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Animated Solution for Mathematics - Sequence and Series: The common difference of the A.P. $b_{1}, b_{2}, ..., b_{m}$ is 2 more than the common difference of A.P. $a_{1}, a_{2}, ..., a_{n}$ , If $a_{40} = - 159, a_{100} = - 399$ and $b_{100} = a_{70}$ , then $b_{1}$ is equal to:

Q: Is there a faster way to find $d_a$ without solving the full system?

Yes! You can use the property that $a_n - a_m = (n-m)d$. So, $a_{100} - a_{40} = (100-40)d_a$. This gives $60d_a = -399 - (-159) = -240$. It's a great shortcut to save precious seconds!

Select Answer:

Visualized Solution

Define the Two A.P.s

Let the first A.P. be $a_{n}$ with first term $a_{1}$ and common difference $d_{a}$ .
Let the second A.P. be $b_{m}$ with first term $b_{1}$ and common difference $d_{b}$ .

Relationship Between $d_{a}$ and $d_{b}$

Given: The common difference of $b_{m}$ is $2$ more than that of $a_{n}$ .
Equation: $d_{b} = d_{a} + 2$

Equation for $a_{40}$

General term of an A.P.: $a_{n} = a_{1} + (n - 1) d_{a}$
Given $a_{40} = - 159$
$a_{1} + 39 d_{a} = - 159$ ... (Equation 1)

Equation for $a_{100}$

Given $a_{100} = - 399$
$a_{1} + 99 d_{a} = - 399$ ... (Equation 2)

Solving for $d_{a}$

Subtract Equation 1 from Equation 2:
$(a_{1} + 99 d_{a}) - (a_{1} + 39 d_{a}) = - 399 - (- 159)$
$60 d_{a} = - 240$

Value of $d_{a}$

$60 d_{a} = - 240$
$d_{a} = \frac{- 240}{60}$
$d_{a} = - 4$

Finding First Term $a_{1}$

Substitute $d_{a} = - 4$ into Equation 1:
$a_{1} + 39 (- 4) = - 159$
$a_{1} - 156 = - 159$
$a_{1} = - 3$

Calculate $d_{b}$

Recall: $d_{b} = d_{a} + 2$
Substitute $d_{a} = - 4$ :
$d_{b} = - 4 + 2$
$d_{b} = - 2$

Calculate $a_{70}$

We need $a_{70}$ to link with the second A.P.
$a_{70} = a_{1} + 69 d_{a}$
$a_{70} = - 3 + 69 (- 4)$
$a_{70} = - 3 - 276 = - 279$

Equation for $b_{1}$

Given: $b_{100} = a_{70}$
$b_{100} = - 279$
Formula for $b_{100}$ : $b_{1} + 99 d_{b} = - 279$

Solving for $b_{1}$

Substitute $d_{b} = - 2$ :
$b_{1} + 99 (- 2) = - 279$
$b_{1} - 198 = - 279$
$b_{1} = - 279 + 198$
$b_{1} = - 81$

Conclusion

Final Answer: $b_{1} = - 81$
Matches Option B.
Key Takeaway: Linking two sequences requires systematically finding the parameters $(a, d)$ of one to unlock the other.

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

Analyzing the Setup

Arithmetic Progressions (APs) are the heartbeat of sequences. They are predictable, elegant, and deeply structured.

We define the first sequence,

a_{n}

, with a first term

a_{1}

and a common difference

d_{a}

. Similarly, the second sequence,

b_{m}

, has a first term

b_{1}

and a common difference

d_{b}

The problem provides a vital piece of intelligence: the common difference of the second sequence is

2

more than the first. Mathematically, we capture this as:

d_{b} = d_{a} + 2

Decoding the First Sequence

We are given two data points for the first sequence:

a_{40} = - 159

and

a_{100} = - 399

. Using the general term formula

a_{n} = a_{1} + (n - 1) d_{a}

, we construct the following system of linear equations:

a_{1} + 39 d_{a} = - 159

a_{1} + 99 d_{a} = - 399

To solve this, we subtract the first equation from the second:

(a_{1} + 99 d_{a}) - (a_{1} + 39 d_{a}) = - 399 - (- 159)

60 d_{a} = - 240

Dividing by

60

, we find

d_{a} = - 4

. Substituting this back into the first equation:

a_{1} + 39 (- 4) = - 159

a_{1} - 156 = - 159

a_{1} = - 3

Crossing the Bridge

Now, we return to our bridge equation. Since

d_{b} = d_{a} + 2

and

d_{a} = - 4

, it follows that:

d_{b} = - 4 + 2 = - 2

The problem states that

b_{100} = a_{70}

. First, we calculate the value of

a_{70}

using our parameters for the first sequence:

a_{70} = a_{1} + 69 d_{a}

a_{70} = - 3 + 69 (- 4) = - 3 - 276 = - 279

The Final Reveal

We equate this result to the hundredth term of the second sequence:

b_{100} = b_{1} + 99 d_{b} = - 279

Substituting

d_{b} = - 2

into the equation:

b_{1} + 99 (- 2) = - 279

b_{1} - 198 = - 279

Solving for

b_{1}

, we arrive at the final result:

b_{1} = - 279 + 198 = - 81

The final value is $b_{1} = - 81$ .

