JEE Main 2018 (Paper 1)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: 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 :

Q: Wait, why does the summation $\sum_{k=0}^{12} a_{4k+1}$ have 13 terms and not 12?

This is a classic trap! When we sum from $k=0$ to $k=12$, we are counting the integers $0, 1, 2, \dots, 12$. If you count them, there are exactly $12 - 0 + 1 = 13$ values. Always remember: the number of terms in a summation $\sum_{k=m}^{n}$ is $n - m + 1$.

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Look closely at the summation

\sum_{k = 0}^{12} a_{4 k + 1}

. This is not a random collection of terms; it is a sub-sequence of the Arithmetic Progression. Write out the first few terms: for

k = 0

, you get

a_{1}

; for

k = 1

, you get

a_{5}

; for

k = 2

, you get

a_{9}

. Notice the indices are

1, 5, 9, \dots, 49

. How many terms are in this sequence from

k = 0

12

Hint 2

Hint 2 (The Arsenal): You have two unknowns: the first term

a_{1}

and the common difference

d

. Use the general term formula

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

to convert both given conditions into a system of two linear equations. The first condition

\sum_{k = 0}^{12} a_{4 k + 1} = 416

will simplify to a linear equation in

a_{1}

and

d

after you evaluate the sum of the arithmetic series.

Hint 3

Hint 3 (The Execution): Once you have

a_{1}

and

d

, express the general term

a_{r}

as a function of

r

. The sum of squares

\sum_{r = 1}^{17} a_{r}^{2}

then becomes a sum of the form

\sum_{r = 1}^{17} (r + c)^{2}

. Expand this using the identity

(r + c)^{2} = r^{2} + 2 r c + c^{2}

and apply the standard summation formulas for

\sum r^{2}

\sum r

, and

\sum 1

Visualized Solution

Problem Setup

Sequence: $a_{1}, a_{2}, a_{3}, \dots, a_{49}$ is an A.P.
First term = $a_{1}$ , Common difference = $d$
Condition 1: $\sum_{k = 0}^{12} a_{4 k + 1} = 416$
Condition 2: $a_{9} + a_{43} = 66$

Expanding Condition 1

$\sum_{k = 0}^{12} a_{4 k + 1} = a_{1} + a_{5} + a_{9} + \dots + a_{49} = 416$
General term: $a_{n} = a_{1} + (n - 1) d$
$a_{4 k + 1} = a_{1} + 4 k d$

Simplifying Condition 1

$\sum_{k = 0}^{12} (a_{1} + 4 k d) = 416$
$13 a_{1} + 4 d \sum_{k = 0}^{12} k = 416$
$13 a_{1} + 4 d (\frac{12 \times 13}{2}) = 416$
$13 a_{1} + 312 d = 416 ⟹ a_{1} + 24 d = 32$

Analyzing Condition 2

$a_{9} + a_{43} = 66$
$(a_{1} + 8 d) + (a_{1} + 42 d) = 66$
$2 a_{1} + 50 d = 66 ⟹ a_{1} + 25 d = 33$

Solving for $a_{1}$ and $d$

Eq 2: $a_{1} + 25 d = 33$
Eq 1: $a_{1} + 24 d = 32$
Subtracting: $d = 1$
Substituting $d = 1$ : $a_{1} = 8$

Setting up the Sum of Squares

Required: $S = \sum_{r = 1}^{17} a_{r}^{2} = 140 m$
$a_{r} = a_{1} + (r - 1) d = 8 + (r - 1) (1) = r + 7$
$S = \sum_{r = 1}^{17} (r + 7)^{2}$

Expanding the Square

$S = \sum_{r = 1}^{17} (r^{2} + 14 r + 49)$
$S = \sum_{r = 1}^{17} r^{2} + 14 \sum_{r = 1}^{17} r + \sum_{r = 1}^{17} 49$

Applying Summation Formulas

$\sum r^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$
$\sum r = \frac{n ( n + 1 )}{2}$
For $n = 17$ :
$\sum r^{2} = \frac{17 \times 18 \times 35}{6} = 1785$

Calculating Remaining Terms

$14 \sum r = 14 \times \frac{17 \times 18}{2} = 2142$
$\sum 49 = 17 \times 49 = 833$
Total Sum $S = 1785 + 2142 + 833 = 4760$

Finding $m$

$140 m = 4760$
$m = \frac{4760}{140} = 34$
Final Answer: $m = 34$

00:00 / 00:00

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

Analyzing the Setup

We are given an Arithmetic Progression (AP) with terms

a_{1}, a_{2}, a_{3}, \dots, a_{49}

and a common difference

d

. The general term is defined as

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

Our goal is to determine the value of

m

given the sum of the squares of the first seventeen terms, expressed as

\sum_{r = 1}^{17} a_{r}^{2} = 140 m

Decoding the Hidden Summation

The first condition is

\sum_{k = 0}^{12} a_{4 k + 1} = 416

. Expanding this summation, we have terms

a_{1}, a_{5}, a_{9}, \dots, a_{49}

Using the general term formula

a_{4 k + 1} = a_{1} + 4 k d

, the summation becomes:

k = 0 \sum 12 (a_{1} + 4 k d) = 13 a_{1} + 4 d k = 0 \sum 12 k = 416

Since

\sum_{k = 0}^{12} k = \frac{12 \times 13}{2} = 78

, the equation simplifies to:

