GSEB Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.1

Question 1. In which of the following situations does the list of numbers involved make an arithmetic progression and why?
(i) The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional kilometre.
(ii) The amount of air present in a cylinder when a vacuum pump removes \( \frac { 1 }{ 4 } \) of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it cost Rs. 150, for the first metre and rises by Rs. 50 for each subsequent metre.
(iv) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at 8% per annum.
Answer:
(i) The taxi fare for the initial kilometre is Rs. 15. For the next 2 kilometres, the total fare becomes Rs. \( 15 + 8 = 23 \). For the third kilometre, it's Rs. \( 15 + (2 \times 8) = 31 \). Similarly, for the fourth kilometre, the total cost is Rs. \( 15 + (3 \times 8) = 39 \). So, the amounts are 15, 23, 31, 39, and so on. This sequence forms an Arithmetic Progression (AP) because each subsequent term increases by a fixed amount of Rs. 8 compared to the previous term.
In simple words: The taxi charges start at Rs. 15 for the first kilometre, then Rs. 8 is added for each extra kilometre. This forms a list of numbers where each new number is Rs. 8 more than the last, so it is an AP.

Exam Tip: Always clearly state the first term and the common difference to justify if a sequence forms an AP.

(ii) Let's assume the starting air volume in the cylinder is \( v \) litres. With each pump stroke, \( \frac { 1 }{ 4 } \) of the existing air is removed. After the initial stroke, the remaining air is \( v - \frac {1}{4} v = \frac { 3v }{ 4 } \). Following the second stroke, the air left is \( \frac { 3v }{ 4 } - \frac { 1 }{ 4 } \left( \frac { 3v }{ 4 } \right) = \frac { 3v }{ 4 } - \frac { 3v }{ 16 } = \frac { 12v-3v }{ 16 } = \frac { 9v }{ 16 } \). After the third stroke, the air remaining will be \( \frac { 9v }{16} - \frac {1}{4} \times \frac { 9v }{ 16 } = \frac { 9v }{ 16 } \left( 1-\frac { 1 }{ 4 } \right) = \frac { 9v }{ 16 } \times \frac { 3 }{ 4 } = \frac { 27v }{ 64 } \). For the fourth stroke, the air left is \( \frac { 27v }{ 64 } – \frac { 1 }{ 4 } \left( \frac { 27v }{ 64 } \right) = \frac { 27v }{ 64 } \left( 1-\frac { 1 }{ 4 } \right) = \frac { 27v }{ 64 } \times \frac { 3 }{ 4 } = \frac { 81v }{ 256 } \). So, the series of air volumes is \( v, \frac {3v }{ 4 }, \frac { 9v }{ 16 }, \frac { 27v }{ 64 }, \frac { 81v }{ 256 }, \dots \). Now, let's find the differences between consecutive terms: \( a_2 – a_1 = \frac { 3v }{ 4 } - v = \frac { -v }{ 4 } \). Also, \( a_3 – a_2 = \frac { 9v }{ 16 } – \frac { 3v }{ 4 } = \frac { 9v-12v }{ 16 } = \frac { -3v }{ 16 } \). Since \( a_2 – a_1 \ne a_3 - a_2 \), the difference between successive terms is not constant. Therefore, this sequence does not form an AP.
In simple words: If you remove a quarter of the air each time, the amount of air left gets smaller, but not by the same amount each time. This means the list of numbers doesn't follow a steady pattern, so it's not an AP.

Exam Tip: For problems involving percentage or fractional removal/addition, remember that the common difference is often not constant, indicating it's likely not an AP.

(iii) It is stated that the cost to dig the first metre of a well is Rs. 150. For the second metre, the total cost becomes Rs. \( 150 + 50 = 200 \). For the third metre, the cost is Rs. \( 150 + (2 \times 50) = 250 \). And for the fourth metre, the total cost will be Rs. \( 150 + (3 \times 50) = 300 \). We can see that the sequence of costs 150, 200, 250, 300 forms an Arithmetic Progression (AP) because each new term is exactly Rs. 50 higher than the one before it.
In simple words: The initial cost to dig a well is Rs. 150, and then Rs. 50 is added for each extra metre. This creates a list of costs where each new cost is Rs. 50 more than the last, which means it is an AP.

