JEE Main 2023 (13 Apr Shift 1)

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Animated Solution for Mathematics - Sequence and Series: Let $s_{1}, s_{2}, s_{3}, \dots, s_{10}$ respectively be the sum of 12 terms of 10 A.Ps whose first terms are $1, 2, 3, \dots, 10$ and the common differences are $1, 3, 5, \dots, 19$ respectively. Then $\sum_{i = 1}^{10} s_{i}$ is equal to

Q: How did we get $d_i = 2i - 1$ for the common difference?

Look at the sequence of common differences: $1, 3, 5, \dots, 19$. These are consecutive odd numbers. The $i$-th odd number is always given by the formula $2i - 1$. For $i=1$, $2(1)-1 = 1$. For $i=2$, $2(2)-1 = 3$. It is a perfect match! Recognizing these patterns is a core skill for solving sequence problems.

Select Answer:

Visualized Solution

Understanding the 10 A.P.s

We are given 10 different Arithmetic Progressions (A.P.s).
Each A.P. has exactly $n = 12$ terms.
Let $s_{i}$ denote the sum of the $i$ -th A.P., where $i = 1, 2, 3, \dots, 10$ .
Our ultimate goal is to find the total sum: $\sum_{i = 1}^{10} s_{i}$ .

Identifying the General First Term $a_{i}$

Look at the first terms of the given A.P.s: $1, 2, 3, \dots, 10$ .
The first term of the 1st A.P. is 1, the 2nd is 2, and so on.
Therefore, for the general $i$ -th A.P., the first term is simply $a_{i} = i$ .

Identifying the General Common Difference $d_{i}$

Now observe the common differences: $1, 3, 5, \dots, 19$ .
These are consecutive odd numbers.
The $i$ -th odd number can be written as $2 i - 1$ .
So, for the $i$ -th A.P., the common difference is $d_{i} = 2 i - 1$ .

The A.P. Sum Formula

Recall the standard formula for the sum of $n$ terms of an A.P.
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
In our problem, every A.P. has exactly $n = 12$ terms.

Setting up the Expression for $s_{i}$

Let's substitute our general terms into the sum formula for the $i$ -th A.P.
We plug in $a_{i} = i$ , $d_{i} = 2 i - 1$ , and $n = 12$ .
$s_{i} = \frac{12}{2} [2 (i) + (12 - 1) (2 i - 1)]$

Simplifying the General Sum $s_{i}$

First, simplify the constants outside and inside the bracket.
$\frac{12}{2} = 6$ and $(12 - 1) = 11$ .
This gives: $s_{i} = 6 [2 i + 11 (2 i - 1)]$

Expanding the Bracket

Now, distribute the 11 into the inner bracket.
$11 \times 2 i = 22 i$ and $11 \times (- 1) = - 11$ .
The expression becomes: $s_{i} = 6 [2 i + 22 i - 11]$

Final Simplified Form of $s_{i}$

Combine the like terms: $2 i + 22 i = 24 i$ .
So, $s_{i} = 6 [24 i - 11]$ .
Multiply by 6: $s_{i} = 144 i - 66$ .

Applying the Total Summation

We need the sum of all 10 A.P. sums: $\sum_{i = 1}^{10} s_{i}$ .
Substitute our simplified $s_{i}$ : $\sum_{i = 1}^{10} (144 i - 66)$ .
Using the linearity of summation: $144 \sum_{i = 1}^{10} i - \sum_{i = 1}^{10} 66$ .

Evaluating the Standard Summations

The sum of the first 10 natural numbers is $\sum_{i = 1}^{10} i = \frac{10 \times 11}{2} = 55$ .
The sum of a constant 66 added 10 times is $\sum_{i = 1}^{10} 66 = 10 \times 66 = 660$ .

Final Calculation

Substitute these values back into our total sum expression.
Total Sum = $144 (55) - 660$ .
First, calculate $144 \times 55 = 7920$ .
Finally, subtract: $7920 - 660 = 7260$ .

Conclusion and Takeaway

Key Takeaway: When dealing with multiple sequences, always find the general term for the parameters ( $a_{i}$ , $d_{i}$ ).
Simplify the general expression before applying the final summation to save time and avoid errors.
The correct option is 7260.

