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Animated Solution for Mathematics - Sequence and Series: 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

Q: Wait, why do we subtract the terms instead of solving for $b_1$ first?

You absolutely could solve for $b_1$ first, but that adds an unnecessary step! In competitive exams like the JEE, time is your most precious resource. By subtracting $b_{31}$ from $b_{43}$, the $b_1$ term vanishes instantly, allowing you to find $d_2$ in a single algebraic stroke. It is all about efficiency and reducing the chance of arithmetic errors.

Q: Is there a shortcut for finding $a_1$ once I have $d_1$?

While there is no 'magic' shortcut, the formula $a_{78} = a_1 + 77d_1$ is the most direct path. The key is to trust the formula. Once you have $d_1 = 4$, the equation becomes $327 = a_1 + 77 \times 4$. The multiplication $77 \times 4 = 308$ is straightforward. Just keep your focus on the final isolation of $a_1$.

Select Answer:

Progressive Hints

Hint 1

The Spark: Do not rush to find the first term of the second arithmetic progression. Instead, focus on the relationship between the two given terms,

b_{31}

and

b_{43}

. Notice that the difference between these two terms is directly related to the common difference

d_{2}

by the number of steps between them.

Hint 2

The Arsenal: Remember the general term formula for an arithmetic progression:

T_{n} = a + (n - 1) d

. When you subtract two terms

T_{n}

and

T_{m}

, the first term

a

cancels out, leaving you with

(n - m) d

. Use this to isolate

d_{2}

quickly.

Hint 3

The Execution: Start by calculating

d_{2}

using the equation

b_{43} - b_{31} = (43 - 31) d_{2}

. Once you have

d_{2}

, use the given relation

d_{1} = d_{2} + 13

to find

d_{1}

. Finally, apply the general term formula to the first progression to solve for

a_{1}

using

a_{78} = a_{1} + 77 d_{1}

Visualized Solution

Defining the Two Arithmetic Progressions

Let the common difference of the first A.P. ( $a_{1}, a_{2}, \dots$ ) be $d_{1}$ .
Let the common difference of the second A.P. ( $b_{1}, b_{2}, \dots$ ) be $d_{2}$ .
We are given the relation: $d_{1} = d_{2} + 13$ .
We also have specific terms: $b_{31} = - 277$ , $b_{43} = - 385$ , and $a_{78} = 327$ .

The General Term of an A.P.

Recall the formula for the $n$ -th term of an A.P.: $T_{n} = a + (n - 1) d$ .
We will apply this to the second A.P. to find its common difference, $d_{2}$ .
For the $b$ sequence, the formula is $b_{n} = b_{1} + (n - 1) d_{2}$ .

Setting up Equations for the Second A.P.

For $n = 31$ : $b_{31} = b_{1} + (31 - 1) d_{2} \Rightarrow b_{1} + 30 d_{2} = - 277$ .
For $n = 43$ : $b_{43} = b_{1} + (43 - 1) d_{2} \Rightarrow b_{1} + 42 d_{2} = - 385$ .

Eliminating $b_{1}$ to find $d_{2}$

To find $d_{2}$ , we subtract the 31st term from the 43rd term.
$b_{43} - b_{31} = (b_{1} + 42 d_{2}) - (b_{1} + 30 d_{2})$ .
This simplifies to: $b_{43} - b_{31} = 12 d_{2}$ .

Calculating the value of $d_{2}$

Substitute the given values: $- 385 - (- 277) = 12 d_{2}$ .
$- 385 + 277 = 12 d_{2}$ .
$- 108 = 12 d_{2}$ .
Dividing by $12$ gives: $d_{2} = - 9$ .

Finding the Common Difference $d_{1}$

Now, we use the given relation between the common differences: $d_{1} = d_{2} + 13$ .
Substitute $d_{2} = - 9$ into the equation.
$d_{1} = - 9 + 13$ .
Therefore, $d_{1} = 4$ .

Setting up the Equation for the First A.P.

We need to find the first term, $a_{1}$ .
We know the 78th term: $a_{78} = 327$ .
Using the general term formula for the first A.P.: $a_{78} = a_{1} + (78 - 1) d_{1}$ .
$a_{78} = a_{1} + 77 d_{1}$ .

Substituting Known Values for $a_{78}$

We know $a_{78} = 327$ and we found $d_{1} = 4$ .
Substitute these into our equation: $327 = a_{1} + 77 (4)$ .

Solving for $a_{1}$

First, calculate the product: $77 \times 4 = 308$ .
The equation becomes: $327 = a_{1} + 308$ .
Isolate $a_{1}$ : $a_{1} = 327 - 308$ .
$a_{1} = 19$ .

Final Answer and Conclusion

We successfully found the first term of the first A.P., $a_{1} = 19$ .
Key Takeaway: When given terms of an A.P., subtracting them directly eliminates the first term and allows quick calculation of the common difference.
The correct option is 19.

00:00 / 00:00

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

Analyzing the Setup

Imagine you are standing at the edge of two parallel paths, each defined by a sequence of numbers. These are our arithmetic progressions (A.P.s).

