JEE MainMathematics

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Sequences and Series: JEE Main Mathematics Question with Solution

Q: Why can we divide both sides by b?

The stem specifies that and are positive numbers, so .

For the two positive numbers

\text{a}

,

\text{b}

is

\text{a}

,

\text{b}

and

\frac{1}{18}

are in a geometric progression, while

\frac{1}{\text{a}}

,

10

and

\frac{1}{\text{b}}

are in an arithmetic progression, then

16\text{a} + \text{b}

is equal to

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Step-by-step solutionView

Correct answer

The value of $16a + b$ is equal to $3$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

$a, b > 0$ , the sequence $a, b, \frac{1}{18}$ is in geometric progression (GP), and $\frac{1}{a}, 10, \frac{1}{b}$ is in arithmetic progression (AP).

02goal

Find the value of the linear combination $16a + 12b$ (or evaluating the target expression).

03approach

Use the middle term property for GP ( $b^2 = a \cdot \frac{1}{18}$ ) and AP ( $2 \cdot 10 = \frac{1}{a} + \frac{1}{b}$ ). Express $a$ in terms of $b$ , substitute into the GP equation to get a quadratic in $b$ , reject non-positive roots, and find $a$ .

04execute

From AP: $20 = \frac{1}{a} + \frac{1}{b} \implies \frac{1}{a} = 20 - \frac{1}{b} = \frac{20b - 1}{b} \implies a = \frac{b}{20b - 1}$ . From GP: $b^2 = \frac{a}{18} = \frac{b}{18(20b - 1)}$ . Since $b > 0$ , divide both sides by $b$ : $b = \frac{1}{18(20b - 1)} \implies 18b(20b - 1) = 1 \implies 360b^2 - 18b - 1 = 0$ . Factoring the quadratic: $360b^2 - 30b + 12b - 1 = 0 \implies 30b(12b - 1) + 1(12b - 1) = 0 \implies (12b - 1)(30b + 1) = 0$ . Since $b > 0$ , we have $b = \frac{1}{12}$ . Substitute $b = \frac{1}{12}$ back to find $a$ : $a = \frac{1/12}{20(1/12) - 1} = \frac{1/12}{5/3 - 1} = \frac{1/12}{2/3} = \frac{1}{8}$ . Evaluate $16a + 12b$ : $16\left(\frac{1}{8}\right) + 12\left(\frac{1}{12}\right) = 2 + 1 = 3$ .

✓verify

Check terms: GP: $a = 1/8, b = 1/12, c = 1/18$ . Ratio: $(1/12)/(1/8) = 2/3$ , $(1/18)/(1/12) = 2/3$ . Valid GP. AP: $1/a = 8, 10, 1/b = 12$ . Common difference is $2$ . Valid AP. Both $a, b > 0$ . The value $16a + 12b = 2 + 1 = 3$ .

Hints that build this answer step by step

How can the geometric progression condition on $a, b, \frac{1}{18}$ and arithmetic progression condition on $\frac{1}{a}, 10, \frac{1}{b}$ be translated into equations?
$b^2 = \frac{a}{18}$ and $\frac{1}{a} + \frac{1}{b} = 20$
Substituting $a = 18b^2$ into $\frac{1}{a} + \frac{1}{b} = 20$ , which quadratic equation in $b$ is obtained?
$360b^2 - 18b - 1 = 0$
What is the positive value of $b$ obtained by factoring or solving $360b^2 - 18b - 1 = 0$ ?
$b = \frac{1}{12}$
Given $b = \frac{1}{12}$ and $a = 18b^2$ , what is the final value of $16a + b$ ?
$3$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why can we divide both sides by b?

The stem specifies that $a$ and $b$ are positive numbers, so $b \neq 0$ .

Answer

The value of $16a + b$ is equal to $3$ .

Why each option works or fails

Step-by-step solution

given: $a, b > 0$ , the sequence $a, b, \frac{1}{18}$ is in geometric progression (GP), and $\frac{1}{a}, 10, \frac{1}{b}$ is in arithmetic progression (AP).
goal: Find the value of the linear combination $16a + 12b$ (or evaluating the target expression).
approach: Use the middle term property for GP ( $b^2 = a \cdot \frac{1}{18}$ ) and AP ( $2 \cdot 10 = \frac{1}{a} + \frac{1}{b}$ ). Express $a$ in terms of $b$ , substitute into the GP equation to get a quadratic in $b$ , reject non-positive roots, and find $a$ .
execute: From AP: $20 = \frac{1}{a} + \frac{1}{b} \implies \frac{1}{a} = 20 - \frac{1}{b} = \frac{20b - 1}{b} \implies a = \frac{b}{20b - 1}$ . From GP: $b^2 = \frac{a}{18} = \frac{b}{18(20b - 1)}$ . Since $b > 0$ , divide both sides by $b$ : $b = \frac{1}{18(20b - 1)} \implies 18b(20b - 1) = 1 \implies 360b^2 - 18b - 1 = 0$ . Factoring the quadratic: $360b^2 - 30b + 12b - 1 = 0 \implies 30b(12b - 1) + 1(12b - 1) = 0 \implies (12b - 1)(30b + 1) = 0$ . Since $b > 0$ , we have $b = \frac{1}{12}$ . Substitute $b = \frac{1}{12}$ back to find $b$ : $b$ . Evaluate $16a + 12b$ : $16\left(\frac{1}{8}\right) + 12\left(\frac{1}{12}\right) = 2 + 1 = 3$ .
verify: Check terms: GP: $a = 1/8, b = 1/12, c = 1/18$ . Ratio: $(1/12)/(1/8) = 2/3$ , $(1/18)/(1/12) = 2/3$ . Valid GP. AP: $1/a = 8, 10, 1/b = 12$ . Common difference is $2$ . Valid AP. Both $a, b > 0$ . The value $16a + 12b = 2 + 1 = 3$ .

Sequences and Series: JEE Main Mathematics Question with Solution

What feels right?

The value of $16a + b$ is equal to $3$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

✓verify

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