NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 5: Arithematic Progressions

Exercise 5.2

Step 1: Recall the definition of an AP.

A sequence is called an Arithmetic Progression (AP) if the difference between any two consecutive terms is always the same. This difference is called the common difference, denoted by \(d\).

(i) \(-1, -1, -1, -1, \ldots\)

All terms are the same. So the difference between any two consecutive terms is \(-1 - (-1) = 0\).

Since the difference is always 0, it is an AP with \(d = 0\).

(ii) \(0, 2, 0, 2, \ldots\)

Find differences:

• \(2 - 0 = 2\)
• \(0 - 2 = -2\)

The differences are not the same (sometimes 2, sometimes -2). So it is not an AP.

(iii) \(1, 1, 2, 2, 3, 3, \ldots\)

Find differences:

• \(1 - 1 = 0\)
• \(2 - 1 = 1\)
• \(2 - 2 = 0\)
• \(3 - 2 = 1\)

The differences are changing (0, 1, 0, 1,...). So it is not an AP.

(iv) \(11, 22, 33, \ldots\)

Find differences:

• \(22 - 11 = 11\)
• \(33 - 22 = 11\)

The difference is always 11. So it is an AP with \(d = 11\).

(v) \(\tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4}, \ldots\)

Find differences:

• \(\tfrac{1}{3} - \tfrac{1}{2} = -\tfrac{1}{6}\)
• \(\tfrac{1}{4} - \tfrac{1}{3} = -\tfrac{1}{12}\)

The differences are not the same (\(-\tfrac{1}{6}, -\tfrac{1}{12}, ...\)). So it is not an AP.

(vi) \(2, 2^2, 2^3, 2^4, \ldots\)

This sequence is: \(2, 4, 8, 16, \ldots\).

Find differences:

• \(4 - 2 = 2\)
• \(8 - 4 = 4\)
• \(16 - 8 = 8\)

The differences are not the same. This is actually a Geometric Progression (terms are multiplied by 2 each time). So it is not an AP.

(vii) \(\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \ldots\)

Simplify square roots:

• \(\sqrt{12} = 2\sqrt{3}\)
• \(\sqrt{27} = 3\sqrt{3}\)
• \(\sqrt{48} = 4\sqrt{3}\)

So the sequence becomes: \(\sqrt{3}, 2\sqrt{3}, 3\sqrt{3}, 4\sqrt{3}, \ldots\).

Find differences:

• \(2\sqrt{3} - \sqrt{3} = \sqrt{3}\)
• \(3\sqrt{3} - 2\sqrt{3} = \sqrt{3}\)

The difference is always \(\sqrt{3}\). So it is an AP with \(d = \sqrt{3}\).

Step 1: Write the first three terms clearly.

\(a_1 = -1, \; a_2 = -\tfrac{3}{2}, \; a_3 = -2\).

Step 2: Find the difference between the 2nd and 1st term.

\(a_2 - a_1 = -\tfrac{3}{2} - (-1) = -\tfrac{3}{2} + 1\).

\(-\tfrac{3}{2} + 1 = -\tfrac{3}{2} + \tfrac{2}{2} = -\tfrac{1}{2}\).

Step 3: Find the difference between the 3rd and 2nd term.

\(a_3 - a_2 = -2 - (-\tfrac{3}{2}) = -2 + \tfrac{3}{2}\).

\(-2 = -\tfrac{4}{2}\), so \(-\tfrac{4}{2} + \tfrac{3}{2} = -\tfrac{1}{2}\).

Step 4: Both differences are equal.

\(a_2 - a_1 = a_3 - a_2 = -\tfrac{1}{2}\).

Step 5: If the difference between consecutive terms is the same, the sequence is an Arithmetic Progression (AP).

So, the given sequence is an AP with common difference \(d = -\tfrac{1}{2}\).

Step 1: Recall the formula for the (n)-th term of an AP: \(a_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference.

Step 2: To find the difference between two terms of an AP, say \(a_n - a_m\), we can subtract: \(a_n - a_m = [a + (n-1)d] - [a + (m-1)d]\).

Step 3: Simplify the subtraction: \(a_n - a_m = (n-1)d - (m-1)d = (n-m)d\).

Step 4: Apply this to our problem: \(a_{30} - a_{20} = (30 - 20) d\).

Step 5: The common difference of the given AP is \(d = -7 - (-3) = -4\).

Step 6: Substitute the values: \(a_{30} - a_{20} = (10)(-4) = -40\).

So, we do not need to calculate \(a_{30}\) and \(a_{20}\) separately. The answer is -40.

Step 1: Recall the formula for the \(n\)th term of an AP:

\(a_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference.

Step 2: Write the general term for both APs.

• For the first AP (first term = 2): \(T_n = 2 + (n-1)d\)
• For the second AP (first term = 7): \(T'_n = 7 + (n-1)d\)

Step 3: Find the difference between the two corresponding terms.

\(T_n - T'_n = [2 + (n-1)d] - [7 + (n-1)d]\)

Step 4: Simplify:

\(T_n - T'_n = 2 - 7 = -5\)

Step 5: Notice that \((n-1)d\) cancels out, so the difference does not depend on \(n\). That means the difference will always be \(-5\), whether \(n=10\), \(n=21\), or any other \(n\).

