JEE MainMathematics

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Sequences and Series: JEE Main Mathematics Question with Solution

Q: Do we need to use the full sum formula ?

No, because is by definition just the single term , which simplifies directly without dealing with fractions in the sum formula.

The

4^{\text{th}}

term of GP is

500

and its common ratio is

\frac{1}{m}, m \in \mathbb{N}

. Let

\mathrm{S}_n

denote the sum of the first

n

terms of this GP. If

\mathrm{S}_6 > \mathrm{S}_5 + 1

and

\mathrm{S}_7 < \mathrm{S}_6 + \frac{1}{2}

, then the number of possible values of

m

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Step-by-step solutionView

Correct answer

There are 12 possible natural number values for m, specifically m ∈ {11, 12, ..., 22}.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

A GP has 4th term $T_4 = 500$ , common ratio $r = \frac{1}{m}$ with $m \in \mathbb{N}$ . $S_n$ is the sum of the first $n$ terms. Inequalities: $S_6 > S_5 + 1$ and $S_7 < S_6 + \frac{1}{2}$ .

02goal

Find the number of possible natural values of $m$ .

03approach

Recall that $S_n - S_{n-1} = T_n = a r^{n-1}$ . Express the inequalities directly in terms of the individual terms $T_6$ and $T_7$ , then relate them to $T_4 = 500$ and $r = \frac{1}{m}$ .

04execute

From the first condition, $S_6 - S_5 > 1 \implies T_6 > 1$ . Since $T_6 = T_4 \cdot r^2 = 500 \left(\frac{1}{m}\right)^2 = \frac{500}{m^2}$ , we have: $\frac{500}{m^2} > 1 \implies m^2 < 500$ . Since $22^2 = 484 < 500$ and $23^2 = 529 > 500$ , for natural number $m$ , we have $m \le 22$ .

05execute

From the second condition, $S_7 - S_6 < \frac{1}{2} \implies T_7 < \frac{1}{2}$ . Since $T_7 = T_4 \cdot r^3 = 500 \left(\frac{1}{m}\right)^3 = \frac{500}{m^3}$ , we have: $\frac{500}{m^3} < \frac{1}{2} \implies m^3 > 1000$ . Since $10^3 = 1000$ , we have $m > 10$ .

06execute

Combining both bounds for $m \in \mathbb{N}$ : $10 < m \le 22$ . The integers in this range are $m \in \{11, 12, \ldots, 22\}$ . The count of such integers is $22 - 11 + 1 = 12$ .

✓verify

Check endpoints: For $m=10$ , $T_7 = 500/1000 = 0.5$ , but the condition requires $T_7 < 0.5$ (strict), so $m=10$ is excluded. For $m=11$ , $T_7 = 500/1331 < 0.5$ and $T_6 = 500/121 > 1$ . For $m=22$ , $T_6 = 500/484 > 1$ . For $m=23$ , $T_6 = 500/529 < 1$ , excluded. Thus $m \in [11, 22]$ gives exactly 12 values.

Hints that build this answer step by step

How can the given conditions $\mathrm{S}_6 > \mathrm{S}_5 + 1$ and $\mathrm{S}_7 < \mathrm{S}_6 + \frac{1}{2}$ be expressed directly in terms of individual terms of the GP?
$T_6 > 1$ and $T_7 < \frac{1}{2}$
Given $T_4 = 500$ and common ratio $r = \frac{1}{m}$ , what expressions represent $T_6$ and $T_7$ in terms of $m$ ?
$T_6 = \frac{500}{m^2}$ and $T_7 = \frac{500}{m^3}$
Applying the inequalities $\frac{500}{m^2} > 1$ and $\frac{500}{m^3} < \frac{1}{2}$ , what range of values for $m \in \mathbb{N}$ is obtained?
$10 < m < \sqrt{500}$ , which gives $11 \le m \le 22$
How many integers satisfy $11 \le m \le 22$ ?
12

✓ 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

Do we need to use the full sum formula

S_n = a(1-r^n)/(1-r)

No, because $S_n - S_{n-1}$ is by definition just the single term $T_n$ , which simplifies directly without dealing with fractions in the sum formula.

Answer

There are 12 possible natural number values for m, specifically m ∈ {11, 12, ..., 22}.

Why each option works or fails

Step-by-step solution

given: A GP has 4th term $T_4 = 500$ , common ratio $r = \frac{1}{m}$ with $m \in \mathbb{N}$ . $S_n$ is the sum of the first $n$ terms. Inequalities: $S_6 > S_5 + 1$ and $S_7 < S_6 + \frac{1}{2}$ .
goal: Find the number of possible natural values of $m$ .
approach: Recall that $S_n - S_{n-1} = T_n = a r^{n-1}$ . Express the inequalities directly in terms of the individual terms $T_6$ and $T_7$ , then relate them to $T_4 = 500$ and $r = \frac{1}{m}$ .
execute: From the first condition, $S_6 - S_5 > 1 \implies T_6 > 1$ . Since $T_6 = T_4 \cdot r^2 = 500 \left(\frac{1}{m}\right)^2 = \frac{500}{m^2}$ , we have: $\frac{500}{m^2} > 1 \implies m^2 < 500$ . Since $22^2 = 484 < 500$ and $23^2 = 529 > 500$ , for natural number $m$ , we have $m \le 22$ .
execute: From the second condition, $S_7 - S_6 < \frac{1}{2} \implies T_7 < \frac{1}{2}$ . Since $T_7 = T_4 \cdot r^3 = 500 \left(\frac{1}{m}\right)^3 = \frac{500}{m^3}$ , we have: $\frac{500}{m^3} < \frac{1}{2} \implies m^3 > 1000$ . Since $10^3 = 1000$ , we have $m > 10$ .
execute: Combining both bounds for $m \in \mathbb{N}$ : $10 < m \le 22$ . The integers in this range are $m \in \{11, 12, \ldots, 22\}$ . The count of such integers is $22 - 11 + 1 = 12$ .
verify: Check endpoints: For $m=10$ , $T_7 = 500/1000 = 0.5$ , but the condition requires $T_7 < 0.5$ (strict), so $m=10$ is excluded. For $m=11$ , $T_7 = 500/1331 < 0.5$ and $T_6 = 500/121 > 1$ . For $m=22$ , $T_6 = 500/484 > 1$ . For $m=23$ , $T_6 = 500/529 < 1$ , excluded. Thus $m \in [11, 22]$ gives exactly 12 values.

Sequences and Series: JEE Main Mathematics Question with Solution

What feels right?

There are 12 possible natural number values for m, specifically m ∈ {11, 12, ..., 22}.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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