Exam Tip: Always identify the first term (a) and the common difference (d) clearly when determining if a sequence is an AP.

(iv) Let \( P \) represent the initial amount deposited, \( r \) be the annual compound interest rate, and \( n \) be the number of years. The money accumulated after \( n \) years is calculated as \( P\left(1+\frac { r }{ 100 } \right)^n \). Here, the principal amount \( P = 10000 \) Rs. and the interest rate \( r = 8\% \). So, the amount after 1 year is \( 10000 \left( 1+\frac { 8 }{ 100 } \right)^1 \). After 2 years, it becomes \( 10000\left(1+\frac { 8 }{ 100 } \right)^2 \). And after 3 years, the amount is \( 10000\left( 1+\frac { 8 }{ 100 } \right)^3 \). This generates a sequence like \( 10000\left(1+\frac { 8 }{ 100 } \right)^1, 10000\left( 1+\frac { 8 }{ 100 } \right)^2, 10000\left(1+\frac { 8 }{ 100 } \right)^3, \dots \). We can observe that the difference between any two successive terms is not constant. Therefore, this series of numbers does not form an Arithmetic Progression.
In simple words: When money earns compound interest, it grows by a percentage each year, not a fixed amount. This means the increase isn't the same every year, so the amounts won't form an AP.

Exam Tip: Always remember that compound interest leads to exponential growth, while simple interest or a constant increment creates an AP.

Question 2. Write first four terms of the AP when the first term a and the common difference d are given as follows.
(i) a = 10, d = 10
(ii) a = 4, d = -3
(iii) a = 2, d = 0
(iv) a = -2, d = \( \frac{1}{2} \)
(v) a = -1.25, d = -0.25
Answer:
(i) Given that \( a = 10 \) and \( d = 10 \).
The first four terms of the AP are:
\( a_1 = a = 10 \)
\( a_2 = a + d = 10 + 10 = 20 \)
\( a_3 = a + 2d = 10 + (2 \times 10) = 10 + 20 = 30 \)
\( a_4 = a + 3d = 10 + (3 \times 10) = 10 + 30 = 40 \)
Therefore, the first four terms of the AP are: 10, 20, 30, 40.
In simple words: Start with 10, then keep adding 10 to get the next number. The first four numbers are 10, 20, 30, and 40.

Exam Tip: The nth term of an AP is given by \( a_n = a + (n-1)d \). Use this formula to avoid calculation errors.

(ii) Given that \( a = 4 \) and \( d = -3 \).
The first four terms of the AP are:
\( a_1 = a = 4 \)
\( a_2 = a + d = 4 + (-3) = 1 \)
\( a_3 = a + 2d = 4 + (2 \times -3) = 4 - 6 = -2 \)
\( a_4 = a + 3d = 4 + (3 \times -3) = 4 - 9 = -5 \)
So, the first four terms of the AP are: 4, 1, -2, -5.
In simple words: Begin with 4, then repeatedly subtract 3. The first four numbers are 4, 1, -2, and -5.

Exam Tip: Be careful with negative common differences; ensure you correctly subtract or add the negative value.

(iii) Given that \( a = -2 \) and \( d = 0 \).
The first four terms of the AP are:
\( a_1 = a = -2 \)
\( a_2 = a + d = -2 + 0 = -2 \)
\( a_3 = a + 2d = -2 + (2 \times 0) = -2 \)
\( a_4 = a + 3d = -2 + (3 \times 0) = -2 \)
Therefore, the first four terms of the AP are: -2, -2, -2, -2.
In simple words: Start with 2, then always add 0. The first four numbers will all be 2.

Exam Tip: An AP with a common difference of zero means all terms are identical to the first term.

(iv) Given that \( a = -2 \) and \( d = \frac{1}{2} \).
The first four terms of the AP are:
\( a_1 = a = -2 \)
\( a_2 = a + d = -2 + \frac{1}{2} = -\frac{4}{2} + \frac{1}{2} = -\frac{3}{2} \)
\( a_3 = a + 2d = -2 + (2 \times \frac{1}{2}) = -2 + 1 = -1 \)
\( a_4 = a + 3d = -2 + (3 \times \frac{1}{2}) = -2 + \frac{3}{2} = -\frac{4}{2} + \frac{3}{2} = -\frac{1}{2} \)
Hence, the first four terms of the AP are: \( -2, -\frac{3}{2}, -1, -\frac{1}{2} \).
In simple words: Start with -2, then keep adding \( \frac{1}{2} \). The first four numbers are -2, \( -\frac{3}{2} \), -1, and \( -\frac{1}{2} \).

Exam Tip: When working with fractions, convert all numbers to a common denominator to simplify calculations.

(v) Given that \( a = -1.25 \) and \( d = -0.25 \).
The first four terms of the AP are:
\( a_1 = a = -1.25 \)
\( a_2 = a + d = -1.25 + (-0.25) = -1.50 \)
\( a_3 = a + 2d = -1.25 + (2 \times -0.25) = -1.25 - 0.50 = -1.75 \)
\( a_4 = a + 3d = -1.25 + (3 \times -0.25) = -1.25 - 0.75 = -2.00 \)
Hence, the first four terms of the AP are: -1.25, -1.50, -1.75, -2.00.
In simple words: Start with -1.25, then continuously subtract 0.25. The first four numbers are -1.25, -1.50, -1.75, and -2.00.

Exam Tip: Decimal calculations must be precise, especially when dealing with negative numbers and multiple operations.

Question 3. For the following APs, write the first term and the common difference.
(i) 3, 1, -1, -3, ...
(ii) -5, -1, 3, 7, ...
(iii) \( \frac { 1 }{ 3 }, \frac { 5 }{ 3 }, \frac { 9 }{ 3 }, \frac { 13 }{ 3 } \)
(iv) 0.6, 1.7, 2.8, 3.9, ...
Answer:
(i) As the given AP is 3, 1, -1, -3, .....
The first term \( a = 3 \).
The common difference \( d = a_2 - a_1 = 1 - 3 = -2 \).
In simple words: The first term is 3. To find the common difference, subtract the first term from the second, which gives -2.

Exam Tip: The common difference is always \( a_n - a_{n-1} \). Always check at least two pairs to ensure it's truly an AP.

(ii) As the given AP is -5, -1, 3, 7, .....
The first term \( a = -5 \).
The common difference \( d = a_2 - a_1 = -1 - (-5) = -1 + 5 = 4 \).
In simple words: The starting term is -5. The common difference is found by subtracting -5 from -1, resulting in 4.

Exam Tip: Be careful with signs when subtracting negative numbers; \( -(-x) \) becomes \( +x \).

(iii) As the given AP is \( \frac { 1 }{ 3 }, \frac { 5 }{ 3 }, \frac { 9 }{ 3 }, \frac { 13 }{ 3 } \).
The first term \( a = \frac{1}{3} \).
The common difference \( d = a_2 - a_1 = \frac{5}{3} - \frac{1}{3} = \frac{4}{3} \).
In simple words: The first term is \( \frac{1}{3} \). To find the common difference, subtract \( \frac{1}{3} \) from \( \frac{5}{3} \), which gives \( \frac{4}{3} \).

Exam Tip: When terms are fractions with the same denominator, calculating the common difference is straightforward by subtracting the numerators.

(iv) As the given AP is 0.6, 1.7, 2.8, 3.9, .....
The first term \( a = 0.6 \).
The common difference \( d = a_2 - a_1 = 1.7 - 0.6 = 1.1 \).
In simple words: The first term is 0.6. The common difference is 1.1, found by subtracting 0.6 from 1.7.