Final Result: The difference between the 10th terms is also \(-5\), the difference between the 21st terms is also \(-5\), and in fact the difference between any corresponding terms is always \(-5\).

Step 1: In an Arithmetic Progression (AP), the first term is written as \(a\) and the common difference as \(d\).

Here, the first term \(a = 31\).

The common difference \(d = 28 - 31 = -3\).

Step 2: The formula for the \(n\)-th term of an AP is:

\(a_n = a + (n - 1) \times d\).

Step 3: We want to check if \(0\) is a term in this AP. So, put \(a_n = 0\).

\(0 = 31 + (n - 1)(-3)\)

Step 4: Simplify the equation:

\(0 = 31 - 3(n - 1)\)

\(0 = 31 - 3n + 3\)

\(0 = 34 - 3n\)

Step 5: Solve for \(n\):

\(3n = 34\)

\(n = \dfrac{34}{3}\)

Step 6: Since \(n\) must be a whole number (like 1, 2, 3, …) for the term to exist, and \(\tfrac{34}{3}\) is not a whole number, it means there is no such term.

Therefore, 0 is not a term of the given AP.

Let us carefully calculate the fare step by step.

Step 1: For the first kilometre, the fare is Rs 15. So, the total after 1 km = 15.

Step 2: For the second kilometre, we add Rs 8 more. So, total after 2 km = 15 + 8 = 23.

Step 3: For the third kilometre, again add Rs 8. So, total after 3 km = 23 + 8 = 31.

Step 4: For the fourth kilometre, again add Rs 8. So, total after 4 km = 31 + 8 = 39.

So the total fares after each km are: \(15, 23, 31, 39, \ldots\).

Step 5: Now check if this is an Arithmetic Progression (AP). In an AP, the difference between two consecutive terms must be the same.

Here: \(23 - 15 = 8, \; 31 - 23 = 8, \; 39 - 31 = 8\). The difference is always 8.

Conclusion: The total fares form an AP with first term \(a = 15\) and common difference \(d = 8\). So the given statement is False.

(i) The fee is the same every month: 400, 400, 400, ...

Here, the difference between any two months is 0. Since the difference is the same each time (0), this is an AP with common difference \(d=0\).

(ii) The fees are: 250 for Class I, 300 for Class II, 350 for Class III, and so on.

So the list is: 250, 300, 350, ...

The difference between each term is \(300 - 250 = 50\), \(350 - 300 = 50\), and so on. Since the difference is always the same (50), this is an AP with common difference \(d=50\).

(iii) Simple interest formula is \(A = P(1 + rt)\).

Here, \(P=1000\), \(r = 10\% = 0.1\), \(t\) is time in years.

For 1 year: \(1000(1 + 0.1 \times 1) = 1100\).

For 2 years: \(1000(1 + 0.1 \times 2) = 1200\).

For 3 years: \(1000(1 + 0.1 \times 3) = 1300\).

So the sequence is 1100, 1200, 1300, ... Each year it increases by 100. This is an AP with common difference \(d=100\).

(iv) The bacteria double every second.

Starting from 2, we get 2, 4, 8, 16, ...

Here, the change is not by addition but by multiplication (each time multiply by 2). This is called a Geometric Progression (GP), not an AP.

(i) Expression: \(2n - 3\)

Step 1: Write first few terms by putting values of \(n\).

For \(n=1\): \(2(1) - 3 = -1\)

For \(n=2\): \(2(2) - 3 = 1\)

For \(n=3\): \(2(3) - 3 = 3\)

So the terms are: \(-1, 1, 3, 5, ...\)

Step 2: Find the difference between consecutive terms.

\(1 - (-1) = 2,\; 3 - 1 = 2,\; 5 - 3 = 2\)

The difference is always 2 (same each time). So it is an AP.

(ii) Expression: \(3n^2 + 5\)

Step 1: Write first few terms.

For \(n=1\): \(3(1)^2 + 5 = 8\)

For \(n=2\): \(3(2)^2 + 5 = 17\)

For \(n=3\): \(3(3)^2 + 5 = 32\)

For \(n=4\): \(3(4)^2 + 5 = 53\)

So the terms are: \(8, 17, 32, 53, ...\)

Step 2: Find the differences.

\(17 - 8 = 9,\; 32 - 17 = 15,\; 53 - 32 = 21\)

The difference keeps changing (9, 15, 21, ...). Since it is not the same, it is not an AP.

(iii) Expression: \(1 + n + n^2\)

Step 1: Write first few terms.

For \(n=1\): \(1 + 1 + 1^2 = 3\)

For \(n=2\): \(1 + 2 + 2^2 = 7\)

For \(n=3\): \(1 + 3 + 3^2 = 13\)

For \(n=4\): \(1 + 4 + 4^2 = 21\)

So the terms are: \(3, 7, 13, 21, ...\)

Step 2: Find the differences.

\(7 - 3 = 4,\; 13 - 7 = 6,\; 21 - 13 = 8\)

The difference is not the same (4, 6, 8, ...). So it is not an AP.