OP Malhotra Class 11 Maths Solutions Chapter 14 Sequence and Series Exercise 14 (A)

Question 1. Write the first five terms of the sequence using the given rule. In each case, the initial value of the index is 1.
(i) \( a_n = 2n \)
(ii) \( a_n = 3n - 2 \)
(iii) \( a_n = n^2 + 5 \)
(iv) \( a_n = \frac{(-1)^{n-1}}{n^3} \)
(v) \( a_n = \) nth prime number for all natural numbers \( n \)
Answer:
(i) Given \( a_n = 2n \). To find the terms, we put \( n = 1, 2, 3, 4, 5 \) into the rule.
For \( n = 1 \), \( a_1 = 2 \times 1 = 2 \)
For \( n = 2 \), \( a_2 = 2 \times 2 = 4 \)
For \( n = 3 \), \( a_3 = 2 \times 3 = 6 \)
For \( n = 4 \), \( a_4 = 2 \times 4 = 8 \)
For \( n = 5 \), \( a_5 = 2 \times 5 = 10 \)
So, the first five terms of the sequence are 2, 4, 6, 8, and 10. This sequence is a simple arithmetic progression.
(ii) Given \( a_n = 3n - 2 \). To find the terms, we put \( n = 1, 2, 3, 4, 5 \) into the rule.
For \( n = 1 \), \( a_1 = 3 \times 1 - 2 = 1 \)
For \( n = 2 \), \( a_2 = 3 \times 2 - 2 = 4 \)
For \( n = 3 \), \( a_3 = 3 \times 3 - 2 = 7 \)
For \( n = 4 \), \( a_4 = 3 \times 4 - 2 = 10 \)
For \( n = 5 \), \( a_5 = 3 \times 5 - 2 = 13 \)
Therefore, the first five terms of the sequence are 1, 4, 7, 10, and 13. This is another arithmetic progression.
(iii) Given \( a_n = n^2 + 5 \). To find the terms, we put \( n = 1, 2, 3, 4, 5 \) into the rule.
For \( n = 1 \), \( a_1 = 1^2 + 5 = 1 + 5 = 6 \)
For \( n = 2 \), \( a_2 = 2^2 + 5 = 4 + 5 = 9 \)
For \( n = 3 \), \( a_3 = 3^2 + 5 = 9 + 5 = 14 \)
For \( n = 4 \), \( a_4 = 4^2 + 5 = 16 + 5 = 21 \)
For \( n = 5 \), \( a_5 = 5^2 + 5 = 25 + 5 = 30 \)
So, the first five terms of the sequence are 6, 9, 14, 21, and 30. Each term shows a quadratic relationship with its position.
(iv) Given \( a_n = \frac{(-1)^{n-1}}{n^3} \). To find the terms, we put \( n = 1, 2, 3, 4, 5 \) into the rule.
For \( n = 1 \), \( a_1 = \frac{(-1)^{1-1}}{1^3} = \frac{(-1)^0}{1} = \frac{1}{1} = 1 \)
For \( n = 2 \), \( a_2 = \frac{(-1)^{2-1}}{2^3} = \frac{(-1)^1}{8} = \frac{-1}{8} = -\frac{1}{8} \)
For \( n = 3 \), \( a_3 = \frac{(-1)^{3-1}}{3^3} = \frac{(-1)^2}{27} = \frac{1}{27} \)
For \( n = 4 \), \( a_4 = \frac{(-1)^{4-1}}{4^3} = \frac{(-1)^3}{64} = \frac{-1}{64} = -\frac{1}{64} \)
For \( n = 5 \), \( a_5 = \frac{(-1)^{5-1}}{5^3} = \frac{(-1)^4}{125} = \frac{1}{125} \)
So, the first five terms of the sequence are \( 1, -\frac{1}{8}, \frac{1}{27}, -\frac{1}{64}, \) and \( \frac{1}{125} \). Notice how the terms alternate in sign due to the \( (-1)^{n-1} \) part.
(v) Given \( a_n = \) nth prime number. The terms of this sequence are the prime numbers in order.
For \( n = 1 \), \( a_1 = \) 1st prime number \( = 2 \)
For \( n = 2 \), \( a_2 = \) 2nd prime number \( = 3 \)
For \( n = 3 \), \( a_3 = \) 3rd prime number \( = 5 \)
For \( n = 4 \), \( a_4 = \) 4th prime number \( = 7 \)
For \( n = 5 \), \( a_5 = \) 5th prime number \( = 11 \)
Thus, the first five terms of the sequence are 2, 3, 5, 7, and 11. Prime numbers are only divisible by 1 and themselves.
In simple words: For each rule, replace 'n' with the numbers 1 through 5 to find each term. The type of sequence changes based on the rule given.

🎯 Exam Tip: When finding terms of a sequence, carefully substitute the value of 'n' into the given rule and follow the order of operations. For prime numbers, ensure you list them in ascending order starting from 2.

Question 2. Write the first four terms of the sequence whose nth term is given
(i) \( a_n = \frac{2n+1}{2n-1} \)
(ii) \( a_n = \frac{n^2+1}{n} \)
(iii) \( a_n = \frac{2^n}{n^2} \)
(iv) \( a_n = \sin^n 30^\circ \)
(v) \( a_n = (-1)^n \sin\frac{n \pi}{2} \)
(vi) \( a_n = (-1)^{n-1} \cos\frac{n \pi}{4} \)
Answer:
(i) Given \( a_n = \frac{2n+1}{2n-1} \). To find the first four terms, we substitute \( n = 1, 2, 3, 4 \) into the rule.
For \( n = 1 \), \( a_1 = \frac{2(1)+1}{2(1)-1} = \frac{2+1}{2-1} = \frac{3}{1} = 3 \)
For \( n = 2 \), \( a_2 = \frac{2(2)+1}{2(2)-1} = \frac{4+1}{4-1} = \frac{5}{3} \)
For \( n = 3 \), \( a_3 = \frac{2(3)+1}{2(3)-1} = \frac{6+1}{6-1} = \frac{7}{5} \)
For \( n = 4 \), \( a_4 = \frac{2(4)+1}{2(4)-1} = \frac{8+1}{8-1} = \frac{9}{7} \)
The first four terms are \( 3, \frac{5}{3}, \frac{7}{5}, \) and \( \frac{9}{7} \). These terms show a pattern of increasing numerator and denominator.
(ii) Given \( a_n = \frac{n^2+1}{n} \). To find the first four terms, we substitute \( n = 1, 2, 3, 4 \) into the rule.
For \( n = 1 \), \( a_1 = \frac{1^2+1}{1} = \frac{1+1}{1} = 2 \)
For \( n = 2 \), \( a_2 = \frac{2^2+1}{2} = \frac{4+1}{2} = \frac{5}{2} \)
For \( n = 3 \), \( a_3 = \frac{3^2+1}{3} = \frac{9+1}{3} = \frac{10}{3} \)
For \( n = 4 \), \( a_4 = \frac{4^2+1}{4} = \frac{16+1}{4} = \frac{17}{4} \)
The first four terms are \( 2, \frac{5}{2}, \frac{10}{3}, \) and \( \frac{17}{4} \). Notice that these terms can be written as \( n + \frac{1}{n} \).
(iii) Given \( a_n = \frac{2^n}{n^2} \). To find the first four terms, we substitute \( n = 1, 2, 3, 4 \) into the rule.
For \( n = 1 \), \( a_1 = \frac{2^1}{1^2} = \frac{2}{1} = 2 \)
For \( n = 2 \), \( a_2 = \frac{2^2}{2^2} = \frac{4}{4} = 1 \)
For \( n = 3 \), \( a_3 = \frac{2^3}{3^2} = \frac{8}{9} \)
For \( n = 4 \), \( a_4 = \frac{2^4}{4^2} = \frac{16}{16} = 1 \)
The first four terms are \( 2, 1, \frac{8}{9}, \) and \( 1 \). This sequence shows how exponential growth in the numerator competes with polynomial growth in the denominator.
(iv) Given \( a_n = \sin^n 30^\circ \). To find the first four terms, we substitute \( n = 1, 2, 3, 4 \) into the rule. We know that \( \sin 30^\circ = \frac{1}{2} \).
For \( n = 1 \), \( a_1 = \sin^1 30^\circ = \sin 30^\circ = \frac{1}{2} \)
For \( n = 2 \), \( a_2 = \sin^2 30^\circ = (\sin 30^\circ)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)
For \( n = 3 \), \( a_3 = \sin^3 30^\circ = (\sin 30^\circ)^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \)
For \( n = 4 \), \( a_4 = \sin^4 30^\circ = (\sin 30^\circ)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \)
The first four terms are \( \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \) and \( \frac{1}{16} \). This sequence is a geometric progression with a common ratio of \( \frac{1}{2} \).
(v) Given \( a_n = (-1)^n \sin\frac{n \pi}{2} \). To find the first four terms, we substitute \( n = 1, 2, 3, 4 \) into the rule.
For \( n = 1 \), \( a_1 = (-1)^1 \sin\frac{1 \pi}{2} = (-1) \times \sin(\frac{\pi}{2}) = (-1) \times 1 = -1 \)
For \( n = 2 \), \( a_2 = (-1)^2 \sin\frac{2 \pi}{2} = (1) \times \sin(\pi) = (1) \times 0 = 0 \)
For \( n = 3 \), \( a_3 = (-1)^3 \sin\frac{3 \pi}{2} = (-1) \times \sin(\frac{3\pi}{2}) = (-1) \times (-1) = 1 \)
For \( n = 4 \), \( a_4 = (-1)^4 \sin\frac{4 \pi}{2} = (1) \times \sin(2\pi) = (1) \times 0 = 0 \)
The first four terms are \( -1, 0, 1, 0 \). The sine component of this sequence causes terms to be zero for even \( n \).
(vi) Given \( a_n = (-1)^{n-1} \cos\frac{n \pi}{4} \). To find the first four terms, we substitute \( n = 1, 2, 3, 4 \) into the rule.
For \( n = 1 \), \( a_1 = (-1)^{1-1} \cos\frac{1 \pi}{4} = (-1)^0 \cos(\frac{\pi}{4}) = 1 \times \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \)
For \( n = 2 \), \( a_2 = (-1)^{2-1} \cos\frac{2 \pi}{4} = (-1)^1 \cos(\frac{\pi}{2}) = -1 \times 0 = 0 \)
For \( n = 3 \), \( a_3 = (-1)^{3-1} \cos\frac{3 \pi}{4} = (-1)^2 \cos(\frac{3\pi}{4}) = 1 \times (-\frac{1}{\sqrt{2}}) = -\frac{1}{\sqrt{2}} \)
For \( n = 4 \), \( a_4 = (-1)^{4-1} \cos\frac{4 \pi}{4} = (-1)^3 \cos(\pi) = -1 \times (-1) = 1 \)
The first four terms are \( \frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}}, \) and \( 1 \). This sequence demonstrates the interplay of alternating signs and cosine values at various angles.
In simple words: For each given rule, put 'n' as 1, 2, 3, and 4 to find the first four terms. Be careful with calculations involving fractions, powers, and trigonometric functions.

🎯 Exam Tip: Pay close attention to the definition of the \( n^{th} \) term for each sequence. Use the exact value of trigonometric ratios and correctly handle powers and negative signs.

Question 3. Find the first 4 terms and the 20th term of the sequence whose \( S_n = \frac{3}{2} (3^n – 1) \).
Answer: We are given the sum of the first \( n \) terms, \( S_n = \frac{3}{2} (3^n - 1) \). To find the \( n^{th} \) term \( T_n \), we use the formula \( T_n = S_n - S_{n-1} \).
First, we find \( S_{n-1} \):
\( S_{n-1} = \frac{3}{2} (3^{n-1} - 1) \)
Now, we calculate \( T_n \):
\( T_n = S_n - S_{n-1} \)
\( \implies T_n = \frac{3}{2} (3^n - 1) - \frac{3}{2} (3^{n-1} - 1) \)
\( \implies T_n = \frac{3}{2} [(3^n - 1) - (3^{n-1} - 1)] \)
\( \implies T_n = \frac{3}{2} [3^n - 1 - 3^{n-1} + 1] \)
\( \implies T_n = \frac{3}{2} [3^n - 3^{n-1}] \)
\( \implies T_n = \frac{3}{2} [3 \cdot 3^{n-1} - 3^{n-1}] \)
\( \implies T_n = \frac{3}{2} \cdot 3^{n-1} (3 - 1) \)
\( \implies T_n = \frac{3}{2} \cdot 3^{n-1} \cdot 2 \)
\( \implies T_n = 3 \cdot 3^{n-1} = 3^n \)
Now we find the first four terms using \( T_n = 3^n \):
For \( n = 1 \), \( T_1 = 3^1 = 3 \)
For \( n = 2 \), \( T_2 = 3^2 = 9 \)
For \( n = 3 \), \( T_3 = 3^3 = 27 \)
For \( n = 4 \), \( T_4 = 3^4 = 81 \)
Next, we find the 20th term, \( T_{20} \):
For \( n = 20 \), \( T_{20} = 3^{20} \)
The terms are powers of 3, making this a geometric progression. The first four terms are 3, 9, 27, 81. The 20th term is \( 3^{20} \).
In simple words: We first find a general formula for the \( n^{th} \) term by subtracting the sum of \( (n-1) \) terms from the sum of \( n \) terms. This gives us \( T_n = 3^n \). Then we use this new formula to find the first four terms and the 20th term.

🎯 Exam Tip: Remember the key relationship \( T_n = S_n - S_{n-1} \) to find the general term of a sequence when the sum to \( n \) terms is given. Be careful with algebraic simplifications and exponent rules.

Question 4. Find the 10th term of the sequence whose sum to n terms is 6n² + 7.
Answer: We are given the sum of the first \( n \) terms, \( S_n = 6n^2 + 7 \). To find the \( n^{th} \) term \( T_n \), we use the formula \( T_n = S_n - S_{n-1} \).
First, we find \( S_{n-1} \):
\( S_{n-1} = 6(n - 1)^2 + 7 \)
\( \implies S_{n-1} = 6(n^2 - 2n + 1) + 7 \)
\( \implies S_{n-1} = 6n^2 - 12n + 6 + 7 = 6n^2 - 12n + 13 \)
Now, we calculate \( T_n \):
\( T_n = S_n - S_{n-1} \)
\( \implies T_n = (6n^2 + 7) - (6n^2 - 12n + 13) \)
\( \implies T_n = 6n^2 + 7 - 6n^2 + 12n - 13 \)
\( \implies T_n = 12n - 6 \)
Finally, we find the 10th term by putting \( n = 10 \) into the formula for \( T_n \):
\( T_{10} = 12(10) - 6 = 120 - 6 = 114 \)
The 10th term of this sequence is 114. This method allows us to find any term directly if the sum formula is known.
In simple words: We find the general rule for the \( n^{th} \) term by subtracting the sum up to \( (n-1) \) terms from the sum up to \( n \) terms. This gives us \( T_n = 12n - 6 \). Then, we simply put \( n = 10 \) into this rule to get the 10th term.

🎯 Exam Tip: When using \( T_n = S_n - S_{n-1} \), be very careful with expanding \( (n-1)^2 \) and distributing the negative sign across all terms of \( S_{n-1} \) to avoid calculation errors.