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

Exercise 5.3

Step 1: For (A₁)

The sequence is: 2, −2, −6, −10, …

To find the common difference, subtract any term from the next term:

−2 − 2 = −4, −6 − (−2) = −4, and so on.

So, the common difference \(d = -4\).

This matches with (B₄).

---

Step 2: For (A₂)

We are told: first term \(a = -18\), 10th term \(a_{10} = 0\).

General formula for the nth term of an AP is:

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

So, \(a_{10} = -18 + (10-1)d = -18 + 9d\).

It is given that \(a_{10} = 0\).

So, \(0 = -18 + 9d\).

⇒ \(9d = 18\).

⇒ \(d = 2\).

This matches with (B₅).

---

Step 3: For (A₃)

We are told: \(a = 0\), 10th term \(a_{10} = 6\).

Again, use the nth term formula:

\(a_{10} = a + (10-1)d = 0 + 9d = 9d\).

But we know \(a_{10} = 6\).

So, \(9d = 6\).

⇒ \(d = \dfrac{6}{9} = \dfrac{2}{3}\).

This matches with (B₁).

---

Step 4: For (A₄)

We are told: second term \(a_2 = 13\), fourth term \(a_4 = 3\).

General term formula: \(a_n = a + (n-1)d\).

So, \(a_2 = a + d = 13\). … (1)

Also, \(a_4 = a + 3d = 3\). … (2)

Subtract (1) from (2):

(a + 3d) − (a + d) = 3 − 13

⇒ 2d = −10

⇒ d = −5

This matches with (B₂).

---

Final Matches:

(A₁) → (B₄), (A₂) → (B₅), (A₃) → (B₁), (A₄) → (B₂).

Step 1: Recall the rule of an Arithmetic Progression (AP)

A sequence is called an AP if the difference between any two consecutive terms is always the same. This difference is called the common difference (d).

(i)

Terms: \(0, \; \tfrac{1}{4}, \; \tfrac{1}{2}, \; \tfrac{3}{4}\)

Find differences:

• \(\tfrac{1}{4} - 0 = \tfrac{1}{4}\)
• \(\tfrac{1}{2} - \tfrac{1}{4} = \tfrac{1}{4}\)
• \(\tfrac{3}{4} - \tfrac{1}{2} = \tfrac{1}{4}\)

Since all differences are equal, it is an AP with \(d = \tfrac{1}{4}\).

Next three terms: add \(\tfrac{1}{4}\) each time → \(1, \tfrac{5}{4}, \tfrac{3}{2}\).

(ii)

Terms: \(5, \; \tfrac{14}{3}, \; \tfrac{13}{3}, \; 4\)

Find differences:

• \(\tfrac{14}{3} - 5 = \tfrac{14}{3} - \tfrac{15}{3} = -\tfrac{1}{3}\)
• \(\tfrac{13}{3} - \tfrac{14}{3} = -\tfrac{1}{3}\)
• \(4 - \tfrac{13}{3} = \tfrac{12}{3} - \tfrac{13}{3} = -\tfrac{1}{3}\)

All differences are \(-\tfrac{1}{3}\), so it is an AP with \(d = -\tfrac{1}{3}\).

Next three terms: add \(-\tfrac{1}{3}\) each time → \(\tfrac{11}{3}, \tfrac{10}{3}, 3\).

(iii)

Terms: \(\sqrt{3}, \; 2\sqrt{3}, \; 3\sqrt{3}\)

Find differences:

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

All differences are equal to \(\sqrt{3}\), so it is an AP with \(d = \sqrt{3}\).

Next three terms: add \(\sqrt{3}\) each time → \(4\sqrt{3}, 5\sqrt{3}, 6\sqrt{3}\).

(iv)

Terms: \(a+b, \; (a+1)+b, \; (a+1)+(b+1)\)

Write them clearly:

• First term: \(a+b\)
• Second term: \(a+1+b = a+b+1\)
• Third term: \(a+1+b+1 = a+b+2\)

Differences:

• Second – First = \((a+b+1) - (a+b) = 1\)
• Third – Second = \((a+b+2) - (a+b+1) = 1\)

So the common difference \(d = 1\).

Next three terms: \(a+b+3, a+b+4, a+b+5\).

(v)

Terms: \(a, \; 2a+1, \; 3a+2, \; 4a+3\)

Differences:

• Second – First = \((2a+1) - a = a+1\)
• Third – Second = \((3a+2) - (2a+1) = a+1\)
• Fourth – Third = \((4a+3) - (3a+2) = a+1\)

All are equal, so common difference \(d = a+1\).

Next three terms: add \(a+1\) each time → \(5a+4, 6a+5, 7a+6\).

Conclusion: In each case, the differences are constant, so they are APs and the next terms are found by repeatedly adding the common difference.

An Arithmetic Progression (AP) is a list of numbers where each term is obtained by adding the common difference (\(d\)) to the previous term.

The first three terms are always:

\(T_1 = a\) (the first term)

\(T_2 = a + d\) (first term + common difference)

\(T_3 = a + 2d\) (first term + twice the common difference)

(i) \(a = \tfrac{1}{2}, d = -\tfrac{1}{6}\)

\(T_1 = a = \tfrac{1}{2}\)

\(T_2 = a + d = \tfrac{1}{2} - \tfrac{1}{6} = \tfrac{3}{6} - \tfrac{1}{6} = \tfrac{2}{6} = \tfrac{1}{3}\)

\(T_3 = a + 2d = \tfrac{1}{2} + 2\left(-\tfrac{1}{6}\right) = \tfrac{1}{2} - \tfrac{2}{6} = \tfrac{3}{6} - \tfrac{2}{6} = \tfrac{1}{6}\)

(ii) \(a = -5, d = -3\)

\(T_1 = a = -5\)

\(T_2 = a + d = -5 + (-3) = -8\)

\(T_3 = a + 2d = -5 + 2(-3) = -5 - 6 = -11\)

(iii) \(a = \sqrt{2}, d = \tfrac{1}{\sqrt{2}}\)

\(T_1 = a = \sqrt{2}\)

\(T_2 = a + d = \sqrt{2} + \tfrac{1}{\sqrt{2}}\)

\(T_3 = a + 2d = \sqrt{2} + 2\left(\tfrac{1}{\sqrt{2}}\right) = \sqrt{2} + \tfrac{2}{\sqrt{2}} = \sqrt{2} + \sqrt{2} = 2\sqrt{2}\)

So, the first three terms of each AP are as written in the answer.

