GSEB Class 11 Maths Solutions Chapter 9 Sequences and Series Exercise 9.1

Write the first five terms of each of the sequences of questions 1 to 6 whose nth terms are:

Question 1. \( a_n = n(n + 2) \)
Answer: We need to find the first five terms by putting \( n = 1, 2, 3, 4, \) and \( 5 \).
\( a_1 = 1(1 + 2) = 1 \cdot 3 = 3 \)
\( a_2 = 2(2 + 2) = 2 \cdot 4 = 8 \)
\( a_3 = 3(3 + 2) = 3 \cdot 5 = 15 \)
\( a_4 = 4(4 + 2) = 4 \cdot 6 = 24 \)
\( a_5 = 5(5 + 2) = 5 \cdot 7 = 35 \)
Therefore, the first five terms of the sequence, where the nth term is \( a_n = n(n + 2) \), are 3, 8, 15, 24, and 35.
In simple words: To get the terms, just put \( n \) values from 1 to 5 into the given formula for \( a_n \). Then, list out each answer you get.

Exam Tip: Remember to clearly show the substitution for each term and calculate the result carefully. A simple calculation error can lead to incorrect terms.

Question 2. \( a_n = \frac{n}{n+1} \)
Answer: We need to find the first five terms by putting \( n = 1, 2, 3, 4, \) and \( 5 \).
\( a_1 = \frac{1}{1+1} = \frac{1}{2} \)
\( a_2 = \frac{2}{2+1} = \frac{2}{3} \)
\( a_3 = \frac{3}{3+1} = \frac{3}{4} \)
\( a_4 = \frac{4}{4+1} = \frac{4}{5} \)
\( a_5 = \frac{5}{5+1} = \frac{5}{6} \)
Therefore, the first five terms of the sequence where the nth term is \( a_n = \frac{n}{n+1} \) are \( \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \) and \( \frac{5}{6} \).
In simple words: Replace \( n \) with 1, 2, 3, 4, and 5 in the fraction formula. Calculate each fraction to get the first five terms of the sequence.

Exam Tip: For fractions, always simplify them if possible. Also, ensure the denominator is calculated correctly before performing the division.

Question 3. \( a_n = 2^n \)
Answer: We need to find the first five terms by putting \( n = 1, 2, 3, 4, \) and \( 5 \).
\( a_1 = 2^1 = 2 \)
\( a_2 = 2^2 = 4 \)
\( a_3 = 2^3 = 8 \)
\( a_4 = 2^4 = 16 \)
\( a_5 = 2^5 = 32 \)
The required first five terms of the sequence are 2, 4, 8, 16, and 32.
In simple words: Take the number 2 and raise it to the power of 1, then 2, then 3, then 4, and finally 5. Write down each result.

Exam Tip: Be careful with exponents; \( 2^3 \) means \( 2 \times 2 \times 2 \), not \( 2 \times 3 \). Simple multiplications are crucial here.

Question 4. \( a_n = \frac{2n-3}{6} \)
Answer: We need to find the first five terms by putting \( n = 1, 2, 3, 4, \) and \( 5 \).
\( a_1 = \frac{2 \times 1 - 3}{6} = \frac{2 - 3}{6} = \frac{-1}{6} \)
\( a_2 = \frac{2 \times 2 - 3}{6} = \frac{4 - 3}{6} = \frac{1}{6} \)
\( a_3 = \frac{2 \times 3 - 3}{6} = \frac{6 - 3}{6} = \frac{3}{6} = \frac{1}{2} \)
\( a_4 = \frac{2 \times 4 - 3}{6} = \frac{8 - 3}{6} = \frac{5}{6} \)
\( a_5 = \frac{2 \times 5 - 3}{6} = \frac{10 - 3}{6} = \frac{7}{6} \)
The first five terms of the sequence are \( \frac{-1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \) and \( \frac{7}{6} \).
In simple words: Substitute \( n = 1, 2, 3, 4, 5 \) into the given formula. Perform the multiplication, subtraction, and then division for each term, simplifying if possible.

Exam Tip: Pay close attention to the order of operations (PEMDAS/BODMAS) when calculating the numerator, especially with negative numbers. Simplify fractions when they can be reduced, like \( \frac{3}{6} \) to \( \frac{1}{2} \).

Question 5. \( a_n = (-1)^{n-1} 5^{n+1} \)
Answer: We need to find the first five terms by putting \( n = 1, 2, 3, 4, \) and \( 5 \).
\( a_1 = (-1)^{1-1} 5^{1+1} = (-1)^0 \cdot 5^2 = 1 \cdot 25 = 25 \)
\( a_2 = (-1)^{2-1} 5^{2+1} = (-1)^1 \cdot 5^3 = -1 \cdot 125 = -125 \)
\( a_3 = (-1)^{3-1} 5^{3+1} = (-1)^2 \cdot 5^4 = 1 \cdot 625 = 625 \)
\( a_4 = (-1)^{4-1} 5^{4+1} = (-1)^3 \cdot 5^5 = -1 \cdot 3125 = -3125 \)
\( a_5 = (-1)^{5-1} 5^{5+1} = (-1)^4 \cdot 5^6 = 1 \cdot 15625 = 15625 \)
The first five terms of the given sequence are 25, -125, 625, -3125, and 15625.
In simple words: Replace \( n \) with 1, 2, 3, 4, and 5 in the formula. Remember that \( (-1) \) to an even power is 1, and \( (-1) \) to an odd power is -1. Calculate the powers of 5 and then multiply.

Exam Tip: Be very careful with the signs introduced by \( (-1)^{n-1} \). The exponent for \( (-1) \) determines if the term will be positive or negative. The exponent for 5 is \( n+1 \).

Question 6. \( a_n = \frac{n(n^2+5)}{4} \)
Answer: We need to find the first five terms by putting \( n = 1, 2, 3, 4, \) and \( 5 \).
\( a_1 = \frac{1(1^2+5)}{4} = \frac{1(1+5)}{4} = \frac{1 \cdot 6}{4} = \frac{6}{4} = \frac{3}{2} \)
\( a_2 = \frac{2(2^2+5)}{4} = \frac{2(4+5)}{4} = \frac{2 \cdot 9}{4} = \frac{18}{4} = \frac{9}{2} \)
\( a_3 = \frac{3(3^2+5)}{4} = \frac{3(9+5)}{4} = \frac{3 \cdot 14}{4} = \frac{42}{4} = \frac{21}{2} \)
\( a_4 = \frac{4(4^2+5)}{4} = \frac{4(16+5)}{4} = \frac{4 \cdot 21}{4} = 21 \)
\( a_5 = \frac{5(5^2+5)}{4} = \frac{5(25+5)}{4} = \frac{5 \cdot 30}{4} = \frac{150}{4} = \frac{75}{2} \)
Therefore, the first five terms are \( \frac{3}{2}, \frac{9}{2}, \frac{21}{2}, 21, \) and \( \frac{75}{2} \).
In simple words: Substitute \( n \) values from 1 to 5 into the given formula. Calculate the square, add 5, then multiply by \( n \), and finally divide by 4. Simplify each fraction you get.

Exam Tip: Remember to calculate \( n^2 \) first, then add 5, then multiply by \( n \), and finally divide by 4, following the correct order of operations. Simplify all fractions to their lowest terms.

Find the indicated terms in each of the following sequences in questions 7 to 10 whose nth terms are:

Question 7. \( a_n = 4n - 3; a_{17}, a_{24} \)
Answer: The nth term is given as \( a_n = 4n - 3 \). We need to find \( a_{17} \) and \( a_{24} \).
To find \( a_{17} \), we put \( n = 17 \):
\( a_{17} = 4 \cdot 17 - 3 = 68 - 3 = 65 \)
To find \( a_{24} \), we put \( n = 24 \):
\( a_{24} = 4 \cdot 24 - 3 = 96 - 3 = 93 \)
Thus, the 17th term is 65 and the 24th term is 93.
In simple words: Use the formula \( a_n = 4n - 3 \). To get \( a_{17} \), replace \( n \) with 17. To get \( a_{24} \), replace \( n \) with 24. Then, solve the math problem for each.

Exam Tip: Ensure you substitute the correct value of \( n \) for each required term. Double-check your multiplication and subtraction steps.

Question 8. \( a_n = \frac{n^2}{2^n}; a_7 \)
Answer: The nth term is given as \( a_n = \frac{n^2}{2^n} \). We need to find \( a_7 \).
To find \( a_7 \), we put \( n = 7 \):
\( a_7 = \frac{7^2}{2^7} = \frac{49}{128} \)
Thus, the 7th term is \( \frac{49}{128} \).
In simple words: Replace \( n \) with 7 in the formula. Calculate 7 squared for the top part and 2 raised to the power of 7 for the bottom part. Write the result as a fraction.