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

Solution Diagram

Analyzing the Setup

Imagine you are standing on the edge of a vast landscape of numbers, looking at ten different paths, each an Arithmetic Progression (A.P.). We are tasked with finding the sum of ten different A.P.s, each containing

n = 12

terms.

The first terms are

a_{i} = 1, 2, 3, \dots, 10

, and the common differences are

d_{i} = 1, 3, 5, \dots, 19

. Instead of brute force, we will utilize the power of the general term to map these sequences.

Decoding the DNA of the Sequences

Every sequence has a DNA—its first term

a

and its common difference

d

. For our

i

-th A.P., the first term is simply

a_{i} = i

The common differences are the odd numbers

1, 3, 5, \dots, 19

. We know that the

i

-th odd number is represented by:

d_{i} = 2 i - 1

We now have the building blocks for any of the ten A.P.s. We view them as a single, generalized sequence defined by the index

i

The Master Equation

We recall the sum formula for an A.P. with

n

terms:

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

Here,

n = 12

for every one of our ten sequences. By substituting our generalized

a_{i}

and

d_{i}

into this formula, we create a master expression for

s_{i}

s_{i} = \frac{12}{2} [2 (i) + (12 - 1) (2 i - 1)]

This is the moment where the complexity collapses. We simplify the expression:

s_{i} = 6 [2 i + 11 (2 i - 1)]

Expanding the inner term gives us

s_{i} = 6 [2 i + 22 i - 11]

, which simplifies further to

s_{i} = 6 [24 i - 11]

. Finally, distributing the

6

, we arrive at the elegant general form:

s_{i} = 144 i - 66

Final Calculation

Now, we reach the grand finale. We need the sum of all these

s_{i}

values from

i = 1

10

i = 1 \sum 10 (144 i - 66)

Thanks to the linearity of summation, we can split this into two manageable parts:

144 i = 1 \sum 10 i - i = 1 \sum 10 66

The sum of the first

10

natural numbers is a classic result:

i = 1 \sum 10 i = \frac{10 \times 11}{2} = 55

The second part is simply adding

66

ten times, which equals

660

. Our final calculation becomes:

144 (55) - 660 = 7920 - 660 = 7260

The final result of the summation is 7260.

Common Traps & FAQs

Why can't we just calculate each of the 10 sums individually? It seems like it would work.

While calculating each sum individually is technically possible, it is a trap! In JEE Advanced, time is your most precious resource. By finding the general term

s_{i}

first, you reduce ten separate, error-prone calculations into a single, elegant algebraic expression. This approach is not just faster; it is mathematically superior because it reveals the underlying structure of the problem.

How did we get

d_{i} = 2 i - 1

for the common difference?

Look at the sequence of common differences:

1, 3, 5, \dots, 19

. These are consecutive odd numbers. The

i

-th odd number is always given by the formula

2 i - 1

. For

i = 1

2 (1) - 1 = 1

. For

i = 2

2 (2) - 1 = 3

. It is a perfect match! Recognizing these patterns is a core skill for solving sequence problems.

Is there a shortcut using the average of the A.P.s?

You are thinking like a mathematician! Yes, the sum of an A.P. is also

n

times the average of the first and last terms. You could express the

i

-th sum as

s_{i} = 12 \times \frac{a _{i} + l _{i}}{2}

. This would lead you to the same result. However, the standard sum formula

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

is often more direct when you are given the common difference

d

explicitly.

Animated Solution for Mathematics - Sequence and Series: Let $s_{1}, s_{2}, s_{3}, \dots, s_{10}$ respectively be the sum of 12 terms of 10 A.Ps whose first terms are $1, 2, 3, \dots, 10$ and the common differences are $1, 3, 5, \dots, 19$ respectively. Then $\sum_{i = 1}^{10} s_{i}$ is equal to

Select Answer:

Visualized Solution

Understanding the 10 A.P.s

Identifying the General First Term $a_{i}$

Identifying the General Common Difference $d_{i}$

The A.P. Sum Formula

Setting up the Expression for $s_{i}$

Simplifying the General Sum $s_{i}$

Expanding the Bracket

Final Simplified Form of $s_{i}$

Applying the Total Summation

Evaluating the Standard Summations

Final Calculation

Conclusion and Takeaway

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

Analyzing the Setup

Decoding the DNA of the Sequences

The Master Equation

Final Calculation

Similar Questions

If the $1 0^{t h}$ term of an A.P. is $\frac{1}{20}$ and its $2 0^{t h}$ term is $\frac{1}{10}$ , then the sum of its first 200 term is

