JEE MainMathematics

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Sum of Telescoping Radical Fractions in an Arithmetic Progression

Q: 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.

Let a_1, a_2, …, a_n be in A.P. If a_5=2a_7 and a_11=18, then 12(1/(√(a_10)+√(a_11))+1/(√(a_11)+√(a_12))+…+1/(√(a_17)+√(a_18))) 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 $a_{5} = 2 a_{7}$ 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 $a_{5} = 2 a_{7}$ and $a_{11} = 18$ . Then rationalize each term in the sum to create a telescoping series.

04Execute

From $a_{5} = 2 a_{7}$ : $a_{1} + 4 d = 2 (a_{1} + 6 d) ⟹ a_{1} + 8 d = 0$ . Given $a_{11} = a_{1} + 10 d = 18$ . Subtracting gives $(a_{1} + 10 d) - (a_{1} + 8 d) = 2 d = 18 ⟹ d = 9$ . Then $a_{1} = - 8 d = - 72$ .

05Execute

Evaluate key terms: $a_{10} = a_{1} + 9 d = - 72 + 81 = 9$ , so $\sqrt{a_{10}} = 3$ . $a_{18} = a_{1} + 17 d = - 72 + 153 = 81$ , 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 $k \geq 10$ 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 $a_{5} = 2 a_{7}$ , what is the relationship between the first term $a$ and the common difference $d$ ?
$a = - 8 d$
Given $a_{11} = 18$ and $a = - 8 d$ , 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
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $a_{5} = 2 a_{7}$ 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 $a_{5} = 2 a_{7}$ and $a_{11} = 18$ . Then rationalize each term in the sum to create a telescoping series.
Execute: From $a_{5} = 2 a_{7}$ : $a_{1} + 4 d = 2 (a_{1} + 6 d) ⟹ a_{1} + 8 d = 0$ . Given $a_{11} = a_{1} + 10 d = 18$ . Subtracting gives $(a_{1} + 10 d) - (a_{1} + 8 d) = 2 d = 18 ⟹ d = 9$ . Then $a_{1} = - 8 d = - 72$ .
Execute: Evaluate key terms: $a_{10} = a_{1} + 9 d = - 72 + 81 = 9$ , so $\sqrt{a_{10}} = 3$ . $a_{18} = a_{1} + 17 d = - 72 + 153 = 81$ , 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 $k \geq 10$ are strictly positive ( $a_{10} = 9, a_{11} = 18, \dots$ ), so square roots are well-defined real numbers. Telescoping is exact.

Sum of Telescoping Radical Fractions in an Arithmetic Progression

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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