Step 1: Recall what an Arithmetic Progression (AP) means.

In an AP, the difference between two consecutive terms is always the same.

Another useful fact: Any middle term in an AP is the average of the terms just before and just after it.

Step 2: Use this idea for \(b\).

Here, \(b\) lies between \(7\) and \(23\).

So, \(b = \dfrac{7 + 23}{2} = \dfrac{30}{2} = 15\).

Step 3: Now find \(a\).

Here, \(7\) lies between \(a\) and \(b\).

So, \(7 = \dfrac{a + b}{2}\).

Substitute \(b = 15\): \(7 = \dfrac{a + 15}{2}\).

Multiply both sides by 2: \(14 = a + 15\).

So, \(a = 14 - 15 = -1\).

Step 4: Finally, find \(c\).

Here, \(23\) lies between \(b\) and \(c\).

So, \(23 = \dfrac{b + c}{2}\).

Substitute \(b = 15\): \(23 = \dfrac{15 + c}{2}\).

Multiply both sides by 2: \(46 = 15 + c\).

So, \(c = 46 - 15 = 31\).

Final Answer: \(a = -1,\; b = 15,\; c = 31\).

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

Step 1: Write formula for nth term
The nth term of an AP is given by: \(a_n = a + (n-1)d\).

Step 2: Use the condition for the 5th term
5th term = 19 \(a_5 = a + (5-1)d = a + 4d\). So, \(a + 4d = 19\). (Equation 1)

Step 3: Use the condition \(a_{13} - a_8 = 20\)
13th term = \(a + 12d\)
8th term = \(a + 7d\)
Difference = \((a + 12d) - (a + 7d) = 5d\). So, \(5d = 20\). Therefore, \(d = 4\).

Step 4: Substitute value of d in Equation (1)
From Equation (1): \(a + 4d = 19\). Put \(d = 4\): \(a + 16 = 19\). So, \(a = 3\).

Final Answer:
First term \(a = 3\), common difference \(d = 4\). Hence, the AP is: 3, 7, 11, …

Step 1: Write the formula for the \(n\)-th term of an AP.

In an arithmetic progression (AP), the \(n\)-th term is:

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

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

Step 2: Use the information about the 26th term.

The 26th term is given as 0.

So, \(a + 25d = 0\) … (1)

Step 3: Use the information about the 11th term.

The 11th term is given as 3.

So, \(a + 10d = 3\) … (2)

Step 4: Subtract the two equations to find \(d\).

(1) – (2):

\((a + 25d) - (a + 10d) = 0 - 3\)

\(15d = -3\)

\(d = -\dfrac{1}{5}\)

Step 5: Find the first term \(a\).

Put \(d = -\dfrac{1}{5}\) in equation (2):

\(a + 10(-\dfrac{1}{5}) = 3\)

\(a - 2 = 3\)

\(a = 5\)

Step 6: Use the last term to find the number of terms.

Last term is \(-\dfrac{1}{5}\).

So, \(a + (n-1)d = -\dfrac{1}{5}\)

\(5 + (n-1)(-\dfrac{1}{5}) = -\dfrac{1}{5}\)

Multiply through by 5 to clear the fraction:

\(25 - (n-1) = -1\)

\(25 + 1 = n - 1\)

\(n - 1 = 26\)

\(n = 27\)

Final Answer: Common difference \(d = -\dfrac{1}{5}\), Number of terms \(n = 27\).

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 expressions for the required terms.

• \(a_5 = a + (5-1)d = a + 4d\)
• \(a_7 = a + (7-1)d = a + 6d\)

Step 3: Use the condition \(a_5 + a_7 = 52\).

\((a + 4d) + (a + 6d) = 52\)

\(2a + 10d = 52\)

Divide both sides by 2: \(a + 5d = 26\) … (Equation 1)

Step 4: Use the condition \(a_{10} = 46\).

\(a_{10} = a + (10-1)d = a + 9d = 46\) … (Equation 2)

Step 5: Solve the two equations.

From Equation (1): \(a + 5d = 26\)

From Equation (2): \(a + 9d = 46\)

Subtract (1) from (2):

\((a + 9d) - (a + 5d) = 46 - 26\)

\(4d = 20 \Rightarrow d = 5\)

Step 6: Find \(a\).

Put \(d = 5\) in Equation (1): \(a + 5(5) = 26\)

\(a + 25 = 26 \Rightarrow a = 1\)

Step 7: Write the AP.

The first term is \(1\) and common difference is \(5\).

So, the AP is: 1, 6, 11, 16, …

Step 1: Recall the general formula for the \(n^{th}\) term of an arithmetic progression (AP):

\(a_n = a_1 + (n-1)\,d\)

Here, \(a_1 = 12\) and \(d\) is the common difference (unknown).

Step 2: Write expressions for \(a_7\) and \(a_{11}\):

\(a_7 = a_1 + (7-1)\,d = 12 + 6d\)

\(a_{11} = a_1 + (11-1)\,d = 12 + 10d\)

Step 3: We are told that \(a_7\) is 24 less than \(a_{11}\). That means:

\(a_{11} - a_7 = 24\)

Step 4: Substitute the values we found:

\((12 + 10d) - (12 + 6d) = 24\)

\(12 + 10d - 12 - 6d = 24\)

\(4d = 24\)

Step 5: Solve for \(d\):

\(d = \dfrac{24}{4} = 6\)

Step 6: Now find \(a_{20}\):

\(a_{20} = a_1 + (20-1)\,d\)

\(= 12 + 19 \times 6\)

\(= 12 + 114\)

\(= 126\)

Final Answer: \(a_{20} = 126\)

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

\(a_n = a + (n-1)d\), where:

• \(a\) = first term
• \(d\) = common difference
• \(n\) = position of the term

Step 2: For the 9th term, put \(n=9\):

\(a_9 = a + (9-1)d = a + 8d\).

Step 3: We are given that \(a_9 = 0\). So,

\(a + 8d = 0 \; \Rightarrow \; a = -8d\).

Step 4: Now, find the 29th term:

\(a_{29} = a + (29-1)d = a + 28d\).

Substitute \(a = -8d\):

\(a_{29} = -8d + 28d = 20d\).

Step 5: Next, find the 19th term:

\(a_{19} = a + (19-1)d = a + 18d\).

Substitute \(a = -8d\):

\(a_{19} = -8d + 18d = 10d\).

Step 6: Compare the results:

\(a_{29} = 20d\) and \(a_{19} = 10d\).

So, \(a_{29} = 2 \times 10d = 2a_{19}\).

Conclusion: Hence, it is proved that \(a_{29} = 2a_{19}\).

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

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

where \(a\) = first term, \(d\) = common difference, and \(n\) = term number.

Step 2: Identify values from the given AP \(7, 10, 13, \ldots\):

• First term \(a = 7\)
• Common difference \(d = 10 - 7 = 3\)
• We need to check if 55 is a term.

Step 3: Put values in the formula:

\(55 = 7 + (n-1) \times 3\)

Step 4: Simplify the equation:

\(55 - 7 = (n-1) \times 3\)

\(48 = (n-1) \times 3\)

Step 5: Solve for \(n-1\):

\(n-1 = 48 \div 3 = 16\)

Step 6: Find \(n\):

\(n = 16 + 1 = 17\)

Final Answer: 55 is the 17th term of the AP.

Step 1: Recall the property of an Arithmetic Progression (AP).

For three numbers \(A, B, C\) to be in AP, the difference between consecutive terms must be equal:

That is, \(B - A = C - B\).

Step 2: Identify the given terms:

\(A = k^2 + 4k + 8\)

\(B = 2k^2 + 3k + 6\)

\(C = 3k^2 + 4k + 4\)

Step 3: Find \(B - A\):

\(B - A = (2k^2 + 3k + 6) - (k^2 + 4k + 8)\)

= \(2k^2 - k^2 + 3k - 4k + 6 - 8\)

= \(k^2 - k - 2\)

Step 4: Find \(C - B\):

\(C - B = (3k^2 + 4k + 4) - (2k^2 + 3k + 6)\)

= \(3k^2 - 2k^2 + 4k - 3k + 4 - 6\)

= \(k^2 + k - 2\)

Step 5: Use the AP condition \(B - A = C - B\):

\(k^2 - k - 2 = k^2 + k - 2\)

Step 6: Simplify the equation:

Cancel \(k^2\) on both sides:

\(-k - 2 = k - 2\)

Add 2 to both sides:

\(-k = k\)

Add \(k\) to both sides:

\(0 = 2k\)

So, \(k = 0\).

Final Answer: \(k = 0\).

Step 1: Let the three parts in Arithmetic Progression (AP) be \((a - d), a, (a + d)\).

This means the middle term is \(a\), the first term is smaller by \(d\), and the last term is larger by \(d\).

Step 2: According to the question, the sum of the three parts is 207.

So, \((a - d) + a + (a + d) = 207\).

Simplify: \(3a = 207\).

Therefore, \(a = \dfrac{207}{3} = 69\).

Step 3: Now the three numbers are \((69 - d), 69, (69 + d)\).

Step 4: The product of the two smaller parts is given as 4623.

The two smaller parts are \((69 - d)\) and 69.

So, \(69 \times (69 - d) = 4623\).

Step 5: Expand the equation:

\(69 \times 69 - 69d = 4623\).

\(4761 - 69d = 4623\).

Step 6: Bring terms together:

\(4761 - 4623 = 69d\).

\(138 = 69d\).

Step 7: Solve for \(d\):

\(d = \dfrac{138}{69} = 2\).

Step 8: Substitute \(d = 2\):

First part = \(69 - 2 = 67\).

Second part = \(69\).

Third part = \(69 + 2 = 71\).

Final Answer: The three parts are 67, 69, 71.

Step 1: In an Arithmetic Progression (AP), three terms can be written as:

\(a - d, \, a, \, a + d\)

Here, \(a\) is the middle angle, and \(d\) is the common difference.

Step 2: It is given that the greatest angle is twice the least angle.

So, \(a + d = 2(a - d)\).

Step 3: Simplify:

\(a + d = 2a - 2d\)

\(a - 3d = 0 \Rightarrow a = 3d\)

Step 4: The sum of angles in a triangle is \(180^{\circ}\).

\((a - d) + a + (a + d) = 180^{\circ}\)

That is, \(3a = 180^{\circ}\).

Step 5: Solve for \(a\):

\(a = \frac{180^{\circ}}{3} = 60^{\circ}\).

Step 6: Since \(a = 3d\),

\(60^{\circ} = 3d \Rightarrow d = 20^{\circ}\).

Step 7: Now find the three angles:

• Least angle = \(a - d = 60^{\circ} - 20^{\circ} = 40^{\circ}\)
• Middle angle = \(a = 60^{\circ}\)
• Greatest angle = \(a + d = 60^{\circ} + 20^{\circ} = 80^{\circ}\)

Final Answer: The angles are \(40^{\circ}, 60^{\circ}, 80^{\circ}\).

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: For the first AP \(9, 7, 5, \ldots\):

First term \(a_1 = 9\), common difference \(d = 7 - 9 = -2\).

So, \(n\)th term = \(9 + (n-1)(-2) = 9 - 2n + 2 = 11 - 2n\).

Step 3: For the second AP \(24, 21, 18, \ldots\):

First term \(a_1 = 24\), common difference \(d = 21 - 24 = -3\).

So, \(n\)th term = \(24 + (n-1)(-3) = 24 - 3n + 3 = 27 - 3n\).

Step 4: Since the two \(n\)th terms are equal, we set them equal:

\(11 - 2n = 27 - 3n\)

Step 5: Solve for \(n\):

\(11 - 2n = 27 - 3n\)

Bring terms together: \(-2n + 3n = 27 - 11\)

\(n = 16\)

Step 6: Find the term by substituting \(n=16\):

In first AP: \(11 - 2(16) = 11 - 32 = -21\)

In second AP: \(27 - 3(16) = 27 - 48 = -21\)

Final Answer: \(n = 16\) and the common term is \(-21\).

Step 1: Recall the formula for the \(n^{th}\) term of an arithmetic progression (AP):

\(a_n = a + (n-1)d\), where:

• \(a\) = first term
• \(d\) = common difference

Step 2: Write expressions for \(a_3\) and \(a_8\):

\(a_3 = a + (3-1)d = a + 2d\)

\(a_8 = a + (8-1)d = a + 7d\)

Step 3: Add them because the question says \(a_3 + a_8 = 7\):

\((a + 2d) + (a + 7d) = 7\)

\(2a + 9d = 7\)   …(Equation 1)

Step 4: Write expressions for \(a_7\) and \(a_{14}\):

\(a_7 = a + (7-1)d = a + 6d\)

\(a_{14} = a + (14-1)d = a + 13d\)

Step 5: Add them because the question says \(a_7 + a_{14} = -3\):

\((a + 6d) + (a + 13d) = -3\)

\(2a + 19d = -3\)   …(Equation 2)

Step 6: Solve the two equations:

Equation (1): \(2a + 9d = 7\)

Equation (2): \(2a + 19d = -3\)

Subtract Equation (1) from Equation (2):

\((2a + 19d) - (2a + 9d) = -3 - 7\)

\(10d = -10\)

\(d = -1\)

Step 7: Put \(d = -1\) into Equation (1):

\(2a + 9(-1) = 7\)

\(2a - 9 = 7\)

\(2a = 16\)

\(a = 8\)

Step 8: Find \(a_{10}\):

\(a_{10} = a + (10-1)d = a + 9d\)

\(a_{10} = 8 + 9(-1) = 8 - 9 = -1\)

Final Answer: \(a_{10} = -1\)

Step 1: First, note that the sequence is an Arithmetic Progression (AP): \(-2, -4, -6, ...\).

Here, the first term \(a = -2\), and the common difference \(d = -4 - (-2) = -2\).

Step 2: Find the total number of terms in the AP.

We know the last term \(l = -100\).

Formula for the \(n^{th}\) term: \(a_n = a + (n-1) \times d\).

Put the values: \(-100 = -2 + (n-1)(-2)\).

\(-100 = -2 - 2(n-1)\).

\(-100 = -2 - 2n + 2\).

\(-100 = -2n\).

So, \(n = 50\).

This means there are 50 terms in the AP.

Step 3: To find the 12th term from the end, we use:

Position from start = \(n - 12 + 1 = 50 - 12 + 1 = 39\).

So, the 12th term from the end is the 39th term from the beginning.

Step 4: Find the 39th term.

Formula: \(a_{39} = a + (39-1) \times d\).

\(a_{39} = -2 + 38 \times (-2)\).

\(a_{39} = -2 - 76\).

\(a_{39} = -78\).

Final Answer: The 12th term from the end is \(-78\).

Step 1: Recall the formula for the \(n^{th}\) term of an Arithmetic Progression (AP):

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

• Here, \(a = 53\) (the first term).
• \(d = 48 - 53 = -5\) (the common difference).

Step 2: We want the first term that is negative, i.e., \(a_n < 0\).

So, \(53 + (n - 1)(-5) < 0\).

Step 3: Simplify the inequality:

\(53 - 5(n - 1) < 0\)

\(53 - 5n + 5 < 0\)

\(58 - 5n < 0\)

Step 4: Solve for \(n\):

\(58 < 5n\)

\(n > \dfrac{58}{5}\)

\(n > 11.6\)

Step 5: Since \(n\) must be a whole number (term number), the smallest integer greater than 11.6 is \(12\).

Therefore, the 12th term is the first negative term.

Step 1: Understand the condition.

A number leaves remainder 3 when divided by 4 if it looks like:

\(4k + 3\), where \(k\) is an integer (whole number).

Step 2: Find the smallest number greater than 10 of this form.

Check numbers just after 10:

• 11 ÷ 4 = 2 remainder 3 ✅
• So, first number = 11.

Step 3: Find the largest number less than 300 of this form.

Check numbers near 300:

• 299 ÷ 4 = 74 remainder 3 ✅
• So, last number = 299.

Step 4: Notice the sequence.

The numbers are: 11, 15, 19, 23, …, 299.

This is an arithmetic progression (AP) with:

• First term \(a = 11\)
• Common difference \(d = 4\)
• Last term \(l = 299\)

Step 5: Use the formula to find the number of terms in AP:

\(n = \dfrac{l - a}{d} + 1\)

\(n = \dfrac{299 - 11}{4} + 1\)

\(n = \dfrac{288}{4} + 1 = 72 + 1 = 73\)

Final Answer: There are 73 numbers between 10 and 300 that leave remainder 3 when divided by 4.

Step 1: Identify first term, common difference and last term

The first term \(a = -\dfrac{4}{3}\).

The common difference is \(d = -1 - \left(-\dfrac{4}{3}\right) = -1 + \dfrac{4}{3} = \dfrac{1}{3}\).

The last term is given as \(l = 4\dfrac{1}{3} = \dfrac{13}{3}\).