If the $1 0^{t h}$ term of an A.P. is $1/20$ and its $2 0^{t h}$ term is $1/10$ , then the sum of its first 200 terms is:

Let $2^{n d}, 8^{t h}$ and $4 4^{t h}$ terms of a non-constant A.P. be respectively the $1^{s t}, 2^{n d}$ and $3^{r d}$ terms of G.P. If the first term of A.P. is 1 then the sum of first 20 terms 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 :

Suppose $a_{1}, a_{2}, \dots, a_{n}, \dots$ be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms to the sum of first nine terms of the progression is $5 : 17$ and $110 < a_{15} < 120$ , then the sum of the first ten terms of the progression is equal to -

Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is $6 : 11$ and the seventh term lies in between $130$ and $140$ , then the common difference of this A.P. is \dots.

Let $S_{n}$ denote the sum of first $n$ terms an arithmetic progression. If $S_{20} = 790$ and $S_{10} = 145$ , then $S_{15} - S_{5}$ is :

Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is ________.

If the sum of first 11 terms of an A.P., $a_{1}, a_{2}, a_{3} ... i s 0 (a_{1} \neq = 0)$ , then the sum of the A.P., $a_{1}, a_{3}, a_{5}, ..., a_{23}$ is $k a_{1}$ , where $k$ is equal to:

Select Answer:

Visualized Solution

Understanding the 10 A.P.s

Identifying the General First Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ai​

Identifying the General Common Difference .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } di​

The A.P. Sum Formula

Setting up the Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } si​

Simplifying the General Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } si​

Expanding the Bracket

Final Simplified Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } si​

Applying the Total Summation

Evaluating the Standard Summations

Final Calculation

Conclusion and Takeaway

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

Analyzing the Setup

Decoding the DNA of the Sequences

The Master Equation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the 10th term of an A.P. is 201​ and its 20th term is 101​, then the sum of its first 200 term is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the 10th term of an A.P. is 1/20 and its 20th term is 1/10, then the sum of its first 200 terms is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let 2nd,8th and 44th terms of a non-constant A.P. be respectively the 1st,2nd and 3rd terms of G.P. If the first term of A.P. is 1 then the sum of first 20 terms 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 Sn​ denote the sum of first n terms an arithmetic progression. If S20​=790 and S10​=145, then S15​−S5​ is :

Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is ________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the sum of first 11 terms of an A.P., a1​,a2​,a3​...is0(a1​=0), then the sum of the A.P., a1​,a3​,a5​,...,a23​ is ka1​, where k is equal to:

Identifying the General First Term $a_{i}$

Identifying the General Common Difference $d_{i}$

Setting up the Expression for $s_{i}$

Simplifying the General Sum $s_{i}$

Final Simplified Form of $s_{i}$

If the $1 0^{t h}$ term of an A.P. is $\frac{1}{20}$ and its $2 0^{t h}$ term is $\frac{1}{10}$ , then the sum of its first 200 term is

If the $1 0^{t h}$ term of an A.P. is $1/20$ and its $2 0^{t h}$ term is $1/10$ , then the sum of its first 200 terms is:

Let $2^{n d}, 8^{t h}$ and $4 4^{t h}$ terms of a non-constant A.P. be respectively the $1^{s t}, 2^{n d}$ and $3^{r d}$ terms of G.P. If the first term of A.P. is 1 then the sum of first 20 terms 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 $S_{n}$ denote the sum of first $n$ terms an arithmetic progression. If $S_{20} = 790$ and $S_{10} = 145$ , then $S_{15} - S_{5}$ is :

If the sum of first 11 terms of an A.P., $a_{1}, a_{2}, a_{3} ... i s 0 (a_{1} \neq = 0)$ , then the sum of the A.P., $a_{1}, a_{3}, a_{5}, ..., a_{23}$ is $k a_{1}$ , where $k$ is equal to: