Samacheer Kalvi Class 10 Maths Solutions Chapter 2 Numbers and Sequences Exercise 2.4

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.4

Question 1. Find the next three terms of the following sequence.
(i) 8, 24, 72,...
(ii) 5, 1, -3, ...
(iii) \( \frac { 1 }{ 4 } \), \( \frac { 2 }{ 9 } \), \( \frac { 3 }{ 16 } \)
Answer:
(i) For the sequence 8, 24, 72,..., each term is 3 times the previous term. So, the next three terms are obtained by multiplying by 3 repeatedly: 72 \( \times \) 3 = 216, 216 \( \times \) 3 = 648, and 648 \( \times \) 3 = 1944. This is a geometric progression.
(ii) For the sequence 5, 1, -3,..., each term is obtained by subtracting 4 from the previous term. So, the next three terms are: -3 - 4 = -7, -7 - 4 = -11, and -11 - 4 = -15. This forms an arithmetic progression.
(iii) For the sequence \( \frac { 1 }{ 4 } \), \( \frac { 2 }{ 9 } \), \( \frac { 3 }{ 16 } \), the general term follows the pattern \( \frac{n}{(n+1)^{2}} \). So, for the 4th, 5th, and 6th terms:
For n = 4: \( \frac { 4 }{ (4+1)^{2} } = \frac { 4 }{ 5^{2} } = \frac { 4 }{ 25 } \)
For n = 5: \( \frac { 5 }{ (5+1)^{2} } = \frac { 5 }{ 6^{2} } = \frac { 5 }{ 36 } \)
For n = 6: \( \frac { 6 }{ (6+1)^{2} } = \frac { 6 }{ 7^{2} } = \frac { 6 }{ 49 } \)
The next three terms are \( \frac { 4 }{ 25 } \), \( \frac { 5 }{ 36 } \), and \( \frac { 6 }{ 49 } \).
In simple words: Look for the pattern in each number list. For the first one, multiply by 3 to get the next numbers. For the second, subtract 4 each time. For the third, find the rule for how the top and bottom numbers change, then use that rule for the next few terms.

🎯 Exam Tip: Always clearly state the rule or pattern you've identified for each sequence before finding the next terms.

Question 2. Find the first four terms of the sequences whose nth terms are given by
(i) \( a_n = n^3 - 2 \)
(ii) \( a_n = (-1)^{n+1} n(n+1) \)
(iii) \( a_n = 2n^2 - 6 \)
Answer:
(i) For \( a_n = n^3 - 2 \):
\( a_1 = 1^3 - 2 = 1 - 2 = -1 \)
\( a_2 = 2^3 - 2 = 8 - 2 = 6 \)
\( a_3 = 3^3 - 2 = 27 - 2 = 25 \)
\( a_4 = 4^3 - 2 = 64 - 2 = 62 \)
The first four terms are -1, 6, 25, 62.
(ii) For \( a_n = (-1)^{n+1} n(n+1) \):
\( a_1 = (-1)^{1+1} (1)(1+1) = (-1)^2 (1)(2) = 1 \times 2 = 2 \)
\( a_2 = (-1)^{2+1} (2)(2+1) = (-1)^3 (2)(3) = -1 \times 6 = -6 \)
\( a_3 = (-1)^{3+1} (3)(3+1) = (-1)^4 (3)(4) = 1 \times 12 = 12 \)
\( a_4 = (-1)^{4+1} (4)(4+1) = (-1)^5 (4)(5) = -1 \times 20 = -20 \)
The first four terms are 2, -6, 12, -20.
(iii) For \( a_n = 2n^2 - 6 \):
\( a_1 = 2(1)^2 - 6 = 2(1) - 6 = 2 - 6 = -4 \)
\( a_2 = 2(2)^2 - 6 = 2(4) - 6 = 8 - 6 = 2 \)
\( a_3 = 2(3)^2 - 6 = 2(9) - 6 = 18 - 6 = 12 \)
\( a_4 = 2(4)^2 - 6 = 2(16) - 6 = 32 - 6 = 26 \)
The first four terms are -4, 2, 12, 26.
In simple words: To find the terms of a sequence, take the given rule for \( a_n \) and put in \( n=1, 2, 3, 4 \) one by one. Calculate the answer for each \( n \) to get the first four terms.

🎯 Exam Tip: Be careful with signs, especially when \( (-1) \) is raised to different powers, as it alternates between positive and negative results.

Question 3. Find the nth term of the following sequences
(i) 2, 5, 10, 17, ......
(ii) \( 0, \frac { 1 }{ 2 }, \frac { 2 }{ 3 } \)
(iii) 3, 8, 13, 18, .......
Answer:
(i) For the sequence 2, 5, 10, 17, ......
We can see the terms are `\( 1^2 + 1 = 2 \)`, `\( 2^2 + 1 = 5 \)`, `\( 3^2 + 1 = 10 \)`, `\( 4^2 + 1 = 17 \)`.
Thus, the nth term is \( a_n = n^2 + 1 \). This pattern shows a direct relationship between the term number and its square.
(ii) For the sequence \( 0, \frac { 1 }{ 2 }, \frac { 2 }{ 3 } \)
We can write the terms as `\( \frac { 1-1 }{ 1 } = 0 \)`, `\( \frac { 2-1 }{ 2 } = \frac { 1 }{ 2 } \)`, `\( \frac { 3-1 }{ 3 } = \frac { 2 }{ 3 } \)`.
Thus, the nth term is \( a_n = \frac { n-1 }{ n } \). The numerator is one less than the term number, and the denominator is the term number itself.
(iii) For the sequence 3, 8, 13, 18, .......
This is an arithmetic progression where the first term \( a = 3 \) and the common difference \( d = 8 - 3 = 5 \).
The formula for the nth term of an AP is \( a_n = a + (n-1)d \).
So, \( a_n = 3 + (n-1)5 = 3 + 5n - 5 = 5n - 2 \). This is a linear progression.
In simple words: Look for how each number in the list changes. For some, it might be the term number squared plus one. For others, it might be a fraction based on the term number. For a list where the difference between numbers is always the same, use the formula for arithmetic progression.

🎯 Exam Tip: When finding the nth term, first check for simple patterns like `n^2`, `n^2+1`, or `n-1/n`. If not obvious, check if it's an arithmetic progression (constant difference) or geometric progression (constant ratio).

Question 4. Find the indicated terms of the sequences whose nth terms are given by
(i) \( a_n = \frac { 5n }{ n+2 } \); \( a_6 \) and \( a_{13} \)
(ii) \( a_n = - (n^2 - 4) \); \( a_4 \) and \( a_{11} \)
Answer:
(i) For \( a_n = \frac { 5n }{ n+2 } \):
To find \( a_6 \), substitute \( n=6 \):
\( a_6 = \frac { 5(6) }{ 6+2 } = \frac { 30 }{ 8 } = \frac { 15 }{ 4 } \)
To find \( a_{13} \), substitute \( n=13 \):
\( a_{13} = \frac { 5(13) }{ 13+2 } = \frac { 65 }{ 15 } = \frac { 13 }{ 3 } \)
So, \( a_6 = \frac { 15 }{ 4 } \) and \( a_{13} = \frac { 13 }{ 3 } \). These are the specific terms based on the given formula.
(ii) For \( a_n = - (n^2 - 4) \):
To find \( a_4 \), substitute \( n=4 \):
\( a_4 = - (4^2 - 4) = - (16 - 4) = - (12) = -12 \)
To find \( a_{11} \), substitute \( n=11 \):
\( a_{11} = - (11^2 - 4) = - (121 - 4) = - (117) = -117 \)
So, \( a_4 = -12 \) and \( a_{11} = -117 \). It is important to handle the negative sign outside the parenthesis correctly.
In simple words: To find a specific term in a sequence, just put the term number (n) into the given formula and do the math. Remember to simplify fractions if possible.

🎯 Exam Tip: Always double-check your calculations, especially with fractions and negative signs, to avoid simple arithmetic errors.

Question 5. Find \( a_8 \) and \( a_{15} \) whose nth term is \( a_n \)
\( a_n = \begin{cases} \frac{n^2-1}{n+3} & \text{if n is even, n} \in N \\ \frac{n^2}{2n+1} & \text{if n is odd, n} \in N \end{cases} \)
Answer:
We need to find \( a_8 \) and \( a_{15} \).
For \( a_8 \): Since 8 is an even number, we use the formula \( a_n = \frac{n^2-1}{n+3} \).
Substitute \( n=8 \):
\( a_8 = \frac{8^2-1}{8+3} = \frac{64-1}{11} = \frac{63}{11} \)
For \( a_{15} \): Since 15 is an odd number, we use the formula \( a_n = \frac{n^2}{2n+1} \).
Substitute \( n=15 \):
\( a_{15} = \frac{15^2}{2(15)+1} = \frac{225}{30+1} = \frac{225}{31} \)
Thus, \( a_8 = \frac{63}{11} \) and \( a_{15} = \frac{225}{31} \). This type of sequence has different rules depending on whether 'n' is even or odd.
In simple words: This sequence has two rules. If the term number is even (like 8), use the first rule. If it's odd (like 15), use the second rule. Put the number into the correct rule and calculate the answer.

🎯 Exam Tip: When dealing with piecewise sequence definitions, always identify whether 'n' is even or odd first to apply the correct formula.

Question 6. If \( a_1 = 1 \), \( a_2 = 1 \) and \( a_n = 2a_{n-1} + a_{n-2} \), \( n \ge 3 \), \( n \in N \), then find the first six terms of the sequence.
Answer:
We are given the first two terms: \( a_1 = 1 \) and \( a_2 = 1 \).
The recursive formula for the nth term is \( a_n = 2a_{n-1} + a_{n-2} \), for \( n \ge 3 \).
Let's find the next terms:
For \( n=3 \):
\( a_3 = 2a_{3-1} + a_{3-2} = 2a_2 + a_1 = 2(1) + 1 = 2 + 1 = 3 \)
For \( n=4 \):
\( a_4 = 2a_{4-1} + a_{4-2} = 2a_3 + a_2 = 2(3) + 1 = 6 + 1 = 7 \)
For \( n=5 \):
\( a_5 = 2a_{5-1} + a_{5-2} = 2a_4 + a_3 = 2(7) + 3 = 14 + 3 = 17 \)
For \( n=6 \):
\( a_6 = 2a_{6-1} + a_{6-2} = 2a_5 + a_4 = 2(17) + 7 = 34 + 7 = 41 \)
The first six terms of the sequence are 1, 1, 3, 7, 17, 41. Each term depends on the two terms that came before it.
In simple words: Start with the first two numbers given. To get the next number, multiply the previous number by two, then add the number before that. Keep doing this to find the first six terms.

🎯 Exam Tip: In recursive sequences, make sure to use the *already calculated* terms correctly for each step, not the original starting terms repeatedly.