Step 2: Find the number of terms in the AP

We use the formula for the \(n\)-th term of an AP: \(a_n = a + (n-1) d\).

Here \(a_n = l = \dfrac{13}{3}\).

So, \(\dfrac{13}{3} = -\dfrac{4}{3} + (n-1) \cdot \dfrac{1}{3}\).

Multiply through by 3 to remove denominators: \(13 = -4 + (n-1)\).

So, \(13 + 4 = n - 1 \Rightarrow 17 = n - 1 \Rightarrow n = 18\).

Therefore, there are 18 terms in this AP.

Step 3: Find the middle terms

If \(n\) is even, then there are two middle terms. They are the \(\tfrac{n}{2}\)-th term and the \(\left(\tfrac{n}{2} + 1\right)\)-th term.

Here, \(n = 18\), so the middle terms are the 9th term and the 10th term.

Step 4: Calculate the 9th term

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

So, \(a_9 = -\dfrac{4}{3} + (9-1) \cdot \dfrac{1}{3}\).

= \(-\dfrac{4}{3} + 8 \cdot \dfrac{1}{3}\).

= \(-\dfrac{4}{3} + \dfrac{8}{3}\).

= \(\dfrac{4}{3}\).

Step 5: Calculate the 10th term

\(a_{10} = -\dfrac{4}{3} + (10-1) \cdot \dfrac{1}{3}\).

= \(-\dfrac{4}{3} + 9 \cdot \dfrac{1}{3}\).

= \(-\dfrac{4}{3} + \dfrac{9}{3}\).

= \(\dfrac{5}{3}\).

Step 6: Add the two middle terms

\(a_9 + a_{10} = \dfrac{4}{3} + \dfrac{5}{3} = \dfrac{9}{3} = 3\).

Final Answer: The sum of the two middle-most terms is 3.

Step 1: Write down the given values.

• First term, \(a = -5\)
• Last term, \(l = 45\)
• Sum of all terms, \(S = 120\)

Step 2: Use the formula for the sum of an AP.

The sum of \(n\) terms of an AP is:

\(S = \dfrac{n}{2}(a + l)\)

Step 3: Substitute the known values.

\(120 = \dfrac{n}{2}(-5 + 45)\)

\(120 = \dfrac{n}{2} \times 40\)

\(120 = 20n\)

Step 4: Solve for \(n\).

\(20n = 120\)

\(n = \dfrac{120}{20} = 6\)

So, the number of terms is \(n = 6\).

---

Step 5: Use the formula for common difference.

The formula connecting the last term is:

\(l = a + (n - 1)d\)

Step 6: Substitute values.

\(45 = -5 + (6 - 1)d\)

\(45 = -5 + 5d\)

Step 7: Solve for \(d\).

\(45 + 5 = 5d\)

\(50 = 5d\)

\(d = \dfrac{50}{5} = 10\)

Final Answer: Number of terms \(n = 6\), Common difference \(d = 10\).

(i)

We are given the series: \(1, -2, -5, \dots, -236\).

This is an arithmetic progression (AP) because the difference between terms is constant.

First term (\(a\)) = 1.

Common difference (\(d\)) = \(-2 - 1 = -3\).

Last term (\(l\)) = -236.

Formula for last term: \(l = a + (n-1)d\).

Substitute: \(-236 = 1 + (n-1)(-3)\).

\(-236 - 1 = -3(n-1) \Rightarrow -237 = -3(n-1)\).

Divide both sides: \(n-1 = 79 \Rightarrow n = 80\).

Now, sum of \(n\) terms: \(S_n = \dfrac{n}{2}(a + l)\).

\(S_{80} = \dfrac{80}{2}(1 + (-236)) = 40(-235) = -9400\).

So, the sum is \(-9400\).

---

(ii)

The terms are: \(\big(4-\dfrac{1}{n}\big), \big(4-\dfrac{2}{n}\big), \dots, \big(4-\dfrac{n}{n}\big)\).

We can write the sum as:

\(S = \sum_{k=1}^{n} \left(4 - \dfrac{k}{n}\right)\).

Split into two parts:

\(S = \sum_{k=1}^{n} 4 - \sum_{k=1}^{n} \dfrac{k}{n}\).

First part: \(\sum_{k=1}^{n} 4 = 4n\).

Second part: \(\sum_{k=1}^{n} \dfrac{k}{n} = \dfrac{1}{n} \sum_{k=1}^{n} k\).

But \(\sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}\).

So, second part = \(\dfrac{1}{n} \times \dfrac{n(n+1)}{2} = \dfrac{n+1}{2}\).

Therefore, \(S = 4n - \dfrac{n+1}{2}\).

Simplify: \(S = \dfrac{8n - (n+1)}{2} = \dfrac{7n - 1}{2}\).

So, the sum is \(\dfrac{7n-1}{2}\).

---

(iii)

We are asked to find the sum of 11 terms:

\(\dfrac{a-b}{a+b}, \dfrac{3a-2b}{a+b}, \dfrac{5a-3b}{a+b}, \dots\)

Notice that denominator in every term is the same: \((a+b)\).

So, we only need to add the numerators first.

Numerators: \((a-b), (3a-2b), (5a-3b), \dots\)

This forms an arithmetic progression (AP).

First numerator (\(a_1\)) = \(a-b\).

Second numerator = \(3a - 2b\).

So, common difference \(d = (3a - 2b) - (a - b) = 2a - b\).

Number of terms = 11.

Sum of numerators: \(S_{11} = \dfrac{11}{2}[2a_1 + (11-1)d]\).

Substitute: \(S_{11} = \dfrac{11}{2}[2(a-b) + 10(2a-b)]\).

\(= \dfrac{11}{2}[2a - 2b + 20a - 10b]\).

\(= \dfrac{11}{2}[22a - 12b]\).

\(= \dfrac{11}{2} \times 2(11a - 6b)\).

\(= 11(11a - 6b) = 121a - 66b\).

Now divide by denominator \((a+b)\).

So, required sum = \(\dfrac{121a - 66b}{a+b}\).

Step 1: Identify the first term and common difference.

The AP is: \(-2, -7, -12, \ldots\)

So, first term \(a = -2\).

Common difference \(d = -7 - (-2) = -7 + 2 = -5\).

Step 2: Use the nth term formula of an AP.

The nth term is given by: \(a_n = a + (n-1)d\).

We are told that \(a_n = -77\).

So, \(-77 = -2 + (n-1)(-5)\).

Step 3: Solve for n.

\(-77 = -2 - 5(n-1)\)

\(-77 + 2 = -5(n-1)\)

\(-75 = -5(n-1)\)

Divide both sides by -5:

\(n-1 = 15\)

So, \(n = 16\).

Therefore, \(-77\) is the 16th term of the AP.

Step 4: Find the sum up to 16 terms.

Sum of first n terms is given by: \(S_n = \frac{n}{2}[a + a_n]\).

Here, \(n = 16, a = -2, a_{16} = -77\).

So, \(S_{16} = \frac{16}{2}[(-2) + (-77)]\).

\(S_{16} = 8[-79]\)

\(S_{16} = -632\).

Final Answer: The required term is the 16th term and the sum up to it is \(-632\).

Step 1: Write the general term.

We are given \(a_n = 3 - 4n\).

Step 2: Find the first few terms.

For \(n = 1\): \(a_1 = 3 - 4(1) = 3 - 4 = -1\).

For \(n = 2\): \(a_2 = 3 - 4(2) = 3 - 8 = -5\).

For \(n = 3\): \(a_3 = 3 - 4(3) = 3 - 12 = -9\).

Step 3: Check if it is an AP.

Common difference \(d = a_2 - a_1 = -5 - (-1) = -4\).

Also, \(a_3 - a_2 = -9 - (-5) = -4\).

Since the difference is constant, the sequence is an AP with \(d = -4\).

Step 4: Use the formula for sum of first \(n\) terms of an AP.

Formula: \(S_n = \tfrac{n}{2} \big(2a_1 + (n-1)d\big)\).

Step 5: Substitute the values.

Here, \(a_1 = -1\), \(d = -4\), \(n = 20\).

So, \(S_{20} = \tfrac{20}{2}\big(2(-1) + (20-1)(-4)\big)\).

\(= 10 \big(-2 + 19(-4)\big)\).

\(= 10 \big(-2 - 76\big)\).

\(= 10(-78)\).

\(= -780\).

Final Answer: The sequence is an AP with common difference \(d=-4\), and \(S_{20} = -780\).

Step 1: Recall the meaning of \(S_n\)

\(S_n\) represents the sum of the first \(n\) terms of the AP.

Step 2: Find the first term

Put \(n = 1\) in the given formula:

\(S_1 = 1(4(1) + 1) = 1(4 + 1) = 5\).

So, the first term \(a = 5\).

Step 3: Find the second term

Put \(n = 2\):

\(S_2 = 2(4(2) + 1) = 2(8 + 1) = 2 \times 9 = 18\).

But \(S_2\) means the sum of the first two terms:

\(S_2 = a_1 + a_2 = 5 + a_2 = 18\).

So, \(a_2 = 18 - 5 = 13\).

Step 4: Find the common difference

The common difference \(d = a_2 - a_1 = 13 - 5 = 8\).

Step 5: Write the AP

The AP is: \(5, 13, 21, …\).

Step 1: We know that \(S_n\) is the sum of the first \(n\) terms of the AP.

Step 2: The formula to find the \(n^{th}\) term is:

\(a_n = S_n - S_{n-1}\)

This is because the difference between the sum of first \(n\) terms and the sum of first \((n-1)\) terms gives the \(n^{th}\) term.

Step 3: Write \(S_n = 3n^2 + 5n\).

Now find \(S_{n-1}\):

\(S_{n-1} = 3(n-1)^2 + 5(n-1)\)

\(= 3(n^2 - 2n + 1) + 5n - 5\)

\(= 3n^2 - 6n + 3 + 5n - 5\)

\(= 3n^2 - n - 2\)

Step 4: Now find \(a_n = S_n - S_{n-1}\):

\(a_n = (3n^2 + 5n) - (3n^2 - n - 2)\)

\(= 3n^2 + 5n - 3n^2 + n + 2\)

\(= 6n + 2\)

Step 5: The general term is \(a_n = 6n + 2\).

We are told \(a_k = 164\).

So, \(6k + 2 = 164\).

Step 6: Solve for \(k\):

\(6k = 164 - 2 = 162\)

\(k = 162 / 6 = 27\)

Final Answer: \(k = 27\).

Step 1: Recall the formula for the sum of first \(n\) terms of an arithmetic progression (AP):

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

where:

• \(a\) = first term
• \(d\) = common difference
• \(n\) = number of terms

Step 2: Find \(S_{12}\).

\[ S_{12} = \frac{12}{2} \big( 2a + (12-1)d \big) \]

\[ S_{12} = 6(2a + 11d) \]

\[ S_{12} = 12a + 66d \]

Step 3: Find \(S_8\).

\[ S_8 = \frac{8}{2} \big( 2a + (8-1)d \big) \]

\[ S_8 = 4(2a + 7d) \]

Step 4: Find \(S_4\).

\[ S_4 = \frac{4}{2} \big( 2a + (4-1)d \big) \]

\[ S_4 = 2(2a + 3d) \]

Step 5: Calculate \(S_8 - S_4\).

\[ S_8 - S_4 = 4(2a + 7d) - 2(2a + 3d) \]

\[ = (8a + 28d) - (4a + 6d) \]

\[ = 4a + 22d \]

Step 6: Multiply the result by 3.

\[ 3(S_8 - S_4) = 3(4a + 22d) \]

\[ = 12a + 66d \]

Step 7: Compare with \(S_{12}\).