Exam Tip: Remember to calculate both the numerator (\( n^2 \)) and the denominator (\( 2^n \)) correctly. Be careful with powers of 2 (e.g., \( 2^7 = 128 \)).

Question 9. \( a_n = (-1)^{n-1} n^3; a_9 \)
Answer: The nth term is given as \( a_n = (-1)^{n-1} n^3 \). We need to find \( a_9 \).
To find \( a_9 \), we put \( n = 9 \):
\( a_9 = (-1)^{9-1} 9^3 \)
\( a_9 = (-1)^8 \cdot 9^3 \)
Since 8 is an even number, \( (-1)^8 = 1 \).
\( a_9 = 1 \cdot (9 \times 9 \times 9) \)
\( a_9 = 1 \cdot 729 = 729 \)
Thus, the 9th term is 729.
In simple words: Replace \( n \) with 9 in the formula. Calculate \( (-1) \) to the power of \( (9-1) \), and 9 to the power of 3. Multiply these two results together to get the 9th term.

Exam Tip: Remember that any negative number raised to an even power becomes positive. Calculate \( n^3 \) accurately, and then apply the sign from \( (-1)^{n-1} \).

Question 10. \( a_n = \frac{n(n-2)}{n+3}; a_{20} \)
Answer: The nth term is given as \( a_n = \frac{n(n-2)}{n+3} \). We need to find \( a_{20} \).
To find \( a_{20} \), we put \( n = 20 \):
\( a_{20} = \frac{20(20-2)}{20+3} \)
\( a_{20} = \frac{20(18)}{23} \)
\( a_{20} = \frac{360}{23} \)
Thus, the 20th term is \( \frac{360}{23} \).
In simple words: Substitute \( n \) with 20 into the formula. First, solve the math inside the parentheses. Then, do the multiplications on the top and the addition on the bottom. Finally, write the result as a fraction.

Exam Tip: Work carefully through the numerator and denominator separately. Ensure to perform subtraction first in the numerator before multiplication, and addition in the denominator.

Write the first five terms of each of the sequences in questions 11 to 13 and obtain the corresponding series:

Question 11. \( a_1 = 3, a_n = 3a_{n-1} + 2 \) for all \( n > 1 \).
Answer: We are given \( a_1 = 3 \) and the recursive formula \( a_n = 3a_{n-1} + 2 \) for \( n > 1 \). We need to find the first five terms.
Given \( a_1 = 3 \)
For \( n = 2 \): \( a_2 = 3a_{2-1} + 2 = 3a_1 + 2 = 3(3) + 2 = 9 + 2 = 11 \)
For \( n = 3 \): \( a_3 = 3a_{3-1} + 2 = 3a_2 + 2 = 3(11) + 2 = 33 + 2 = 35 \)
For \( n = 4 \): \( a_4 = 3a_{4-1} + 2 = 3a_3 + 2 = 3(35) + 2 = 105 + 2 = 107 \)
For \( n = 5 \): \( a_5 = 3a_{5-1} + 2 = 3a_4 + 2 = 3(107) + 2 = 321 + 2 = 323 \)
The first five terms of the sequence are 3, 11, 35, 107, and 323.
The corresponding series is \( 3 + 11 + 35 + 107 + 323 + \dots \)
In simple words: Start with the first term, \( a_1 = 3 \). Then, use the rule \( a_n = 3 \times a_{n-1} + 2 \) to find the next terms one by one. For example, \( a_2 \) uses \( a_1 \), \( a_3 \) uses \( a_2 \), and so on, until you have five terms. Then, write these terms as a sum for the series.

Exam Tip: When dealing with recursive sequences, accurately calculate each term using the previous one. A mistake in an earlier term will affect all subsequent terms.

Question 12. \( a_1 = 1, a_n = \frac{a_{n-1}}{n}, n \ge 2 \)
Answer: We are given \( a_1 = 1 \) and the recursive formula \( a_n = \frac{a_{n-1}}{n} \) for \( n \ge 2 \). We need to find the first five terms.
Given \( a_1 = 1 \)
For \( n = 2 \): \( a_2 = \frac{a_{2-1}}{2} = \frac{a_1}{2} = \frac{1}{2} \)
For \( n = 3 \): \( a_3 = \frac{a_{3-1}}{3} = \frac{a_2}{3} = \frac{1/2}{3} = \frac{1}{6} \)
For \( n = 4 \): \( a_4 = \frac{a_{4-1}}{4} = \frac{a_3}{4} = \frac{1/6}{4} = \frac{1}{24} \)
For \( n = 5 \): \( a_5 = \frac{a_{5-1}}{5} = \frac{a_4}{5} = \frac{1/24}{5} = \frac{1}{120} \)
The first five terms of the sequence are \( 1, \frac{1}{2}, \frac{1}{6}, \frac{1}{24}, \) and \( \frac{1}{120} \).
The corresponding series is \( 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} + \dots \)
In simple words: Start with \( a_1 = 1 \). To get each new term, take the previous term and divide it by the current term number \( n \). Do this for five terms. Then, write these terms as a sum to show the series.

Exam Tip: When dividing a fraction by a whole number, remember that the whole number effectively multiplies the denominator of the fraction. Carefully carry the fractional values through your calculations.

Question 13. \( a_1 = a_2 = 2, a_n = a_{n-1} - 1, n \ge 2 \)
Answer: We are given \( a_1 = 2, a_2 = 2 \) and the recursive formula \( a_n = a_{n-1} - 1 \) for \( n > 2 \). We need to find the first five terms.
Given \( a_1 = 2 \)
Given \( a_2 = 2 \)
For \( n = 3 \): \( a_3 = a_{3-1} - 1 = a_2 - 1 = 2 - 1 = 1 \)
For \( n = 4 \): \( a_4 = a_{4-1} - 1 = a_3 - 1 = 1 - 1 = 0 \)
For \( n = 5 \): \( a_5 = a_{5-1} - 1 = a_4 - 1 = 0 - 1 = -1 \)
The first five terms of the sequence are 2, 2, 1, 0, and -1.
The corresponding series is \( 2 + 2 + 1 + 0 + (-1) + \dots \)
In simple words: Start with \( a_1 = 2 \) and \( a_2 = 2 \). For any term after the second one, just subtract 1 from the term before it. Do this until you have five terms in total. Then, show them added together to form the series.

Exam Tip: For sequences with an initial two terms, make sure to use the correct starting term for the recursive rule. Pay attention to the range of \( n \) for which the recursive formula applies.

Question 14. The Fibonacci sequence is defined by \( 1 = a_1 = a_2 \) and \( a_n = a_{n-1} + a_{n-2}, n > 2 \). Find \( \frac{a_{n+1}}{a_n} \) for \( n = 1, 2, 3, 4 \) and \( 5 \).
Answer: The Fibonacci sequence is given by \( a_1 = 1, a_2 = 1 \), and \( a_n = a_{n-1} + a_{n-2} \) for \( n > 2 \).
First, let's find the initial terms of the sequence:
\( a_1 = 1 \)
\( a_2 = 1 \)
\( a_3 = a_2 + a_1 = 1 + 1 = 2 \)
\( a_4 = a_3 + a_2 = 2 + 1 = 3 \)
\( a_5 = a_4 + a_3 = 3 + 2 = 5 \)
\( a_6 = a_5 + a_4 = 5 + 3 = 8 \)
Now, we calculate \( \frac{a_{n+1}}{a_n} \) for \( n = 1, 2, 3, 4, 5 \):
For \( n = 1 \): \( \frac{a_{1+1}}{a_1} = \frac{a_2}{a_1} = \frac{1}{1} = 1 \)
For \( n = 2 \): \( \frac{a_{2+1}}{a_2} = \frac{a_3}{a_2} = \frac{2}{1} = 2 \)
For \( n = 3 \): \( \frac{a_{3+1}}{a_3} = \frac{a_4}{a_3} = \frac{3}{2} \)
For \( n = 4 \): \( \frac{a_{4+1}}{a_4} = \frac{a_5}{a_4} = \frac{5}{3} \)
For \( n = 5 \): \( \frac{a_{5+1}}{a_5} = \frac{a_6}{a_5} = \frac{8}{5} \)
The values of \( \frac{a_{n+1}}{a_n} \) for \( n = 1, 2, 3, 4, \) and \( 5 \) are respectively \( 1, 2, \frac{3}{2}, \frac{5}{3}, \) and \( \frac{8}{5} \).
In simple words: First, list the terms of the Fibonacci sequence starting from \( a_1 = 1, a_2 = 1 \) by adding the two previous terms. Then, for each \( n \) from 1 to 5, divide the term \( a_{n+1} \) by \( a_n \).

Exam Tip: Always list out enough terms of the Fibonacci sequence first before attempting to find the ratios. This helps avoid errors and provides a clear path for calculation. Remember that \( a_n \) refers to the previous term and \( a_{n-1} \) refers to the term before that.