Mathematics 11 (Punjab Board)
Solved Examples: Chapter 6 – Sequences and Series
Textbook solved examples for Punjab Board Class 11 Mathematics, Chapter 6: Sequences and Series, worked in full.
Example 1: Find the first four terms of the following sequences whose \(n^{th}\) terms are given: (i) \(a_n = 3n + 1\) (ii) \(a_n = 3n^2 - 3\)
\(n = 1, 2, 3, 4\) rakh kar har term nikalo — yeh sequence ka sab se basic kaam hai.
(i) \(a_n = 3n + 1\)
\[
a_1 = 4, \quad a_2 = 7, \quad a_3 = 10, \quad a_4 = 13
\]
(ii) \(a_n = 3n^2 - 3\)
\[
a_1 = 0, \quad a_2 = 9, \quad a_3 = 24, \quad a_4 = 45
\]
Answer: (i) \(4, 7, 10, 13\), (ii) \(0, 9, 24, 45\)
Example 2: Find the general term and the eleventh term of the A.P. whose first term and the common difference are 2 and −3 respectively. Also write its first four terms.
Formula \(a_n = a_1 + (n-1)d\) mein values rakho aur simplify karo.
\[
a_n = 2 + (n-1)(-3) = 5 - 3n
\]
\[
a_{11} = 5 - 33 = -28, \qquad 2, -1, -4, -7
\]
Answer: \(a_n = 5 - 3n\), \(a_{11} = -28\); first four terms \(2, -1, -4, -7\)
Example 3: If the \(5^{th}\) term of an A.P. is 13 and \(17^{th}\) term is 49, find \(a_n\) and \(a_{13}\).
Do equations banao, subtract karke \(d\) nikalo, phir \(a_1\) aur general term.
\[
d = \frac{49 - 13}{17 - 5} = \frac{36}{12} = 3, \qquad a_1 = 13 - 4(3) = 1
\]
\[
a_n = 1 + (n-1)3 = 3n - 2, \qquad a_{13} = 37
\]
Answer: \(a_n = 3n - 2\), \(a_{13} = 37\)
Example 4: Find the number of terms in the A.P. if \(a_1 = 3\), \(d = 7\) and \(a_n = 59\).
\(a_n = a_1 + (n-1)d\) mein values rakh kar \(n\) ke liye solve karo.
\[
3 + (n-1)7 = 59 \implies 7(n-1) = 56 \implies n = 9
\]
Answer: \(9\) terms
Example 5: If \(a_{n-2} = 3n - 11\), find the \(n^{th}\) term of the sequence.
Index shift karo: \(m = n - 2\) rakho to \(n = m + 2\), phir \(m\) ko wapas \(n\) likh do.
\[
a_m = 3(m+2) - 11 = 3m - 5 \implies a_n = 3n - 5
\]
Answer: \(a_n = 3n - 5\)
Example 6: Find three A.Ms. between \(\sqrt{2}\) and \(3\sqrt{2}\).
3 means hon to \(d = \frac{b-a}{4}\). Phir \(a + d, a + 2d, a + 3d\) nikalo.
\[
d = \frac{3\sqrt{2} - \sqrt{2}}{4} = \frac{\sqrt{2}}{2}
\]
\[
\frac{3\sqrt{2}}{2}, \quad 2\sqrt{2}, \quad \frac{5\sqrt{2}}{2}
\]
Answer: \(\frac{3\sqrt{2}}{2}, 2\sqrt{2}, \frac{5\sqrt{2}}{2}\)
Example 7: Find the sum of the first 100 positive integers.
Yeh A.P. \(1, 2, \dots, 100\) hai — sum ka formula \(\frac{n}{2}(a_1 + a_n)\) lagao.
\[
S_{100} = \frac{100}{2}(1 + 100) = 50 \times 101 = 5050
\]
Answer: \(5050\)
Example 8: Find the \(19^{th}\) term and the partial sum of 19 terms of the arithmetic series: \(2 + \frac{7}{2} + 5 + \frac{13}{2} + \dots\)
Pehle \(d = \frac{3}{2}\) nikalo, phir \(a_{19}\) aur \(S_{19}\) ke formulas lagao.
\[
a_{19} = 2 + 18\left(\frac{3}{2}\right) = 29, \qquad S_{19} = \frac{19}{2}(2 + 29) = \frac{589}{2}
\]
Answer: \(a_{19} = 29\), \(S_{19} = \frac{589}{2}\)
Example 9: Find the arithmetic series if its fifth term is 19 and \(S_4 = a_5 + 1\).
Do equations: \(a + 4d = 19\) aur \(\frac{4}{2}(2a + 3d) = 20\). Solve karke \(a, d\) nikalo.
\[
a + 4d = 19, \quad 2a + 3d = 10 \implies d = \frac{28}{5}, \; a = -\frac{17}{5}
\]
Answer: \(-\frac{17}{5}, \frac{11}{5}, \frac{39}{5}, \dots\)
Example 10: How many terms of the series \(-9 - 6 - 3 + 0 + \dots\) amount to 66?
\(S_n = 66\) rakh kar quadratic banao — positive root hi terms ki tadad hogi.
\[
\frac{n}{2}\big[-18 + 3(n-1)\big] = 66 \implies n^2 - 7n - 44 = 0 \implies n = 11
\]
Answer: \(11\) terms
Example 11: Find the first three terms of an arithmetic series in which \(a_1 = 9\), \(a_n = 105\) and \(S_n = 741\).
\(S_n = \frac{n}{2}(a_1 + a_n)\) se pehle \(n\) nikalo, phir \(d\), phir terms.
\[
\frac{n}{2}(9 + 105) = 741 \implies n = 13, \qquad d = \frac{105-9}{12} = 8
\]
\[
9, \; 17, \; 25
\]
Answer: \(9, 17, 25\)
Example 12: Find the eighth term of a geometric sequence for which \(a_1 = -3\) and \(r = -2\).
Formula \(a_n = a_1 r^{n-1}\) — negative ratio ki odd/even power ka khayal rakho.
\[
a_8 = -3 \cdot (-2)^7 = -3 \cdot (-128) = 384
\]
Answer: \(384\)
Example 13: Find the \(n^{th}\) term of the G.P. \(3, 12, 48, \dots\)
\(r = 4\), isliye \(a_n = 3 \cdot 4^{n-1}\).
\[
a_n = 3 \cdot 4^{n-1}
\]
Answer: \(a_n = 3 \cdot 4^{n-1}\)
Example 14: Find the tenth term of the G.P., for which \(a_4 = 108\) and \(r = 3\).
Pehle \(a_1 = \frac{a_4}{r^3}\) nikalo, phir \(a_{10} = a_1 r^9\).
\[
a_1 = \frac{108}{27} = 4, \qquad a_{10} = 4 \cdot 3^9 = 78732
\]
Answer: \(78732\)
Example 15: Find the \(5^{th}\) term of the G.P., \(3, 6, 12, \dots\)
\(r = 2\): \(a_5 = 3 \cdot 2^4\).
\[
a_5 = 3 \cdot 2^4 = 48
\]
Answer: \(48\)
Example 16: Find \(a_n\) if \(a_4 = \frac{8}{27}\) and \(a_7 = \frac{-64}{729}\) of a G.P.
Dono equations divide karke \(r^3\) nikalo, phir \(r\), phir \(a_1\).
\[
r^3 = \frac{-64/729}{8/27} = -\frac{8}{27} \implies r = -\frac{2}{3}, \qquad a_1 = \frac{8/27}{-8/27} = -1
\]
\[
a_n = -\left(-\frac{2}{3}\right)^{n-1}
\]
Answer: \(a_n = -\left(-\frac{2}{3}\right)^{n-1}\)
Example 17: Insert three G.Ms. between \(2\) and \(\frac{1}{2}\).