We found that:

• \(S_{12} = 12a + 66d\)
• \(3(S_8 - S_4) = 12a + 66d\)

Since both are equal, the given relation is proved.

Step 1: Recall formulas.

• General term of an AP: \(a_n = a + (n-1)d\)
• Sum of first \(n\) terms: \(S_n = \dfrac{n}{2}[2a + (n-1)d]\)

Step 2: Write equations using given terms.

We know:

• \(a_4 = a + 3d = -15\)  ...(i)
• \(a_9 = a + 8d = -30\)  ...(ii)

Step 3: Subtract the equations to find \(d\).

(ii) – (i):

\((a + 8d) - (a + 3d) = -30 - (-15)\)

\(5d = -15\)

So, \(d = -3\).

Step 4: Substitute \(d\) in equation (i) to find \(a\).

\(a + 3(-3) = -15\)

\(a - 9 = -15\)

\(a = -6\).

Step 5: Use the sum formula for \(n = 17\).

\(S_{17} = \dfrac{17}{2}[2a + (17-1)d]\)

\(= \dfrac{17}{2}[2(-6) + 16(-3)]\)

\(= \dfrac{17}{2}[-12 - 48]\)

\(= \dfrac{17}{2}[-60]\)

\(= 17 \times -30\)

\(= -510\).

Final Answer: \(S_{17} = -510\)

Step 1: Recall the formula for the sum of first \(n\) terms of an AP.

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

Here, \(a\) = first term, \(d\) = common difference, \(n\) = number of terms.

Step 2: Write the equation for \(S_6 = 36\).

\[ S_6 = \frac{6}{2}\big(2a + (6-1)d\) = 36 \]

Simplify:

\[ 3(2a + 5d) = 36 \]

\[ 2a + 5d = 12 \quad ...(1) \]

Step 3: Write the equation for \(S_{16} = 256\).

\[ S_{16} = \frac{16}{2}\big(2a + (16-1)d\) = 256 \]

Simplify:

\[ 8(2a + 15d) = 256 \]

\[ 2a + 15d = 32 \quad ...(2) \]

Step 4: Subtract (1) from (2) to find \(d\).

\[ (2a + 15d) - (2a + 5d) = 32 - 12 \]

\[ 10d = 20 \]

\[ d = 2 \]

Step 5: Put the value of \(d\) into (1) to find \(a\).

\[ 2a + 5(2) = 12 \]

\[ 2a + 10 = 12 \]

\[ 2a = 2 \quad \Rightarrow \quad a = 1 \]

Step 6: Use the formula again to find \(S_{10}\).

\[ S_{10} = \frac{10}{2}\big(2a + (10-1)d\) \]

Substitute \(a=1\), \(d=2\):

\[ S_{10} = 5(2(1) + 9(2)) \]

\[ S_{10} = 5(2 + 18) = 5 \times 20 = 100 \]

Final Answer: \(S_{10} = 100\)

Step 1: We are told the AP has 11 terms. So, \(n = 11\).

Step 2: The sum of \(n\) terms of an AP is given by:

\( S_n = \dfrac{n}{2} (\text{first term} + \text{last term}) \).

Step 3: When the number of terms is odd, there is exactly one middle term. For \(n = 11\), the middle term is the \(\dfrac{11+1}{2} = 6^{th}\) term.

Step 4: It is given that this middle term = 30.

Step 5: A property of AP says: The sum of all terms of an AP with an odd number of terms is equal to the number of terms multiplied by the middle term.

So, \( S_{11} = 11 \times 30 \).

Step 6: Multiply: \( 11 \times 30 = 330 \).

Final Answer: \( S_{11} = 330 \).

Step 1: Identify the first term and common difference.

The AP is \(8, 10, 12, \ldots, 126\).

So, first term \(a = 8\) and common difference \(d = 10 - 8 = 2\).

Step 2: Find the total number of terms in the AP.

The last term is 126. Use the nth term formula:

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

\(126 = 8 + (n-1) \times 2\)

\(126 - 8 = (n-1) \times 2\)

\(118 = 2(n-1)\)

\(n-1 = 59\)

\(n = 60\)

So, there are 60 terms in total.

Step 3: Find which terms are the last 10 terms.

The last 10 terms are from 51st to 60th.

So we need \(a_{51}\) and \(a_{60}\).

Step 4: Find the 51st term.

Formula: \(a_n = a + (n-1)d\)

\(a_{51} = 8 + (51-1) \times 2 = 8 + 100 = 108\).

Step 5: Find the 60th term.

\(a_{60} = 8 + (60-1) \times 2 = 8 + 118 = 126\).

Step 6: Use sum formula for an AP.

Sum of \(n\) terms = \(\dfrac{n}{2}(first + last)\).

Here, \(n = 10\), first = 108, last = 126.

So, Sum = \(\dfrac{10}{2}(108 + 126) = 5 \times 234 = 1170\).

Final Answer: The sum of the last 10 terms is 1170.

Step 1: We need numbers that are multiples of both 2 and 9.

The smallest number that is a multiple of both 2 and 9 is their Least Common Multiple (LCM).

LCM of 2 and 9 is:

2 = \(2\)

9 = \(3 \times 3\)

LCM = \(2 \times 3 \times 3 = 18\).

Step 2: So, the first number which is a multiple of both 2 and 9 is 18.

The sequence of such numbers will be: \(18, 36, 54, 72, 90, 108, 126, \ldots\)

Step 3: We only need the first 7 terms:

\(18, 36, 54, 72, 90, 108, 126\).

Step 4: This is an Arithmetic Progression (AP) because the difference between terms is constant.

Common difference (d) = \(36 - 18 = 18\).

Step 5: Formula for the sum of first \(n\) terms of an AP is:

\[ S_n = \frac{n}{2} \times (a + l) \]

where,

\(n = 7\) (number of terms),

\(a = 18\) (first term),

\(l = 126\) (last term).

Step 6: Substitute the values:

\[ S_7 = \frac{7}{2} \times (18 + 126) \]

\[ S_7 = \frac{7}{2} \times 144 \]

\[ S_7 = 7 \times 72 \]

\[ S_7 = 504 \]

Final Answer: The sum of the first seven numbers which are multiples of both 2 and 9 is 504.

Step 1: Write down what is given.

• First term, \(a = -15\)
• Common difference, \(d = -13 - (-15) = 2\)
• Required sum, \(S_n = -55\)

Step 2: Recall the formula for the sum of first \(n\) terms of an AP:

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

Step 3: Substitute the values \(a = -15\) and \(d = 2\):

\[ S_n = \frac{n}{2} \big( 2(-15) + (n-1) \times 2 \big) \]

\[ S_n = \frac{n}{2} ( -30 + 2n - 2 ) \]

\[ S_n = \frac{n}{2} ( 2n - 32 ) \]

\[ S_n = n(n - 16) \]

Step 4: We are told that \(S_n = -55\). So,

\[ n(n - 16) = -55 \]

\[ n^2 - 16n + 55 = 0 \]

Step 5: Solve the quadratic equation.

Factorize: \[ n^2 - 16n + 55 = 0 \]

\[ (n - 5)(n - 11) = 0 \]

So, \(n = 5\) or \(n = 11\).

Step 6: Check both answers.

• For \(n = 5\): First 5 terms are \(-15, -13, -11, -9, -7\). Their sum is \(-55\). ✔
• For \(n = 11\): First 11 terms add up to \(-55\) again because the extra 6 terms after the 5th can be grouped in pairs like \((-5 + 5), (-3 + 3), (-1 + 1)\), each pair giving 0. So the total remains \(-55\). ✔

Final Answer: The required number of terms is 5 or 11.

Step 1: Recall the formula for the sum of first \(n\) terms of an AP:

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

Step 2: Write the sum of the first \(n\) terms of the first AP.

Here, \(a = 8\), \(d = 20\).

\[ S_n = \frac{n}{2}\Big(2(8) + (n-1)(20)\Big) \]

\[ S_n = \frac{n}{2}\Big(16 + 20n - 20\Big) \]

\[ S_n = \frac{n}{2}(20n - 4) \]

\[ S_n = n(10n - 2) \]

Step 3: Write the sum of the first \(2n\) terms of the second AP.

Here, \(a = -30\), \(d = 8\), and the number of terms is \(2n\).

\[ S_{2n} = \frac{2n}{2}\Big(2(-30) + (2n-1)(8)\Big) \]

\[ S_{2n} = n\Big(-60 + (2n-1)(8)\Big) \]

\[ S_{2n} = n(-60 + 16n - 8) \]

\[ S_{2n} = n(16n - 68) \]

Step 4: According to the question,

\[ S_n = S_{2n} \]

So,

\[ n(10n - 2) = n(16n - 68) \]

Step 5: Simplify the equation.

Since \(n \neq 0\), divide both sides by \(n\):

\[ 10n - 2 = 16n - 68 \]

Step 6: Solve for \(n\).

\[ -2 + 68 = 16n - 10n \]

\[ 66 = 6n \]

\[ n = 11 \]

Final Answer: \(n = 11\)

Step 1: Kanika deposits money every day. On day 1 she puts Rs 1, on day 2 she puts Rs 2, … up to day 31 she puts Rs 31. This forms a series:

\(1 + 2 + 3 + \dots + 31\).

Step 2: To find the total of this series, we use the formula for the sum of the first \(n\) natural numbers:

\(S = \dfrac{n(n+1)}{2}\).

Step 3: Here, \(n = 31\). So,

\(S = \dfrac{31 \times (31+1)}{2} = \dfrac{31 \times 32}{2}\).

\(S = \dfrac{992}{2} = 496\).

This means Kanika saved Rs 496 in total.

Step 4: We are told she also spent Rs 204 and still had Rs 100 left with her at the end of the month.

Step 5: So, her total pocket money must be the sum of all three amounts:

Total pocket money = Savings + Amount spent + Amount left

= \(496 + 204 + 100\).

Step 6: Add them: \(496 + 204 = 700\). Then, \(700 + 100 = 800\).

Final Answer: Kanika’s pocket money for the month = Rs 800.

Step 1: The savings form an Arithmetic Progression (AP): 32, 36, 40, ...

Here, first term \(a = 32\), common difference \(d = 4\).

Step 2: To find in how many months Yasmeen saves Rs 2000, we need the formula for the sum of \(n\) terms of an AP:

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

Step 3: Substitute the values of \(a=32\) and \(d=4\):

\[ S_n = \frac{n}{2} [2 \times 32 + (n-1) \times 4] \]

\[ S_n = \frac{n}{2} [64 + 4n - 4] \]

\[ S_n = \frac{n}{2} [4n + 60] \]

\[ S_n = n(2n + 30) \]

Step 4: We want \(S_n = 2000\):

\[ n(2n + 30) = 2000 \]

Step 5: Expand and simplify:

\[ 2n^2 + 30n = 2000 \]

\[ 2n^2 + 30n - 2000 = 0 \]

Divide the whole equation by 2:

\[ n^2 + 15n - 1000 = 0 \]

Step 6: Solve the quadratic equation \(n^2 + 15n - 1000 = 0\):

By quadratic formula:

\[ n = \frac{-15 \pm \sqrt{15^2 - 4(1)(-1000)}}{2} \]

\[ n = \frac{-15 \pm \sqrt{225 + 4000}}{2} \]

\[ n = \frac{-15 \pm \sqrt{4225}}{2} \]

\[ n = \frac{-15 \pm 65}{2} \]

Step 7: Calculate the two roots:

\[ n = \frac{-15 + 65}{2} = \frac{50}{2} = 25 \]

\[ n = \frac{-15 - 65}{2} = -40 \; (\text{not possible since months cannot be negative}) \]

Final Answer: Yasmeen will save Rs 2000 in 25 months.