JEE Main 2021 (01 September Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: 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 :

Q: Why can't we just solve for $a_1$ and $d$ using the two equations?

You absolutely can! You have a system of two equations: $a_1 + a_{21} = 18$ and $a_1 a_{21} = 45$. You could solve for $a_1$ and $a_{21}$, then find $d$. However, using the middle term $a_{11}$ as an anchor is much faster and less prone to calculation errors. It exploits the inherent symmetry of the AP, which is a powerful tool in JEE Advanced problems.

Q: How do I know $a_{11}$ is the middle term?

In an AP with $N$ terms, if $N$ is odd, the middle term is the $\frac{N+1}{2}$-th term. Here, $N=21$, so the middle term is $\frac{21+1}{2} = 11$. The property $a_{11} = \frac{a_1 + a_{21}}{2}$ holds because the AP is symmetric around its center. It is a beautiful property that saves you precious time in the exam hall!

Select Answer:

Visualized Solution

Analyzing the Sequence

Given AP: $a_{1}, a_{2}, \dots, a_{21}$ with common difference $d$ .
We have $21$ terms in total.
Let's visualize the key terms on a number line.

The Telescoping Trick

Consider the general term of the sum: $\frac{1}{a _{n} a _{n + 1}}$ .
Multiply and divide by $d$ : $\frac{1}{d} (\frac{d}{a _{n} a _{n + 1}})$ .
Recall that $a_{n + 1} - a_{n} = d$ .

Rewriting the General Term

Substitute $d = a_{n + 1} - a_{n}$ in the numerator.
$\frac{1}{d} (\frac{a _{n + 1} - a _{n}}{a _{n} a _{n + 1}})$
Split the fraction: $\frac{1}{d} (\frac{1}{a _{n}} - \frac{1}{a _{n + 1}})$

Expanding the Summation

Summation: $\sum_{n = 1}^{20} \frac{1}{d} (\frac{1}{a _{n}} - \frac{1}{a _{n + 1}})$
Expansion: $\frac{1}{d} [(\frac{1}{a _{1}} - \frac{1}{a _{2}}) + (\frac{1}{a _{2}} - \frac{1}{a _{3}}) + \dots + (\frac{1}{a _{20}} - \frac{1}{a _{21}})]$

Simplifying the Result

After cancellation: $\frac{1}{d} (\frac{1}{a _{1}} - \frac{1}{a _{21}}) = \frac{4}{9}$
Take LCM: $\frac{1}{d} (\frac{a _{21} - a _{1}}{a _{1} a _{21}}) = \frac{4}{9}$
Recall that $a_{21} - a_{1} = 20 d$ .

Finding the Product $a_{1} a_{21}$

Substitute $20 d$ : $\frac{1}{d} (\frac{20 d}{a _{1} a _{21}}) = \frac{4}{9}$
The $d$ cancels out: $\frac{20}{a _{1} a _{21}} = \frac{4}{9}$
Cross-multiply: $a_{1} a_{21} = \frac{20 \times 9}{4} = 45$

Using the Sum of AP

We are given the sum of the AP: $S_{21} = 189$ .
Formula for sum of $n$ terms: $S_{n} = \frac{n}{2} (a_{1} + a_{n})$
Substitute $n = 21$ : $\frac{21}{2} (a_{1} + a_{21}) = 189$

Finding the Sum $a_{1} + a_{21}$

Solve for $(a_{1} + a_{21})$ :
$a_{1} + a_{21} = \frac{189 \times 2}{21}$
$a_{1} + a_{21} = 9 \times 2 = 18$

The Middle Term Strategy

In an AP with an odd number of terms, the middle term is the average of the first and last terms.
Middle term index: $\frac{1 + 21}{2} = 11$ .
So, $a_{11} = \frac{a _{1} + a _{21}}{2} = \frac{18}{2} = 9$ .

Expressing Terms via $a_{11}$

We can express $a_{1}$ and $a_{21}$ symmetrically around $a_{11}$ .
Distance from $a_{11}$ to $a_{1}$ is $10$ steps backwards: $a_{1} = a_{11} - 10 d$ .
Distance from $a_{11}$ to $a_{21}$ is $10$ steps forwards: $a_{21} = a_{11} + 10 d$ .

Solving for $d^{2}$

We know $a_{1} a_{21} = 45$ .
Substitute the symmetric expressions: $(a_{11} - 10 d) (a_{11} + 10 d) = 45$ .
Use $(x - y) (x + y) = x^{2} - y^{2}$ : $a_{11}^{2} - 100 d^{2} = 45$ .
Substitute $a_{11} = 9$ : $81 - 100 d^{2} = 45 ⟹ 100 d^{2} = 36 ⟹ d^{2} = \frac{36}{100}$ .

Targeting $a_{6} a_{16}$

The question asks for the product $a_{6} a_{16}$ .
Let's express $a_{6}$ and $a_{16}$ in terms of our anchor $a_{11}$ .
$a_{6}$ is $5$ steps behind $a_{11}$ : $a_{6} = a_{11} - 5 d$ .
$a_{16}$ is $5$ steps ahead of $a_{11}$ : $a_{16} = a_{11} + 5 d$ .