Common Traps & FAQs

Wait, why do we need to find

a_{1}

and

d_{a}

first? Can't we just solve for

b_{1}

directly?

Think of this problem as a locked door. The second sequence is the room you want to enter, but the key is hidden inside the first sequence. Without the parameters

a_{1}

and

d_{a}

, you have no way to evaluate the condition

b_{100} = a_{70}

. In JEE problems, always look for the 'source' of information—here, the first sequence is your source.

I keep messing up the signs. Is

d_{b} = d_{a} + 2

d_{b} = d_{a} - 2

Language is as important as math here. '2 more than' implies addition. If

d_{a}

is negative, adding 2 makes it 'less negative' (closer to zero). Always double-check your arithmetic signs—they are the most common cause of lost marks in the exam hall.

Is there a faster way to find

d_{a}

without solving the full system?

Yes! You can use the property that

a_{n} - a_{m} = (n - m) d

. So,

a_{100} - a_{40} = (100 - 40) d_{a}

. This gives

60 d_{a} = - 399 - (- 159) = - 240

. It's a great shortcut to save precious seconds!

Animated Solution for Mathematics - Sequence and Series: The common difference of the A.P. $b_{1}, b_{2}, ..., b_{m}$ is 2 more than the common difference of A.P. $a_{1}, a_{2}, ..., a_{n}$ , If $a_{40} = - 159, a_{100} = - 399$ and $b_{100} = a_{70}$ , then $b_{1}$ is equal to:

Select Answer:

Visualized Solution

Define the Two A.P.s

Relationship Between $d_{a}$ and $d_{b}$

Equation for $a_{40}$

Equation for $a_{100}$

Solving for $d_{a}$

Value of $d_{a}$

Finding First Term $a_{1}$

Calculate $d_{b}$

Calculate $a_{70}$

Equation for $b_{1}$

Solving for $b_{1}$

Conclusion

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

Analyzing the Setup

Decoding the First Sequence

Crossing the Bridge

The Final Reveal

Similar Questions

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

Let the sum of the first $n$ terms of a non-constant A.P., $a_{1}, a_{2}, a_{3}, \dots$ be $50 n + \frac{n ( n - 7 )}{2} A$ , where $A$ is a constant. If $d$ is the common difference of this A.P., then the ordered pair $(d, a_{50})$ is equal to

If the variance of the terms in an increasing A.P., $b_{1}, b_{2}, b_{3}, \dots, b_{11}$ is 90, then the common difference of this A.P. 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 ________.

Let $a_{1}, a_{2}, .... a_{n}$ be a given A.P. whose common difference is an integer and $S_{n} = a_{1} + a_{2} + .... + a_{n}$ If $a_{1} = 1, a_{n} = 300$ and $15 < n < 50$ , then the ordered pair $(S_{n - 4}, a_{n - 4})$ 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.

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 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

If ${a_{i}}_{i = 1}^{n}$ , where $n$ is an even integer, is an arithmetic progression with common difference 1, and $\sum_{i = 1}^{n} a_{i} = 192$ , $\sum_{i = 1}^{n /2} a_{2 i} = 120$ , then $n$ is equal to:

Select Answer:

Visualized Solution

Define the Two A.P.s

Relationship Between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } da​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } db​

Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a40​

Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a100​

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

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

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

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a70​

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Conclusion

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

Analyzing the Setup

Decoding the First Sequence

Crossing the Bridge

The Final Reveal

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the variance of the terms in an increasing A.P., b1​,b2​,b3​,…,b11​ is 90, then the common difference of this A.P. 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 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 the sum of the first 2n terms of the A.P. 2,5,8,… is equal to the sum of the first n terms of the A.P. 57,59,61,…, then n equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If {ai​}i=1n​, where n is an even integer, is an arithmetic progression with common difference 1, and ∑i=1n​ai​=192, ∑i=1n/2​a2i​=120, then n is equal to:

Relationship Between $d_{a}$ and $d_{b}$

Equation for $a_{40}$

Equation for $a_{100}$

Solving for $d_{a}$

Value of $d_{a}$

Finding First Term $a_{1}$

Calculate $d_{b}$

Calculate $a_{70}$

Equation for $b_{1}$

Solving for $b_{1}$

If the variance of the terms in an increasing A.P., $b_{1}, b_{2}, b_{3}, \dots, b_{11}$ is 90, then the common difference of this A.P. is

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 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

If ${a_{i}}_{i = 1}^{n}$ , where $n$ is an even integer, is an arithmetic progression with common difference 1, and $\sum_{i = 1}^{n} a_{i} = 192$ , $\sum_{i = 1}^{n /2} a_{2 i} = 120$ , then $n$ is equal to: