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Punjab Board Class 11 Mathematics

Exercise 6.1 Solutions — Sequences and Series

Complete worked solutions for Exercise 6.1 from Chapter 6 (Sequences and Series), Punjab Board Class 11 Mathematics. Each question shows every step and names the formula used.

Mathematics 11 (Punjab Board)

Exercise 6.1 Solutions

Worked solutions for Punjab Board Class 11 Mathematics, Chapter 6: Sequences and Series.

Question 1: Find the next four terms of each sequence.

Pehle common difference nikalo (do consecutive terms ka farq), phir usay jor kar agle terms likho.

(i) \(12, 16, 20, \dots\)

\[ d = 4 \implies 24, \; 28, \; 32, \; 36 \]

(ii) \(3, 1, -1, \dots\)

\[ d = -2 \implies -3, \; -5, \; -7, \; -9 \]
Answer: (i) \(24, 28, 32, 36\), (ii) \(-3, -5, -7, -9\)

Question 2: Write down the first three terms of each of the following sequences:

Direct formula ho to \(n = 1, 2, 3\) rakho; recursive ho to pichli term se agli banao.

(i) \(a_n = 3n + 5\)

\[ a_1 = 8, \quad a_2 = 11, \quad a_3 = 14 \]

(ii) \(a_{n+1} = 4a_n - 7,\ a_1 = 3\)

\[ a_1 = 3, \quad a_2 = 4(3)-7 = 5, \quad a_3 = 4(5)-7 = 13 \]

(iii) \(a_n = (n-3)(n+1)\)

\[ a_1 = -4, \quad a_2 = -3, \quad a_3 = 0 \]

(iv) \(a_1 = -1,\ a_{n+1} = \frac{3}{a_n + 2}\)

\[ a_1 = -1, \quad a_2 = \frac{3}{1} = 3, \quad a_3 = \frac{3}{5} \]

(v) \(a_n = 8 - \frac{20}{3+n}\)

\[ a_1 = 8-5 = 3, \quad a_2 = 8-4 = 4, \quad a_3 = 8-\frac{10}{3} = \frac{14}{3} \]

(vi) \(a_1 = 1,\ a_{n+1} = (3a_n + 2)^2\)

\[ a_1 = 1, \quad a_2 = 25, \quad a_3 = 77^2 = 5929 \]

(vii) \(a_n = (-2n)^2\)

\[ a_1 = 4, \quad a_2 = 16, \quad a_3 = 36 \]

(viii) \(a_n = (-1)^n 7n^2\)

\[ a_1 = -7, \quad a_2 = 28, \quad a_3 = -63 \]
Answer: (i) \(8, 11, 14\); (ii) \(3, 5, 13\); (iii) \(-4, -3, 0\); (iv) \(-1, 3, \frac{3}{5}\); (v) \(3, 4, \frac{14}{3}\); (vi) \(1, 25, 5929\); (vii) \(4, 16, 36\); (viii) \(-7, 28, -63\)

Question 3: An expression for the \(n^{th}\) triangular number is \(\frac{n(n+1)}{2}\). Write down the \(15^{th}\) triangular number.

Formula mein \(n = 15\) rakho aur simplify karo.

\[ T_{15} = \frac{15 \times 16}{2} = 15 \times 8 = 120 \]
Answer: \(120\)

Question 4: Write down the \(n^{th}\) term of each of the following sequences:

Terms mein chhupa pattern pehchano — position \(n\) aur term ke darmiyan relation likho.

(i) \(1, 4, 9, \dots\)

\[ a_n = n^2 \]

(ii) \(1, 1+2, 1+2+3, \dots\)

\[ a_n = 1 + 2 + \cdots + n = \frac{n(n+1)}{2} \]

(iii) \(a_1, b_1, a_2, b_2, a_3, b_3, \dots\)

\[ a_n = \begin{cases} a_{(n+1)/2}, & n \text{ odd} \\ b_{n/2}, & n \text{ even} \end{cases} \]

(iv) \(x, 2x^2, 3x^3, \dots\)

\[ a_n = n x^n \]

(v) \(a_1, a_1+d, a_1+2d, \dots\)

\[ a_n = a_1 + (n-1)d \]

(vi) \(a_1, a_1r, a_1r^2, \dots\)

\[ a_n = a_1 r^{n-1} \]

(vii) \(\frac{a_1}{b_1+c_1}, \frac{2a_2}{b_2+c_2}, \frac{3a_3}{b_3+c_3}, \dots\)

\[ a_n = \frac{n a_n}{b_n + c_n} \]

(viii) \(\frac{1}{a_1}, \frac{1}{a_1+d}, \frac{1}{a_1+2d}, \dots\)

\[ a_n = \frac{1}{a_1 + (n-1)d} \]
Answer: (i) \(n^2\), (ii) \(\frac{n(n+1)}{2}\), (iii) \(a_{(n+1)/2}\) for odd \(n\), \(b_{n/2}\) for even \(n\), (iv) \(nx^n\), (v) \(a_1+(n-1)d\), (vi) \(a_1r^{n-1}\), (vii) \(\frac{na_n}{b_n+c_n}\), (viii) \(\frac{1}{a_1+(n-1)d}\)