Solution:

Arithmetic Progression Table

Fill in the blanks in the following table, given that a is the first term, d the common difference, and $a_n$ the nth term of the AP:

(i) $a=7$, $d=3$, $n=8$, $a_n = 7 + (8-1) \times 3 = 7 + 21 = 28$
(ii) $a=-18$, $d = \frac{0 - (-18)}{10 - 1} = 2$
(iii) $-5 = a + (18-1) \times (-3) \Rightarrow a = 46$
(iv) $3.6 = (-18.9) + (n-1) \times 2.5 \Rightarrow n = 10$
(v) $a=3.5$, $d=0$, $n=105$, $a_n = 3.5$

Solution:

To find which term of the arithmetic progression (AP) 3, 8, 13, 18,... is equal to 78, we can use the formula for the \(n\)th term of an AP:

where:
- \(a\) is the first term,
- \(d\) is the common difference,
- \(n\) is the term number.

1. Identify the first term and common difference:

- First term \(a = 3\)
- Common difference \(d = 8 - 3 = 5\)

2. Set up the equation:

We want to find \(n\) such that:

3. Solve for \(n\):

Thus, the term 78 is the 16th term of the AP.

Solution:

Check whether –150 is a term of the AP: 11, 8, 5, 2...

First Term (a):

Common Difference (d):

Formula for the n-th term of an AP:

Setting up the equation:

Solving for n:

Conclusion:

Solution:

Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Let:

Given:

a$_{11}$ = 38
a$_{16}$ = 73

Setting up the equations:

Subtracting equation (1) from (2):

Substituting d back into equation (1):

Finding the 31st term:

a$_{31}$ = a + 30d
a$_{31}$ = -32 + 30(7)
a$_{31}$ = -32 + 210
a$_{31}$ = 178

Thus, the 31st term is:

Solution:

An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Let:

Given:

a$_3$ = 12
a$_{50}$ = 106

Setting up the equations:

Subtracting equation (1) from (2):

Substituting d back into equation (1):

Finding the 29th term:

a$_{29}$ = a + 28d
a$_{29}$ = 8 + 28(2)
a$_{29}$ = 8 + 56
a$_{29}$ = 64

Thus, the 29th term is:

Solution:

If the 3rd and the 9th terms of an AP are 4 and –8 respectively, which term of this AP is zero?

Let:

Given:

Setting up the equations:

Subtracting equation (1) from (2):

Substituting d back into equation (1):

Finding which term is zero:

Thus, the term of this AP that is zero is:

Solution:

The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Let:

Given:

Setting up the equations:

Solving for d:

Thus, the common difference is:

Solution:

Which term of the AP: 3, 15, 27, 39,... will be 132 more than its 54th term?

Let:

Given:

Finding the 54th term:

a$_{54}$ = a + (54 - 1)d
a$_{54}$ = 3 + 53(12)
a$_{54}$ = 3 + 636
a$_{54}$ = 639

Setting up the equation for the term that is 132 more than the 54th term:

a$_n$ = a_{54} + 132
a$_n$ = 639 + 132
a$_n$ = 771

Setting up the equation for a_n:

Solving for n:

Thus, the term of the AP that will be 132 more than its 54th term is:

Solution:

Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Let:

Given:

Using the formula for the n-th term:

Setting up the equation:

Finding the difference between the 1000th terms:

Setting up the equation for the 1000th terms:

Using equation (1):

Thus, the difference between their 1000th terms is:

Solution:

How many three-digit numbers are divisible by 7?

Finding the smallest three-digit number:

Finding the largest three-digit number:

Finding the total count of three-digit numbers divisible by 7:

Using the formula for the n-th term of an AP:

Solving for n:

Thus, the number of three-digit numbers divisible by 7 is:

Solution:

How many multiples of 4 lie between 10 and 250?

Finding the smallest multiple of 4 greater than 10:

Finding the largest multiple of 4 less than 250:

Finding the total count of multiples of 4 between 12 and 248:

Using the formula for the n-th term of an AP:

Solving for n:

Thus, the number of multiples of 4 between 10 and 250 is:

Solution:

Finding the Value of n for Equal nth Terms of Two APs

The first AP is given as: 63, 65, 67,...

Here, the first term (a₁) = 63 and the common difference (d₁) = 2.

The nth term (T_n) of this AP can be expressed as:

T_n = a₁ + (n-1)d₁ = 63 + (n-1) × 2.

The second AP is given as: 3, 10, 17,...

Here, the first term (a₂) = 3 and the common difference (d₂) = 7.