3 means hon to \(r = \left(\frac{1/2}{2}\right)^{1/4} = \frac{1}{\sqrt{2}}\). Phir terms nikalo.
\[
\sqrt{2}, \quad 1, \quad \frac{1}{\sqrt{2}}
\]
Answer: \(\sqrt{2}, 1, \frac{1}{\sqrt{2}}\)
Example 18: Find the sum of \(n\) terms of the geometric series if \(a_n = (-3)\left(\frac{2}{5}\right)^n\).
\(a_1 = -\frac{6}{5}\) aur \(r = \frac{2}{5}\) nikalo, phir \(S_n = \frac{a_1(1-r^n)}{1-r}\) lagao.
\[
S_n = \frac{-\frac{6}{5}\left[1-\left(\frac{2}{5}\right)^n\right]}{1-\frac{2}{5}} = -2\left[1-\left(\frac{2}{5}\right)^n\right]
\]
Answer: \(S_n = -2\left[1-\left(\frac{2}{5}\right)^n\right]\)
Example 19: Sum the series up to \(n\) terms: \(2 \cdot 1 + 3 \cdot 2 + 4 \cdot 2^2 + 5 \cdot 2^3 + \dots\)
Yeh A.G.P. hai (\(a = 2, d = 1, r = 2\)). \(S_n\) likho, \(2S_n\) se ghatao — darmiyan G.P. ban jayegi.
\[
S_n - 2S_n = 2 + (2 + 4 + \cdots + 2^{n-1}) - (n+1)2^n
\]
\[
-S_n = 2 + (2^n - 2) - (n+1)2^n = -n \cdot 2^n \implies S_n = n \cdot 2^n
\]
Answer: \(S_n = n \cdot 2^n\)
Example 20: Sum the series up to \(n\) terms: \(2 + \frac{4}{3} + \frac{6}{9} + \frac{8}{27} + \dots\)
A.G.P. hai (\(a = 2, d = 2, r = \frac{1}{3}\)). Wahi "multiply by \(r\) and subtract" ka tareeqa.
\[
\frac{2}{3}S_n = 2 + \frac{2}{3}\left(1 - \left(\frac{1}{3}\right)^{n-1}\right) - 2n\left(\frac{1}{3}\right)^n
\]
\[
S_n = \frac{9}{2} - \frac{2n+3}{2 \cdot 3^{\,n-1}}
\]
Answer: \(S_n = \frac{9}{2} - \frac{2n+3}{2 \cdot 3^{n-1}}\)
Example 21: Find the sum to \(n\) terms of the series: \(1 + 2x + 3x^2 + 4x^3 + \dots\) where \(x \ne 1\). If \(|x| < 1\), sum the series to infinity.
Standard A.G.P. (\(a = 1, d = 1\)). Subtract karne par \((1-x)S_n = 1 + x + \cdots + x^{n-1} - nx^n\).
\[
(1-x)^2 S_n = 1 - (n+1)x^n + nx^{n+1} \implies S_n = \frac{1-(n+1)x^n + nx^{n+1}}{(1-x)^2}
\]
\[
S_\infty = \frac{1}{(1-x)^2} \quad (|x| < 1)
\]
Answer: \(S_n = \frac{1-(n+1)x^n + nx^{n+1}}{(1-x)^2}\), \(S_\infty = \frac{1}{(1-x)^2}\)
Example 22: Find the \(n^{th}\) and \(8^{th}\) terms of H.P.: \(\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \dots\)
Reciprocals lo — \(2, 5, 8\) A.P. hai (\(d = 3\)). A.P. ka \(n\)th term nikaal kar reciprocal lo.
\[
a_n = \frac{1}{3n-1}, \qquad a_8 = \frac{1}{23}
\]
Answer: \(a_n = \frac{1}{3n-1}\), \(a_8 = \frac{1}{23}\)
Example 23: If the \(4^{th}\) and \(7^{th}\) terms of the H.P. are \(\frac{2}{13}\) and \(\frac{2}{25}\) respectively, find the sequence.