Exam Tip: Decimal subtraction requires aligning decimal points for accurate calculations.

Question 4. Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16,
(ii) \( 2, \frac {5}{2}, 3, \frac { 7 }{ 2 } \),
(iii) -1.2, -3.2, -5.2, -7.2.
(iv) -10, -6, -2, 2,
(v) \( 3, 3+\sqrt { 2 }, 3+2\sqrt {2 }, 3+3\sqrt { 2 } \)
(vi) 0.2, 0.22, 0.222, 0.2222,
(vii) 0, -4, -8, -12,
(viii) \( -\frac {1}{ 2 }, -\frac { -1 }{ 2 }, -\frac { -1 }{ 2 }, -\frac { -1 }{ 2 } \)
(ix) 1, 3, 9, 27,
(x) a, 2a, 3a, 4a,
(xi) \( a, a^2, a^3, a^4 \),
(xii) \( \sqrt { 2 }, \sqrt { 8 }, \sqrt { 18 }, \sqrt { 32 } \)
(xiii) \( \sqrt { 3 }, \sqrt { 6 }, \sqrt { 9 }, \sqrt { 12 } \),
(xiv) \( 1^2, 3^2, 5^2, 7^2 \),
(xv) \( 1^2, 5^2, 7^2, 73 \),
Answer:
(i) The given list of numbers is: 2, 4, 8, 16.
First difference: \( a_2 - a_1 = 4 - 2 = 2 \)
Second difference: \( a_3 - a_2 = 8 - 4 = 4 \)
Here, \( a_2 - a_1 \ne a_3 - a_2 \).
Hence, the given sequence is not an AP.
In simple words: The difference between 4 and 2 is 2, but the difference between 8 and 4 is 4. Since the difference isn't constant, this is not an AP.

Exam Tip: An AP must have a constant difference between consecutive terms. Always check at least two differences.

(ii) Here the given list of numbers is \( 2, \frac {5}{2}, 3, \frac { 7 }{ 2 } \).
First difference: \( a_2 - a_1 = \frac {5}{2} – 2 = \frac {5-4}{2} = \frac { 1 }{ 2 } \)
Second difference: \( a_3 - a_2 = 3 - \frac {5}{2} = \frac {6-5}{2} = \frac { 1 }{ 2 } \)
Third difference: \( a_4 - a_3 = \frac { 7 }{ 2 } – 3 = \frac {7-6}{2} = \frac { 1 }{ 2 } \)
Since the common difference \( d = \frac{1}{2} \) is constant, the given sequence is an AP.
The next three terms are:
\( a_5 = a_4 + d = \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4 \)
\( a_6 = a_5 + d = 4 + \frac{1}{2} = \frac{8+1}{2} = \frac{9}{2} \)
\( a_7 = a_6 + d = \frac{9}{2} + \frac{1}{2} = \frac{10}{2} = 5 \)
Hence, the next three terms of the AP are 4, \( \frac{9}{2} \), 5.
In simple words: The difference between each number is always \( \frac{1}{2} \), so it is an AP. The next three numbers are 4, \( \frac{9}{2} \), and 5.

Exam Tip: Converting mixed numbers or whole numbers to fractions with a common denominator can simplify finding the common difference.

(iii) The given sequence is -1.2, -3.2, -5.2, -7.2.
First difference: \( a_2 - a_1 = -3.2 - (-1.2) = -3.2 + 1.2 = -2 \)
Second difference: \( a_3 - a_2 = -5.2 - (-3.2) = -5.2 + 3.2 = -2 \)
Third difference: \( a_4 - a_3 = -7.2 - (-5.2) = -7.2 + 5.2 = -2 \)
The common difference \( d = -2 \) is constant, so the given sequence is in AP.
The next three terms are:
\( a_5 = a_4 + d = -7.2 + (-2) = -9.2 \)
\( a_6 = a_5 + d = -9.2 + (-2) = -11.2 \)
\( a_7 = a_6 + d = -11.2 + (-2) = -13.2 \)
Hence, the next three terms of the AP are -9.2, -11.2, -13.2.
In simple words: Each number is 2 less than the previous one, so it's an AP. The next three numbers are -9.2, -11.2, and -13.2.

