JEE MainMathematics

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Sequences and Series: JEE Main Mathematics Question with Solution

Q: Why does the denominator become d when rationalizing?

Since is an arithmetic progression, for every consecutive pair of terms.

Let

a_1, a_2, \dots, a_n

be in A.P. If

a_5=2a_7

and

a_{11}=18

, then

12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\dots+\frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)

is equal to

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Step-by-step solutionView

Correct answer

The value of the expression is 8.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

$a_n$ is an A.P. with common difference $d$ . Given $a_5 = 2a_7$ and $a_{11} = 18$ .

02goal

Compute $S = 12\sum_{k=10}^{17} \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}}$ .

03approach

Solve for the first term $a_1$ and common difference $d$ using the linear equations from $a_5 = 2a_7$ and $a_{11} = 18$ . Then rationalize each term in the sum to create a telescoping series.

04execute

From $a_5 = 2a_7$ : $a_1 + 4d = 2(a_1 + 6d) \implies a_1 + 8d = 0$ . Given $a_{11} = a_1 + 10d = 18$ . Subtracting gives $(a_1 + 10d) - (a_1 + 8d) = 2d = 18 \implies d = 9$ . Then $a_1 = -8d = -72$ .

05execute

Evaluate key terms: $a_{10} = a_1 + 9d = -72 + 81 = 9$ , so $\sqrt{a_{10}} = 3$ . $a_{18} = a_1 + 17d = -72 + 153 = 81$ , so $\sqrt{a_{18}} = 9$ .

06execute

Rationalize each term: $\frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{a_{k+1} - a_k} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$ . The sum telescopes to: $\sum_{k=10}^{17} \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d} = \frac{\sqrt{a_{18}} - \sqrt{a_{10}}}{d}$ . Multiplying by 12: $12 \times \frac{9 - 3}{9} = 12 \times \frac{6}{9} = 8$ .

✓verify

Check: $a_9 = 0$ , terms for $k \ge 10$ are strictly positive ( $a_{10}=9, a_{11}=18, \dots$ ), so square roots are well-defined real numbers. Telescoping is exact.

Hints that build this answer step by step

Using the relation $a_5 = 2a_7$ , what is the relationship between the first term $a$ and the common difference $d$ ?
$a = -8d$
Given $a_{11} = 18$ and $a = -8d$ , what are the values of $d$ , $a_{10}$ , and $a_{18}$ ?
$d = 9$ , $a_{10} = 9$ , and $a_{18} = 81$
Rationalizing the terms yields a telescoping sum $\sum_{k=10}^{17} \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d} = \frac{\sqrt{a_{18}} - \sqrt{a_{10}}}{d}$ . What is the final evaluated result when multiplied by 12?
8

✓ 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 does the denominator become d when rationalizing?

Since $a_n$ is an arithmetic progression, $a_{k+1} - a_k = d$ for every consecutive pair of terms.

Answer

The value of the expression is 8.

Why each option works or fails

Step-by-step solution

given: $a_n$ is an A.P. with common difference $d$ . Given $a_5 = 2a_7$ and $a_{11} = 18$ .
goal: Compute $S = 12\sum_{k=10}^{17} \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}}$ .
approach: Solve for the first term $a_1$ and common difference $d$ using the linear equations from $a_5 = 2a_7$ and $a_{11} = 18$ . Then rationalize each term in the sum to create a telescoping series.
execute: From $a_5 = 2a_7$ : $a_1 + 4d = 2(a_1 + 6d) \implies a_1 + 8d = 0$ . Given $a_{11} = a_1 + 10d = 18$ . Subtracting gives $(a_1 + 10d) - (a_1 + 8d) = 2d = 18 \implies d = 9$ . Then $a_1 = -8d = -72$ .
execute: Evaluate key terms: $a_{10} = a_1 + 9d = -72 + 81 = 9$ , so $\sqrt{a_{10}} = 3$ . $a_{18} = a_1 + 17d = -72 + 153 = 81$ , so $\sqrt{a_{18}} = 9$ .
execute: Rationalize each term: $\frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{a_{k+1} - a_k} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$ . The sum telescopes to: $\sum_{k=10}^{17} \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d} = \frac{\sqrt{a_{18}} - \sqrt{a_{10}}}{d}$ . Multiplying by 12: $12 \times \frac{9 - 3}{9} = 12 \times \frac{6}{9} = 8$ .
verify: Check: $a_9 = 0$ , terms for $k \ge 10$ are strictly positive ( $a_{10}=9, a_{11}=18, \dots$ ), so square roots are well-defined real numbers. Telescoping is exact.

Sequences and Series: JEE Main Mathematics Question with Solution

What feels right?

The value of the expression is 8.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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