The nth term (S_n) of this AP can be expressed as:

S_n = a₂ + (n-1)d₂ = 3 + (n-1) × 7.

To find the value of n for which T_n = S_n, we equate:

63 + (n-1) × 2 = 3 + (n-1) × 7.

Expanding both sides:

63 + 2n - 2 = 3 + 7n - 7.

Now simplifying:

61 + 2n = -4 + 7n.

Rearranging gives:

61 + 4 = 7n - 2n.

Thus, we have:

65 = 5n.

Finally, solving for n:

n = 65 / 5 = 13.

Solution:

Determining the Arithmetic Progression (AP)

Let the first term of the AP be a and the common difference be d.

The third term (T₃) is given by:

T₃ = a + 2d = 16.

The seventh term (T₇) is:

T₇ = a + 6d.

The fifth term (T₅) is:

T₅ = a + 4d.

According to the problem, T₇ exceeds T₅ by 12:

T₇ - T₅ = 12.

Substituting the terms:

(a + 6d) - (a + 4d) = 12.

Simplifying gives:

2d = 12.

Thus, we find:

d = 6.

Now substituting d back into the equation for T₃:

a + 2(6) = 16.

This simplifies to:

a + 12 = 16.

Thus:

a = 16 - 12.

So, we have:

a = 4.

The AP can now be expressed as:

4, (4 + 6), (4 + 12),...

Solution:

Given AP: 3, 8, 13, ..., 253

Solution:

Given:

The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. We are required to find the first three terms of the AP.

Let the first term of the AP be $a$ and the common difference be $d$.

General formula for the nth term of an AP is:

$T_n = a + (n - 1) \cdot d$

Using this formula, the 4th term $T_4$ is:

$T_4 = a + (4 - 1) \cdot d = a + 3d$

The 8th term $T_8$ is:

$T_8 = a + (8 - 1) \cdot d = a + 7d$

According to the given information:

The sum of the 4th and 8th terms is 24:

$T_4 + T_8 = (a + 3d) + (a + 7d) = 24$

$2a + 10d = 24$

Or, $a + 5d = 12$ ...(1)

Similarly, the 6th term $T_6$ is:

$T_6 = a + (6 - 1) \cdot d = a + 5d$

The 10th term $T_{10}$ is:

$T_{10} = a + (10 - 1) \cdot d = a + 9d$

According to the given information:

The sum of the 6th and 10th terms is 44:

$T_6 + T_{10} = (a + 5d) + (a + 9d) = 44$

$2a + 14d = 44$

Or, $a + 7d = 22$ ...(2)

Now, subtract equation (1) from equation (2):

$(a + 7d) - (a + 5d) = 22 - 12$

$2d = 10$

$d = 5$

Substitute $d = 5$ in equation (1):

$a + 5 \cdot 5 = 12$

$a + 25 = 12$

$a = 12 - 25 = -13$

Thus, the first term $a = -13$ and the common difference $d = 5$.

The first three terms of the AP are:

First term: $a = -13$

Second term: $a + d = -13 + 5 = -8$

Third term: $a + 2d = -13 + 2 \cdot 5 = -3$

The first three terms of the AP are $-13, -8, -3$.

Solution:

Subba Rao's Income Problem

Let the number of years after 1995 be represented by $n$. Subba Rao's salary in the $n$th year can be expressed as:

Salary in $n$th year = Initial salary + Annual increment $\times$ Number of years

Therefore, the salary in $n$th year = $5000 + 200 \times n$

We need to find the value of $n$ when the salary reaches Rs. 7000. So, we set the equation:

$5000 + 200 \times n = 7000$

Solve for $n$:

$200 \times n = 7000 - 5000$

$200 \times n = 2000$

$n = \frac{2000}{200}$

$n = 10$

This means that Subba Rao's salary will reach Rs. 7000 after 10 years from 1995, i.e., in the year $1995 + 10 = 2005$.

Solution:

Question:

Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs. 20.75, find n.

Solution:

Let the weekly savings in the first week be denoted as $a_1 = 5$. The savings increase by Rs. 1.75 every week, so the common difference is $d = 1.75$.

The nth term of an arithmetic sequence is given by the formula:

$a_n = a_1 + (n - 1) \cdot d$

We are given that $a_n = 20.75$. Substituting the values into the formula:

$20.75 = 5 + (n - 1) \cdot 1.75$

Solving for $n$:

$20.75 - 5 = (n - 1) \cdot 1.75$

$15.75 = (n - 1) \cdot 1.75$

Divide both sides by 1.75:

$\frac{15.75}{1.75} = n - 1$

$9 = n - 1$

So, $n = 10$.