13 a_{1} + 4 d (78) = 13 a_{1} + 312 d = 416

Dividing by

13

, we obtain Equation 1:

a_{1} + 24 d = 32

The System of Equations

The second condition is

a_{9} + a_{43} = 66

. Substituting the general term formula:

(a_{1} + 8 d) + (a_{1} + 42 d) = 2 a_{1} + 50 d = 66

Dividing by

2

, we obtain Equation 2:

a_{1} + 25 d = 33

Subtracting Equation 1 from Equation 2 yields

d = 1

. Substituting

d = 1

into Equation 1 gives

a_{1} + 24 (1) = 32

, resulting in

a_{1} = 8

The Sum of Squares

With

a_{1} = 8

and

d = 1

, the general term is

a_{r} = 8 + (r - 1) (1) = r + 7

. We must calculate the sum:

S = r = 1 \sum 17 (r + 7)^{2} = r = 1 \sum 17 (r^{2} + 14 r + 49)

Applying standard summation formulas for

n = 17

r = 1 \sum 17 r^{2} = \frac{17 ( 18 ) ( 35 )}{6} = 1785

14 r = 1 \sum 17 r = 14 \times \frac{17 \times 18}{2} = 2142

r = 1 \sum 17 49 = 17 \times 49 = 833

Summing these values, we find

S = 1785 + 2142 + 833 = 4760

Final Calculation

We are given the relation

S = 140 m

. Substituting our calculated sum:

140 m = 4760

m = \frac{4760}{140} = 34

The value of

m

is 34.

Common Traps & FAQs

Wait, why does the summation

\sum_{k = 0}^{12} a_{4 k + 1}

have 13 terms and not 12?

This is a classic trap! When we sum from

k = 0

k = 12

, we are counting the integers

0, 1, 2, \dots, 12

. If you count them, there are exactly

12 - 0 + 1 = 13

values. Always remember: the number of terms in a summation

\sum_{k = m}^{n}

n - m + 1

Can I use the property of AP sums for the sum of squares, like

\sum a_{r}^{2} = (\sum a_{r})^{2}

Absolutely not! That is a common algebraic misconception. The sum of squares of an AP is not the square of the sum. You must expand the term

(a_{r})^{2}

into a quadratic expression in

r

and then sum each part individually using the standard formulas for

\sum r^{2}

and

\sum r

Is there a faster way to calculate

\sum_{r = 1}^{17} (r + 7)^{2}

without expanding?

While you could potentially use properties of symmetry or shift the index, expanding is the most robust and error-proof method for JEE Advanced. It turns a complex-looking sum into three simple, standard summations that you likely have memorized. Speed comes from practice, not from avoiding the algebra!

Animated Solution for Mathematics - Sequence and Series: 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 :

Select Answer:

Visualized Solution

Problem Setup

Expanding Condition 1

Simplifying Condition 1

Analyzing Condition 2

Solving for $a_{1}$ and $d$

Setting up the Sum of Squares

Expanding the Square

Applying Summation Formulas

Calculating Remaining Terms

Finding $m$

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

Analyzing the Setup

Decoding the Hidden Summation

The System of Equations

The Sum of Squares

Final Calculation

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 :

If $a_{1}, a_{2}, a_{3}, \dots$ are in A.P. such that $a_{1} + a_{7} + a_{16} = 40$ , then the sum of the first 15 terms of this A.P. 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 $a_{1}, a_{2}, \dots, a_{n}$ 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 $a_{1}, a_{2}, a_{3}, \dots$ be in an arithmetic progression of positive terms. Let $A_{k} = a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} + \dots + a_{2 k - 1}^{2} - a_{2 k}^{2}$ . If $A_{3} = - 153, A_{5} = - 435$ and $a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = 66$ , then $a_{17} - A_{7}$ is equal to

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.

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:

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

The common difference of the A.P.: $a_{1}, a_{2}, \dots, a_{m}$ is 13 more than the common difference of the A.P.: $b_{1}, b_{2}, \dots, b_{n}$ . If $b_{31} = - 277, b_{43} = - 385$ and $a_{78} = 327$ , then $a_{1}$ is equal to

Select Answer:

Visualized Solution

Problem Setup

Expanding Condition 1

Simplifying Condition 1

Analyzing Condition 2

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Setting up the Sum of Squares

Expanding the Square

Applying Summation Formulas

Calculating Remaining Terms

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

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

Analyzing the Setup

Decoding the Hidden Summation

The System of Equations

The Sum of Squares

Final Calculation

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; } If a1​,a2​,a3​,… are in A.P. such that a1​+a7​+a16​=40, then the sum of the first 15 terms of this A.P. is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​,a2​,…,a21​ be an AP such that ∑n=120​an​an+1​1​=94​. If the sum of this AP is 189, then a6​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​… be in an A.P. such that ∑k=112​a2k−1​=−572​a1​,a1​=0. If ∑k=1n​ak​=0, then n is:

.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

Solving for $a_{1}$ and $d$

Finding $m$

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 :

If $a_{1}, a_{2}, a_{3}, \dots$ are in A.P. such that $a_{1} + a_{7} + a_{16} = 40$ , then the sum of the first 15 terms of this A.P. 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 $a_{1}, a_{2}, \dots, a_{n}$ 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 $a_{1}, a_{2}, a_{3} \dots$ 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:

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