Final Calculation

Product: $a_{6} a_{16} = (a_{11} - 5 d) (a_{11} + 5 d)$
Expand using difference of squares: $a_{6} a_{16} = a_{11}^{2} - 25 d^{2}$
Substitute $a_{11} = 9$ and $d^{2} = \frac{36}{100}$ :
$a_{6} a_{16} = 9^{2} - 25 (\frac{36}{100})$

Concluding the Answer

Simplify the expression: $a_{6} a_{16} = 81 - 25 \times \frac{36}{100}$
Cancel $25$ with $100$ : $a_{6} a_{16} = 81 - \frac{36}{4}$
$a_{6} a_{16} = 81 - 9 = 72$
Final Answer: 72

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

Solution Diagram

Analyzing the Setup

The given summation is:

n = 1 \sum 20 \frac{1}{a _{n} a _{n + 1}} = \frac{4}{9}

This is a classic telescoping series. For an Arithmetic Progression (AP), the difference between consecutive terms is constant, defined as

a_{n + 1} - a_{n} = d

By multiplying and dividing the general term by

d

, we can rewrite the expression as:

\frac{1}{d} (\frac{a _{n + 1} - a _{n}}{a _{n} a _{n + 1}}) = \frac{1}{d} (\frac{1}{a _{n}} - \frac{1}{a _{n + 1}})

Expanding this sum causes all intermediate terms to cancel out, leaving only the first and last terms:

\frac{1}{d} (\frac{1}{a _{1}} - \frac{1}{a _{21}}) = \frac{4}{9}

The Anchor of Symmetry

We know that

a_{21} = a_{1} + 20 d

, which implies

a_{21} - a_{1} = 20 d

. Substituting this into our equation yields:

\frac{1}{d} (\frac{a _{21} - a _{1}}{a _{1} a _{21}}) = \frac{1}{d} (\frac{20 d}{a _{1} a _{21}}) = \frac{20}{a _{1} a _{21}} = \frac{4}{9}

Cross-multiplying gives us the product of the boundary terms:

a_{1} a_{21} = 45

Given the sum of the

21

terms is

189

, we use the formula

S_{n} = \frac{n}{2} (a_{1} + a_{n})

\frac{21}{2} (a_{1} + a_{21}) = 189 \Rightarrow a_{1} + a_{21} = 18

In an AP with

21

terms, the middle term

a_{11}

is the arithmetic mean of the first and last terms:

a_{11} = \frac{a _{1} + a _{21}}{2} = \frac{18}{2} = 9

The Final Leap

We express

a_{1}

and

a_{21}

in terms of the anchor

a_{11}

and the common difference

d

a_{1} = a_{11} - 10 d, a_{21} = a_{11} + 10 d

Substituting these into the product

a_{1} a_{21} = 45

(a_{11} - 10 d) (a_{11} + 10 d) = 45 \Rightarrow a_{11}^{2} - 100 d^{2} = 45

Substituting

a_{11} = 9

81 - 100 d^{2} = 45 \Rightarrow 100 d^{2} = 36 \Rightarrow d^{2} = 0.36

To find

a_{6} a_{16}

, we express these terms relative to

a_{11}

a_{6} = a_{11} - 5 d, a_{16} = a_{11} + 5 d

The product is:

a_{6} a_{16} = (a_{11} - 5 d) (a_{11} + 5 d) = a_{11}^{2} - 25 d^{2}

Substituting the known values:

81 - 25 (0.36) = 81 - 9 = 72

The final result is 72.

Common Traps & FAQs

Wait, why do we multiply and divide by

d

in the summation?

Great question! The goal of a telescoping sum is to create a form like

(f (n) - f (n + 1))

. In our case, we have

\frac{1}{a _{n} a _{n + 1}}

. If we write the numerator as

d = a_{n + 1} - a_{n}

, the expression becomes

\frac{1}{d} \frac{a _{n + 1} - a _{n}}{a _{n} a _{n + 1}}

, which simplifies to

\frac{1}{d} (\frac{1}{a _{n}} - \frac{1}{a _{n + 1}})

. Without that

d

, we cannot create the cancellation chain!

Why can't we just solve for

a_{1}

and

d

using the two equations?

You absolutely can! You have a system of two equations:

a_{1} + a_{21} = 18

and

a_{1} a_{21} = 45

. You could solve for

a_{1}

and

a_{21}

, then find

d

. However, using the middle term

a_{11}

as an anchor is much faster and less prone to calculation errors. It exploits the inherent symmetry of the AP, which is a powerful tool in JEE Advanced problems.

How do I know

a_{11}

is the middle term?

In an AP with

N

terms, if

N

is odd, the middle term is the

\frac{N + 1}{2}

-th term. Here,

N = 21

, so the middle term is

\frac{21 + 1}{2} = 11

. The property

a_{11} = \frac{a _{1} + a _{21}}{2}

holds because the AP is symmetric around its center. It is a beautiful property that saves you precious time in the exam hall!