In the first path, the numbers grow or shrink by a steady step we call

d_{1}

. In the second path, the step is

d_{2}

The problem provides a bridge between these two worlds:

d_{1} = d_{2} + 13

Our mission is to uncover the starting point of the first path,

a_{1}

The Mystery of the B-Series

We are given two markers on the second path:

b_{31} = - 277

and

b_{43} = - 385

Using the general term formula

T_{n} = a + (n - 1) d

, we express the second sequence as

b_{n} = b_{1} + (n - 1) d_{2}

If we look at the difference between the forty-third term and the thirty-first term, the

b_{1}

terms vanish:

b_{43} - b_{31} = (b_{1} + 42 d_{2}) - (b_{1} + 30 d_{2}) = 12 d_{2}

Substituting the known values:

- 385 - (- 277) = 12 d_{2}

Simplifying the expression:

- 385 + 277 = 12 d_{2}

- 108 = 12 d_{2}

Dividing by

12

, we find the common difference for the second path:

d_{2} = - 9

Building the Bridge

Now that we have the common difference of the second path, the bridge to the first path is wide open.

We know

d_{1} = d_{2} + 13

. Substituting our value of

d_{2}

d_{1} = - 9 + 13

d_{1} = 4

The first path is climbing by

4

at every step.

The Final Destination

We are told that the seventy-eighth term of the first path is

327

. We use the general term formula one last time:

a_{78} = a_{1} + (78 - 1) d_{1}

a_{78} = a_{1} + 77 d_{1}

Substituting

a_{78} = 327

and

d_{1} = 4

327 = a_{1} + 77 \times 4

Calculating the product:

77 \times 4 = 308

327 = a_{1} + 308

To find

a_{1}

, we subtract

308

from

327

a_{1} = 327 - 308

a_{1} = 19

The starting point of the first path is $19$ .

Common Traps & FAQs

Wait, why do we subtract the terms instead of solving for

b_{1}

first?

You absolutely could solve for

b_{1}

first, but that adds an unnecessary step! In competitive exams like the JEE, time is your most precious resource. By subtracting

b_{31}

from

b_{43}

, the

b_{1}

term vanishes instantly, allowing you to find

d_{2}

in a single algebraic stroke. It is all about efficiency and reducing the chance of arithmetic errors.

I keep getting confused with the negative signs when subtracting. How do I avoid this?

This is a classic trap! When you perform the subtraction, write it out explicitly:

b_{43} - b_{31} = - 385 - (- 277)

. Always use parentheses for the negative number you are subtracting. This turns the operation into

- 385 + 277

, which is much easier to visualize as a simple addition of a negative and a positive number. Stay calm, write it down, and the signs will behave.

Is there a shortcut for finding

a_{1}

once I have

d_{1}

While there is no 'magic' shortcut, the formula

a_{78} = a_{1} + 77 d_{1}

is the most direct path. The key is to trust the formula. Once you have

d_{1} = 4

, the equation becomes

327 = a_{1} + 77 \times 4

. The multiplication

77 \times 4 = 308

is straightforward. Just keep your focus on the final isolation of

a_{1}

Animated Solution for Mathematics - Sequence and Series: 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

Defining the Two Arithmetic Progressions

The General Term of an A.P.

Setting up Equations for the Second A.P.

Eliminating $b_{1}$ to find $d_{2}$

Calculating the value of $d_{2}$

Finding the Common Difference $d_{1}$

Setting up the Equation for the First A.P.

Substituting Known Values for $a_{78}$

Solving for $a_{1}$

Final Answer and Conclusion

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

Analyzing the Setup

The Mystery of the B-Series

Building the Bridge

The Final Destination

Similar Questions

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:

Let $a_{1}, a_{2}, a_{3}, a_{4}$ be an A.P. of four terms such that each term of the A.P. and its common difference $l$ are integers. If $a_{1} + a_{2} + a_{3} + a_{4} = 48$ and $a_{1} a_{2} a_{3} a_{4} + l^{4} = 361$ , then the largest term of the A.P. 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

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 :

The number of terms common the two A.P $3, 7, 11, \dots, 407$ and $2, 9, 16, \dots, 709$ is

Let $T_{r}$ be the $r$ th term of an A.P. whose first term is $a$ and common difference is $d$ . If for some positive integers $m, n, m \neq = n, T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $a - d$ equals

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 $a_{1}, a_{2}, \dots, a_{21}$ 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

Defining the Two Arithmetic Progressions

The General Term of an A.P.

Setting up Equations for the Second A.P.

Eliminating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b1​ to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d2​

Calculating the value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d2​

Finding the Common Difference .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d1​

Setting up the Equation for the First A.P.

Substituting Known Values for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a78​

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

Final Answer and Conclusion

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

Analyzing the Setup

The Mystery of the B-Series

Building the Bridge

The Final Destination

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; } The number of terms common the two A.P 3,7,11,…,407 and 2,9,16,…,709 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Tr​ be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m,n,m=n,Tm​=1/n and Tn​=1/m, then a−d equals

.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; } 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 :

Eliminating $b_{1}$ to find $d_{2}$

Calculating the value of $d_{2}$

Finding the Common Difference $d_{1}$

Substituting Known Values for $a_{78}$

Solving for $a_{1}$

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 :

The number of terms common the two A.P $3, 7, 11, \dots, 407$ and $2, 9, 16, \dots, 709$ is

Let $T_{r}$ be the $r$ th term of an A.P. whose first term is $a$ and common difference is $d$ . If for some positive integers $m, n, m \neq = n, T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $a - d$ equals

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 $a_{1}, a_{2}, \dots, a_{21}$ 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 :