Exam Tip: Practice subtracting negative decimals carefully to avoid sign errors.

(iv) Given sequence is -10, -6, -2, 2.
First difference: \( a_2 - a_1 = -6 - (-10) = -6 + 10 = 4 \)
Second difference: \( a_3 - a_2 = -2 - (-6) = -2 + 6 = 4 \)
Third difference: \( a_4 - a_3 = 2 - (-2) = 2 + 2 = 4 \)
Since the given sequence has the same common difference, it is an AP.
The next three terms of the AP are:
\( a_5 = a_4 + d = 2 + 4 = 6 \)
\( a_6 = a_5 + d = 6 + 4 = 10 \)
\( a_7 = a_6 + d = 10 + 4 = 14 \)
In simple words: Each number is 4 more than the one before it, so it's an AP. The next three numbers are 6, 10, and 14.

Exam Tip: When calculating differences with negative numbers, remember that subtracting a negative is equivalent to adding a positive.

(v) Given sequence is \( 3, 3+\sqrt { 2 }, 3+2\sqrt {2 }, 3+3\sqrt { 2 } \).
First difference: \( a_2 - a_1 = (3+\sqrt{2}) - 3 = \sqrt{2} \)
Second difference: \( a_3 - a_2 = (3+2\sqrt{2}) - (3+\sqrt{2}) = 3+2\sqrt{2} - 3 - \sqrt{2} = \sqrt{2} \)
Third difference: \( a_4 - a_3 = (3+3\sqrt{2}) - (3+2\sqrt{2}) = 3+3\sqrt{2} - 3 - 2\sqrt{2} = \sqrt{2} \)
The difference between each two consecutive terms is the same.
Hence, the given sequence is an AP with \( d = \sqrt{2} \).
The next three terms are:
\( a_5 = a_4 + d = (3+3\sqrt{2}) + \sqrt{2} = 3+4\sqrt{2} \)
\( a_6 = a_5 + d = (3+4\sqrt{2}) + \sqrt{2} = 3+5\sqrt{2} \)
\( a_7 = a_6 + d = (3+5\sqrt{2}) + \sqrt{2} = 3+6\sqrt{2} \)
In simple words: Each number increases by \( \sqrt{2} \) from the previous one, so it is an AP. The next three numbers are \( 3+4\sqrt{2} \), \( 3+5\sqrt{2} \), and \( 3+6\sqrt{2} \).

Exam Tip: Treat \( \sqrt{2} \) as a variable for common difference calculations; combine like terms.

(vi) Given sequence is 0.2, 0.22, 0.222, 0.2222.
First difference: \( a_2 - a_1 = 0.22 - 0.2 = 0.02 \)
Second difference: \( a_3 - a_2 = 0.222 - 0.22 = 0.002 \)
Since \( a_2 - a_1 \ne a_3 - a_2 \), the given sequence does not form an AP.
In simple words: The differences between the numbers are 0.02 and 0.002, which are not the same. So, this is not an AP.

Exam Tip: A sequence where digits are added incrementally (like 0.2, 0.22, 0.222) typically does not form an AP.

(vii) Given sequence is 0, -4, -8, -12.
First difference: \( a_2 - a_1 = -4 - 0 = -4 \)
Second difference: \( a_3 - a_2 = -8 - (-4) = -8 + 4 = -4 \)
Third difference: \( a_4 - a_3 = -12 - (-8) = -12 + 8 = -4 \)
Since the difference between two consecutive terms is the same, the sequence is in AP with \( d = -4 \).
The next three terms are:
\( a_5 = a_4 + d = -12 + (-4) = -16 \)
\( a_6 = a_5 + d = -16 + (-4) = -20 \)
\( a_7 = a_6 + d = -20 + (-4) = -24 \)
In simple words: Each number is 4 less than the previous one, making it an AP. The next three numbers are -16, -20, and -24.

Exam Tip: Confirm the common difference by calculating at least three differences in the sequence.

(viii) Given sequence is \( -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2} \).
First difference: \( a_2 - a_1 = -\frac{1}{2} - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0 \)
Second difference: \( a_3 - a_2 = -\frac{1}{2} - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0 \)
Third difference: \( a_4 - a_3 = -\frac{1}{2} - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0 \)
Since the difference of each two consecutive terms is the same, therefore the sequence is in AP with \( d = 0 \).
The next three terms are:
\( a_5 = a_4 + d = -\frac{1}{2} + 0 = -\frac{1}{2} \)
\( a_6 = a_5 + d = -\frac{1}{2} + 0 = -\frac{1}{2} \)
\( a_7 = a_6 + d = -\frac{1}{2} + 0 = -\frac{1}{2} \)
In simple words: All the numbers are the same, meaning the difference between them is 0. This is an AP. The next three numbers will also be \( -\frac{1}{2} \).

Exam Tip: A sequence with identical terms is a valid AP with a common difference of zero.

(ix) Given sequence is 1, 3, 9, 27.
First difference: \( a_2 - a_1 = 3 - 1 = 2 \)
Second difference: \( a_3 - a_2 = 9 - 3 = 6 \)
Since \( a_2 - a_1 \ne a_3 - a_2 \), the given sequence does not form an AP.
In simple words: The difference between 3 and 1 is 2, but between 9 and 3 it's 6. Since the difference changes, this is not an AP.

Exam Tip: Geometric progressions (where terms are multiplied by a constant ratio) are often confused with APs. Check for a constant difference, not a constant ratio.

(x) Given sequence is a, 2a, 3a, 4a.
First difference: \( a_2 - a_1 = 2a - a = a \)
Second difference: \( a_3 - a_2 = 3a - 2a = a \)
Third difference: \( a_4 - a_3 = 4a - 3a = a \)
Since the common difference is the same, the given sequence is in AP with \( d = a \).
The next three terms of AP are:
\( a_5 = a_4 + d = 4a + a = 5a \)
\( a_6 = a_5 + d = 5a + a = 6a \)
\( a_7 = a_6 + d = 6a + a = 7a \)
In simple words: Each term is \( a \) more than the last one, so it's an AP. The common difference is \( a \), and the next three terms are \( 5a, 6a, 7a \).

Exam Tip: Algebraic expressions can also form APs if the variable (a) represents a constant value in the sequence.

(xi) Given sequence is \( a, a^2, a^3, a^4 \).
First difference: \( a_2 - a_1 = a^2 - a = a(a - 1) \)
Second difference: \( a_3 - a_2 = a^3 - a^2 = a^2(a - 1) \)
Since \( a_2 - a_1 \ne a_3 - a_2 \) (unless \( a=1 \) or \( a=0 \), which makes it a trivial AP), the given sequence does not form an AP.
In simple words: The difference between terms like \( a^2-a \) and \( a^3-a^2 \) changes unless 'a' is 0 or 1. So, this sequence usually doesn't form an AP.

Exam Tip: Sequences involving powers of a variable are typically geometric progressions, not arithmetic progressions.

(xii) Given sequence is \( \sqrt { 2 }, \sqrt { 8 }, \sqrt { 18 }, \sqrt { 32 } \).
First, simplify the terms:
\( \sqrt{2} \)
\( \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \)
\( \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2} \)
\( \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \)
The sequence becomes \( \sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots \).
First difference: \( a_2 - a_1 = 2\sqrt{2} - \sqrt{2} = \sqrt{2} \)
Second difference: \( a_3 - a_2 = 3\sqrt{2} - 2\sqrt{2} = \sqrt{2} \)
Third difference: \( a_4 - a_3 = 4\sqrt{2} - 3\sqrt{2} = \sqrt{2} \)
Since the common difference is the same, the sequence is in AP with \( d = \sqrt{2} \).
Hence, the next three terms are:
\( a_5 = a_4 + d = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{25 \times 2} = \sqrt{50} \)
\( a_6 = a_5 + d = 5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{36 \times 2} = \sqrt{72} \)
\( a_7 = a_6 + d = 6\sqrt{2} + \sqrt{2} = 7\sqrt{2} = \sqrt{49 \times 2} = \sqrt{98} \)
Thus, the next three terms are \( 5\sqrt{2}, 6\sqrt{2}, 7\sqrt{2} \) or \( \sqrt{50}, \sqrt{72}, \sqrt{98} \).
In simple words: When you simplify the square roots, you see that each number increases by \( \sqrt{2} \). So it's an AP. The next three terms are \( \sqrt{50}, \sqrt{72}, \sqrt{98} \).

Exam Tip: Always simplify terms with square roots before calculating differences to easily identify if it's an AP.

(xiii) Given sequence is \( \sqrt { 3 }, \sqrt { 6 }, \sqrt { 9 }, \sqrt { 12 } \).
First difference: \( a_2 - a_1 = \sqrt{6} - \sqrt{3} = \sqrt{3}(\sqrt{2}-1) \)
Second difference: \( a_3 - a_2 = \sqrt{9} - \sqrt{6} = 3 - \sqrt{6} = \sqrt{3}(\sqrt{3}-\sqrt{2}) \)
Since \( a_2 - a_1 \ne a_3 - a_2 \), the given sequence does not form an AP.
In simple words: The differences between the square root terms are not the same. For example, \( \sqrt{6}-\sqrt{3} \) is not equal to \( \sqrt{9}-\sqrt{6} \). Thus, it's not an AP.

Exam Tip: Only if the simplified radical parts are identical can you treat square roots as like terms for common difference calculations.

(xiv) Given sequence is \( 1^2, 3^2, 5^2, 7^2 \).
Convert to numerical values: \( 1^2 = 1 \), \( 3^2 = 9 \), \( 5^2 = 25 \), \( 7^2 = 49 \).
The sequence is 1, 9, 25, 49.
First difference: \( a_2 - a_1 = 9 - 1 = 8 \)
Second difference: \( a_3 - a_2 = 25 - 9 = 16 \)
Since \( a_2 - a_1 \ne a_3 - a_2 \), the given sequence does not form an AP.
In simple words: The squares of odd numbers give 1, 9, 25, 49. The differences between these numbers are 8, then 16, which is not constant. So, it's not an AP.

Exam Tip: When terms are given as powers, calculate the actual numerical values first to find the differences correctly.

(xv) Given sequence is \( 1^2, 5^2, 7^2, 73 \).
Convert to numerical values: \( 1^2 = 1 \), \( 5^2 = 25 \), \( 7^2 = 49 \).
The sequence is 1, 25, 49, 73.
First difference: \( a_2 - a_1 = 25 - 1 = 24 \)
Second difference: \( a_3 - a_2 = 49 - 25 = 24 \)
Third difference: \( a_4 - a_3 = 73 - 49 = 24 \)
Since the common difference is the same, the sequence is in AP with \( d = 24 \).
The next three terms are:
\( a_5 = a_4 + d = 73 + 24 = 97 \)
\( a_6 = a_5 + d = 97 + 24 = 121 \)
\( a_7 = a_6 + d = 121 + 24 = 145 \)
In simple words: The sequence 1, 25, 49, 73 has a constant difference of 24 between its terms. This means it is an AP. The next three terms are 97, 121, and 145.

Exam Tip: Ensure all terms are actual numbers before calculating differences, especially when powers are involved.