Animated Solution for Mathematics - Sequence and Series: 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 :

Select Answer:

Visualized Solution

Analyzing the Sequence

The Telescoping Trick

Rewriting the General Term

Expanding the Summation

Simplifying the Result

Finding the Product $a_{1} a_{21}$

Using the Sum of AP

Finding the Sum $a_{1} + a_{21}$

The Middle Term Strategy

Expressing Terms via $a_{11}$

Solving for $d^{2}$

Targeting $a_{6} a_{16}$

Final Calculation

Concluding the Answer

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

Analyzing the Setup

The Anchor of Symmetry

The Final Leap

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 $a_{1}, a_{2}, \dots, a_{n}$ 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

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 $a_{1}, a_{2}, a_{3} \dots$ 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 $a_{n}$ be the $n^{t h}$ term of an A. P. If $S_{n} = a_{1} + a_{2} + a_{3} + \dots + a_{n} = 700, a_{6} = 7$ and $S_{7} = 7$ , then $a_{n}$ 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:

Consider an A.P.: $a_{1}, a_{2}, \dots, a_{n}; a_{1} > 0$ . If $a_{2} - a_{1} = - 3/4, a_{n} = \frac{1}{4} a_{1}$ , and $\sum_{i = 1}^{n} a_{i} = 525/2$ , then $\sum_{i = 1}^{17} a_{i}$ is equal to

Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its $1 1^{t h}$ term is :

Let $a_{1}, a_{2}, ........., a_{30}$ be an A. P., $S = \sum_{i = 1}^{30} a_{i}$ and $T = \sum_{i = 1}^{15} a_{(2 i - 1)}$ . If $a_{5} = 27$ and $S - 2 T = 75$ , then $a_{10}$ is equal to :

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:

Select Answer:

Visualized Solution

Analyzing the Sequence

The Telescoping Trick

Rewriting the General Term

Expanding the Summation

Simplifying the Result

Finding the Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a1​a21​

Using the Sum of AP

Finding the Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a1​+a21​

The Middle Term Strategy

Expressing Terms via .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a11​

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d2

Targeting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a6​a16​

Final Calculation

Concluding the Answer

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

Analyzing the Setup

The Anchor of Symmetry

The Final Leap

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a1​,a2​,a3​,…,an​ are in A.P. and a1​+a4​+a7​+⋯+a16​=114, then a1​+a6​+a11​+a16​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​,a2​,…,an​ be in A.P. If a5​=2a7​ and a11​=18, then 12(a10​​+a11​​1​+a11​​+a12​​1​+⋯+a17​​+a18​​1​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​,a2​,a3​,…,a49​ be in A.P. such that ∑k=012​a4k+1​=416 and a9​+a43​=66. If a12​+a22​+⋯+a172​=140m, then m is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​,a2​,a3​… be terms on A.P. If a1​+a2​+⋯+aq​a1​+a2​+⋯+ap​​=q2p2​,p=q, then a21​a6​​ equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let an​ be the nth term of an A. P. If Sn​=a1​+a2​+a3​+⋯+an​=700,a6​=7 and S7​=7, then an​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​,a2​,a3​,… be an A.P. If a1​+a2​+⋯+ap​a1​+a2​+⋯+a10​​=p2100​,p=10, then a10​a11​​ is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider an A.P.: a1​,a2​,…,an​;a1​>0. If a2​−a1​=−3/4,an​=41​a1​, and ∑i=1n​ai​=525/2, then ∑i=117​ai​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11th term is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​,a2​,.........,a30​ be an A. P., S=∑i=130​ai​ and T=∑i=115​a(2i−1)​. If a5​=27 and S−2T=75, then a10​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​,a2​,a3​… be in an A.P. such that ∑k=112​a2k−1​=−572​a1​,a1​=0. If ∑k=1n​ak​=0, then n is:

Finding the Product $a_{1} a_{21}$

Finding the Sum $a_{1} + a_{21}$

Expressing Terms via $a_{11}$

Solving for $d^{2}$

Targeting $a_{6} a_{16}$

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 $a_{1}, a_{2}, \dots, a_{n}$ 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

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 $a_{1}, a_{2}, a_{3} \dots$ 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 $a_{n}$ be the $n^{t h}$ term of an A. P. If $S_{n} = a_{1} + a_{2} + a_{3} + \dots + a_{n} = 700, a_{6} = 7$ and $S_{7} = 7$ , then $a_{n}$ 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:

Consider an A.P.: $a_{1}, a_{2}, \dots, a_{n}; a_{1} > 0$ . If $a_{2} - a_{1} = - 3/4, a_{n} = \frac{1}{4} a_{1}$ , and $\sum_{i = 1}^{n} a_{i} = 525/2$ , then $\sum_{i = 1}^{17} a_{i}$ is equal to

Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its $1 1^{t h}$ term is :

Let $a_{1}, a_{2}, ........., a_{30}$ be an A. P., $S = \sum_{i = 1}^{30} a_{i}$ and $T = \sum_{i = 1}^{15} a_{(2 i - 1)}$ . If $a_{5} = 27$ and $S - 2 T = 75$ , then $a_{10}$ is equal to :

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: