Let a₁, a₂, ..., a_n be in A.P. If a₅ = 2a₇ and a₁₁ = 18, then 12 (1/(√a₁₀ + √a₁₁) + 1/(√a₁₁ + √a₁₂)

Q: Wait, why do we rationalize the denominator? Can't we just calculate each term individually?

That is a great question! While you *could* calculate each term, it would be incredibly tedious and prone to arithmetic errors. In JEE Advanced problems, the goal is to find the underlying structure. Rationalizing transforms a complex sum into a 'telescoping sum,' where almost all terms cancel out, leaving you with a simple expression involving only the first and last terms. It is the elegant, professional way to solve it!

Q: How do I know if it's a telescoping sum just by looking at it?

Look for the pattern in the denominator! If you see terms like $\frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}}$, where the terms are consecutive members of a sequence, that is a huge red flag that rationalization will lead to a telescoping sum. Whenever you see a sum of fractions where the denominator is a sum of roots, your first instinct should always be to rationalize and look for that beautiful cancellation.

Q: What if the common difference $d$ was negative? Would the telescoping sum still work?

Absolutely! The math remains perfectly valid. If $d$ is negative, the terms $a_k$ would be decreasing. When you rationalize, the denominator $a_{k+1} - a_k$ would simply be a negative value ($-|d|$). The cancellation pattern would still hold, and you would just need to be careful with the signs when you perform the final subtraction. The beauty of algebra is that it handles these cases automatically if you follow the steps correctly.

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Foundation

Welcome, traveler of the mathematical landscape. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of square roots and indices.

We are dealing with an Arithmetic Progression (A.P.), which is a perfectly uniform staircase. Every step you take, you rise by the exact same height, which we call the common difference, $d$. The ground floor, where we start, is our first term, $a$.

To reach any step $n$, we use the general formula:
$$a_n = a + (n-1)d$$

### Translating the Clues

Our journey begins with two clues: $a_5 = 2a_7$ and $a_{11} = 18$. Let us translate these into the language of algebra.

Using our formula, $a_5 = a + 4d$ and $a_7 = a + 6d$. The condition $a_5 = 2a_7$ becomes:
$$a + 4d = 2(a + 6d)$$

Expanding this, we get $a + 4d = 2a + 12d$, which simplifies to $a + 8d = 0$. This is our first anchor point.

Our second clue, $a_{11} = 18$, gives us:
$$a + 10d = 18$$

Now, we have a system of two linear equations. Subtracting the first from the second, we find $2d = 18$, so $d = 9$. Substituting this back, we find $a = -72$. We have successfully mapped the entire staircase.

### The Art of Rationalization

Now, look at the expression we need to evaluate:
$$12 \left( \frac{1}{\sqrt{a_{10}} + \sqrt{a_{11}}} + \frac{1}{\sqrt{a_{11}} + \sqrt{a_{12}}} + \dots + \frac{1}{\sqrt{a_{17}} + \sqrt{a_{18}}} ight)$$

Whenever you see square roots added in a denominator, you are looking at a prime candidate for rationalization. Let us take a general term from this sum:
$$\frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}}$$

To simplify this, we multiply the numerator and denominator by the conjugate: $(\sqrt{a_{k+1}} - \sqrt{a_k})$. The denominator becomes $(\sqrt{a_{k+1}})^2 - (\sqrt{a_k})^2$, which is simply $a_{k+1} - a_k$.

Since the difference between two consecutive terms in an A.P. is exactly our common difference, $d$, our complex fraction collapses into:
$$\frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$$

### The Telescoping Collapse

This is where the magic happens. Let us rewrite our entire sum using this new, simplified form. We can pull the constant $\frac{12}{d}$ outside the summation.

The sum becomes:
$$\frac{12}{d} [(\sqrt{a_{11}} - \sqrt{a_{10}}) + (\sqrt{a_{12}} - \sqrt{a_{11}}) + \dots + (\sqrt{a_{18}} - \sqrt{a_{17}})]$$

This is a 'telescoping sum'—it collapses like a pirate's telescope. Every intermediate term vanishes, leaving only the very last positive term and the very first negative term:
$$\frac{12}{d} (\sqrt{a_{18}} - \sqrt{a_{10}})$$

### The Grand Finale

We just need the values of $a_{10}$ and $a_{18}$. Using $a = -72$ and $d = 9$:
$$a_{10} = a + 9d = -72 + 9(9) = 9$$
$$a_{18} = a + 17d = -72 + 17(9) = 81$$

Notice how these are perfect squares. Now, we substitute these into our collapsed expression:
$$	ext{Result} = \frac{12}{9} (\sqrt{81} - \sqrt{9}) = \frac{4}{3} (9 - 3) = \frac{4}{3} 	imes 6 = 8$$

The chaos has been tamed, the terms have cancelled, and we are left with the elegant answer of **8**. Remember, in JEE Advanced, the complexity is often just a mask; your job is to peel it away, find the pattern, and let the math do the work for you.

Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, \dots, a_{n}$ be in A.P. If $a_{5} = 2 a_{7}$ and $a_{11} = 18$ , then $12 (\frac{1}{a _{10} + a _{11}} + \frac{1}{a _{11} + a _{12}} + \dots + \frac{1}{a _{17} + a _{18}})$ is equal to

Enter Numerical Value:

Visualized Solution

Setting up the A.P. Parameters

Applying the First Condition: $a_{5} = 2 a_{7}$

Simplifying to Find Equation 1

Applying the Second Condition: $a_{11} = 18$

Solving for the Common Difference $d$

Solving for the First Term $a$

Rationalizing the Terms in the Sum

Expanding the Telescoping Sum

Calculating $a_{10}$ and $a_{18}$

Final Substitution and Answer

Conclusion and Key Takeaways

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Foundation

Translating the Clues

The Art of Rationalization

The Telescoping Collapse

The Grand Finale

Similar Questions

If $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ are in A.P. and $a_{1} + a_{4} + a_{7} + \dots + a_{16} = 114$ , then $a_{1} + a_{6} + a_{11} + a_{16}$ is equal to :

Let $a_{1}, a_{2}, a_{3}, \dots$ be an $A . P .$ If $\frac{a _{1} + a _{2} + \dots + a _{10}}{a _{1} + a _{2} + \dots + a _{p}} = \frac{100}{p ^{2}}, p \neq = 10$ , then $\frac{a _{11}}{a _{10}}$ is equal to:

Let $a_{1}, a_{2}, a_{3} \dots$ be terms on A.P. If $\frac{a _{1} + a _{2} + \dots + a _{p}}{a _{1} + a _{2} + \dots + a _{q}} = \frac{p ^{2}}{q ^{2}}, p \neq = q$ , then $\frac{a _{6}}{a _{21}}$ equals

Let $a_{1}, a_{2}, a_{3} \dots$ be in an A.P. such that $\sum_{k = 1}^{12} a_{2 k - 1} = - \frac{72}{5} a_{1}, a_{1} \neq = 0$ . If $\sum_{k = 1}^{n} a_{k} = 0$ , then $n$ is:

Let $T_{r}$ be the rth term of an A.P. If for some m, $T_{m} = \frac{1}{25}$ and $T_{25} = \frac{1}{20}$ , then $5 m \sum_{r = m}^{2 m} T_{r}$ is equal to Note: $20 \sum_{r = 1}^{25} T_{r} = 13$ is given condition in Hindi version.

If the sum of the first $2 n$ terms of the A.P. $2, 5, 8, \dots$ is equal to the sum of the first $n$ terms of the A.P. $57, 59, 61, \dots$ , then $n$ equals

Let $a_{1} = 8, a_{2}, a_{3}, \dots, a_{n}$ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170, then the product of its middle two terms is

Let $a_{1}, a_{2}, \dots, a_{21}$ be an $A P$ such that $\sum_{n = 1}^{20} \frac{1}{a _{n} a _{n + 1}} = \frac{4}{9}$ . If the sum of this $A P$ is $189$ , then $a_{6} a_{16}$ is equal to :

Let $a_{1}, a_{2}, a_{3}, \dots, a_{49}$ be in A.P. such that $\sum_{k = 0}^{12} a_{4 k + 1} = 416$ and $a_{9} + a_{43} = 66$ . If $a_{1}^{2} + a_{2}^{2} + \dots + a_{17}^{2} = 140 m$ , then $m$ is equal to :

Let $T_{r}$ be the $r$ th term of an A.P., for $r = 1, 2, 3, \dots$ . If for some positive integers $m, n$ we have $T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $T_{mn}$ equals

Enter Numerical Value:

Visualized Solution

Setting up the A.P. Parameters

Applying the First Condition: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a5​=2a7​

Simplifying to Find Equation 1

Applying the Second Condition: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a11​=18

Solving for the Common Difference .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Solving for the First Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a