_{4} b_{4}$ is equal to

The roots of the quadratic equation $3 x^{2} - p x + q = 0$ are $1 0^{t h}$ and $1 1^{t h}$ terms of an arithmetic progression with common difference $\frac{3}{2}$ . If the sum of the first 11 terms of this arithmetic progression is 88, then $q - 2 p$ is equal to

For three positive integers $p, q, r$ , $x^{p q^{2}} = y^{q r} = z^{p^{2} r}$ and $r = pq + 1$ such that $3, 3 lo g_{y} x, 3 lo g_{z} y, 7 lo g_{x} z$ are in A.P. with common difference $\frac{1}{2}$ . Then $r - p - q$ is equal to

Suppose $a, b, c$ are in A.P. and $a^{2}, b^{2}, c^{2}$ are in G.P. if $a < b < c$ and $a + b + c = 3/2$ , then the value of $a$ is

Suppose $a, b, c$ are in A.P. and $a^{2}, 2 b^{2}, c^{2}$ are in G.P. If $a < b < c$ and $a + b + c = 1$ , then $9 (a^{2} + b^{2} + c^{2})$ is equal to ......... .

Let $a_{1}, a_{2}, a_{3} \dots$ be terms on A.P. If $\frac{a _{1} + a _{2} + \dots + a _{p}}{a _{1} + a _{2} + \dots + a _{q}} = \frac{p ^{2}}{q ^{2}}, p \neq = q$ , then $\frac{a _{6}}{a _{21}}$ equals

Let $a_{1}, a_{2}, a_{3} \dots$ be in an A.P. such that $\sum_{k = 1}^{12} a_{2 k - 1} = - \frac{72}{5} a_{1}, a_{1} \neq = 0$ . If $\sum_{k = 1}^{n} a_{k} = 0$ , then $n$ is:

Enter Numerical Value:

Visualized Solution

Analyze the First Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2−8ax+2a=0

Calculate Sum of Reciprocals .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p1​+r1​

Analyze the Second Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2+12bx+6b=0

Calculate Sum of Reciprocals .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q1​+s1​

Define the Arithmetic Progression (A.P.)

Solve for A.P. Terms: Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q1​

Solve for A.P. Terms: Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r1​

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

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a−1

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b−1

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a−1−b−1

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Quadratic Foundation

The Reciprocal Shortcut

The A.P

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the arithmetic mean of a1​ and b1​ be 165​,a>2. If α is such that a,4,α,b are in A.P., then the equation αx2−ax+2(α−2b)=0 has:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​,a2​,…,an​ be in A.P. If a5​=2a7​ and a11​=18, then 12(a10​​+a11​​1​+a11​​+a12​​1​+⋯+a17​​+a18​​1​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a1​,a2​,a3​,… and b1​,b2​,b3​,… are A.P. and a1​=2,a10​=3,a1​b1​=1=a10​b10​ then a4​b4​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For three positive integers p,q,r, xpq2=yqr=zp2r and r=pq+1 such that 3,3logy​x,3logz​y,7logx​z are in A.P. with common difference 21​. Then r−p−q is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Suppose a,b,c are in A.P. and a2,b2,c2 are in G.P. if a<b<c and a+b+c=3/2, then the value of a is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Suppose a,b,c are in A.P. and a2,2b2,c2 are in G.P. If a<b<c and a+b+c=1, then 9(a2+b2+c2) is equal to ......... .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​,a2​,a3​… be terms on A.P. If a1​+a2​+⋯+aq​a1​+a2​+⋯+ap​​=q2p2​,p=q, then a21​a6​​ equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​,a2​,a3​… be in an A.P. such that ∑k=112​a2k−1​=−572​a1​,a1​=0. If ∑k=1n​ak​=0, then n is:

Analyze the First Equation $x^{2} - 8 a x + 2 a = 0$

Calculate Sum of Reciprocals $\frac{1}{p} + \frac{1}{r}$

Analyze the Second Equation $x^{2} + 12 b x + 6 b = 0$

Calculate Sum of Reciprocals $\frac{1}{q} + \frac{1}{s}$

Solve for A.P. Terms: Finding $\frac{1}{q}$

Solve for A.P. Terms: Finding $\frac{1}{r}$

Find Common Difference $d$ and Remaining Terms

Calculate $a^{- 1}$

Calculate $b^{- 1}$

Final Calculation: $a^{- 1} - b^{- 1}$

Let the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$ be $\frac{5}{16}, a > 2$ . If $α$ is such that $a, 4, α, b$ are in A.P., then the equation $α x^{2} - a x + 2 (α - 2 b) = 0$ has:

Let $a_{1}, a_{2}, \dots, a_{n}$ be in A.P. If $a_{5} = 2 a_{7}$ and $a_{11} = 18$ , then $12 (\frac{1}{a _{10} + a _{11}} + \frac{1}{a _{11} + a _{12}} + \dots + \frac{1}{a _{17} + a _{18}})$ is equal to

If $a_{1}, a_{2}, a_{3}, \dots$ and $b_{1}, b_{2}, b_{3}, \dots$ are A.P. and $a_{1} = 2, a_{10} = 3, a_{1} b_{1} = 1 = a_{10} b_{10}$ then $a_{4} b_{4}$ is equal to

For three positive integers $p, q, r$ , $x^{p q^{2}} = y^{q r} = z^{p^{2} r}$ and $r = pq + 1$ such that $3, 3 lo g_{y} x, 3 lo g_{z} y, 7 lo g_{x} z$ are in A.P. with common difference $\frac{1}{2}$ . Then $r - p - q$ is equal to

Suppose $a, b, c$ are in A.P. and $a^{2}, b^{2}, c^{2}$ are in G.P. if $a < b < c$ and $a + b + c = 3/2$ , then the value of $a$ is

Suppose $a, b, c$ are in A.P. and $a^{2}, 2 b^{2}, c^{2}$ are in G.P. If $a < b < c$ and $a + b + c = 1$ , then $9 (a^{2} + b^{2} + c^{2})$ is equal to ......... .

Let $a_{1}, a_{2}, a_{3} \dots$ be terms on A.P. If $\frac{a _{1} + a _{2} + \dots + a _{p}}{a _{1} + a _{2} + \dots + a _{q}} = \frac{p ^{2}}{q ^{2}}, p \neq = q$ , then $\frac{a _{6}}{a _{21}}$ equals

Let $a_{1}, a_{2}, a_{3} \dots$ be in an A.P. such that $\sum_{k = 1}^{12} a_{2 k - 1} = - \frac{72}{5} a_{1}, a_{1} \neq = 0$ . If $\sum_{k = 1}^{n} a_{k} = 0$ , then $n$ is: