JEE MainMathematics

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+4 marks−1 if incorrectNumericalPrevious-year question

Constraining the Common Ratio of a Geometric Progression Using Partial Sum Inequalities

Q: 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.

The 4^th term of GP is 500 and its common ratio is 1/m, m ∈ ℕ. Let S_n denote the sum of the first n terms of this GP. If S_6 > S_5 + 1 and S_7 < S_6 + 1/2, then the number of possible values of m is

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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 $m \in N$ . $S_{n}$ 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, $S_{6} - S_{5} > 1 ⟹ T_{6} > 1$ . Since $T_{6} = T_{4} \cdot r^{2} = 500 {(\frac{1}{m})}^{2} = \frac{500}{m^{2}}$ , we have: $\frac{500}{m^{2}} > 1 ⟹ m^{2} < 500$ . Since $22^{2} = 484 < 500$ and $23^{2} = 529 > 500$ , for natural number $m$ , we have $m \leq 22$ .

05Execute

From the second condition, $S_{7} - S_{6} < \frac{1}{2} ⟹ T_{7} < \frac{1}{2}$ . Since $T_{7} = T_{4} \cdot r^{3} = 500 {(\frac{1}{m})}^{3} = \frac{500}{m^{3}}$ , we have: $\frac{500}{m^{3}} < \frac{1}{2} ⟹ m^{3} > 1000$ . Since $10^{3} = 1000$ , we have $m > 10$ .

06Execute

Combining both bounds for $m \in N$ : $10 < m \leq 22$ . The integers in this range are $m \in {11, 12, \dots, 22}$ . 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 $S_{6} > S_{5} + 1$ and $S_{7} < S_{6} + \frac{1}{2}$ 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 $m \in N$ is obtained?
$10 < m < \sqrt{500}$ , which gives $11 \leq m \leq 22$
How many integers satisfy $11 \leq m \leq 22$ ?
12

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $m \in N$ . $S_{n}$ 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, $S_{6} - S_{5} > 1 ⟹ T_{6} > 1$ . Since $T_{6} = T_{4} \cdot r^{2} = 500 {(\frac{1}{m})}^{2} = \frac{500}{m^{2}}$ , we have: $\frac{500}{m^{2}} > 1 ⟹ m^{2} < 500$ . Since $22^{2} = 484 < 500$ and $23^{2} = 529 > 500$ , for natural number $m$ , we have $m \leq 22$ .
Execute: From the second condition, $S_{7} - S_{6} < \frac{1}{2} ⟹ T_{7} < \frac{1}{2}$ . Since $S_{n} - S_{n - 1} = T_{n} = a r^{n - 1}$ , we have: $\frac{500}{m^{3}} < \frac{1}{2} ⟹ m^{3} > 1000$ . Since $10^{3} = 1000$ , we have $m > 10$ .
Execute: Combining both bounds for $m \in N$ : $10 < m \leq 22$ . The integers in this range are $m \in {11, 12, \dots, 22}$ . 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.

Constraining the Common Ratio of a Geometric Progression Using Partial Sum Inequalities

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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