Reciprocals ki A.P. mein \(b_4 = \frac{13}{2}, b_7 = \frac{25}{2}\) — \(d\) aur \(b_1\) nikalo, phir H.P. likho.
\[
d = 2, \quad b_1 = \frac{1}{2} \implies \text{H.P.: } 2, \frac{2}{5}, \frac{2}{9}, \dots, \frac{2}{4n-3}, \dots
\]
Answer: \(2, \frac{2}{5}, \frac{2}{9}, \dots\)
Example 24: Find three harmonic means between \(\frac{1}{5}\) and \(\frac{1}{17}\).
Reciprocals \(5\) aur \(17\) ke darmiyan 3 A.M.s (\(d = 3\)), phir unke reciprocals lo.
\[
\frac{1}{8}, \quad \frac{1}{11}, \quad \frac{1}{14}
\]
Answer: \(\frac{1}{8}, \frac{1}{11}, \frac{1}{14}\)
Example 25: Find the sum of the series \(1^3 + 3^3 + 5^3 + \dots\) to \(n\) terms.
\(T_n = (2n-1)^3 = 8n^3 - 12n^2 + 6n - 1\) — phir standard sums lagao.
\[
S_n = 8\sum n^3 - 12\sum n^2 + 6\sum n - n = 2n^4 - n^2 = n^2(2n^2-1)
\]
Answer: \(S_n = n^2(2n^2-1)\)
Example 26: Find the sum of \(n\) terms of the series whose \(n^{th}\) term is \(n^3 + \frac{3}{2}n^2 + \frac{1}{2}n + 1\).
Har part ka standard sum alag lagao aur \(\frac{n}{4}\) common nikaal kar simplify karo.
\[
S_n = \sum n^3 + \frac{3}{2}\sum n^2 + \frac{1}{2}\sum n + n = \frac{n}{4}\big[(n+1)^2(n+2) + 4\big]
\]
Answer: \(S_n = \frac{n}{4}\big[(n+1)^2(n+2) + 4\big]\)
Example 27: Vehicles arrive at a toll booth at a rate of 4 cars every 5 minutes. Represent the number of cars arriving over time as a sequence and predict the total number of cars after an hour.
Sequence \(4, 8, 12, \dots\) A.P. hai (\(d = 4\)). Aik ghante mein 12 intervals — \(a_{12}\) nikalo.
\[
a_{12} = 4 + 11(4) = 48
\]
Answer: \(48\) cars
Example 28: To buy furniture for a new apartment Tayyab borrowed Rs. 50,000 at 8% simple interest for 11 years. How much interest will he pay?
Simple interest: \(I = P \times r \times t\).
\[
I = 50000 \times 0.08 \times 11 = 44000
\]
Answer: Rs. \(44000\)
Example 29: Amna invests Rs. 200,000 at 5% interest compounded annually. What total amount will she get after 10 years?
Compound amount: \(A = P(1 + r)^n\).
\[
A = 200000(1.05)^{10} = 200000 \times 1.6289 = 325778.9
\]
Answer: Rs. \(325779\) (approx.)
Example 30: A web designer is using a 12-column grid system where each column increases in width by 10px from the previous one. The first column width is 50px wide. Find the total width occupied by all 12 columns.
A.P. hai (\(a_1 = 50, d = 10, n = 12\)) — sum ka formula lagao.
\[
S_{12} = \frac{12}{2}\big[2(50) + 11(10)\big] = 6 \times 210 = 1260
\]
Answer: \(1260\) px
Example 31: Motor Vehicle Leasing Using Arithmetic Sequence. A company leases a motor vehicle with the following terms: the first monthly payment is Rs. 15,000; each subsequent payment increases by Rs. 500; the lease term is 24 months. Find: (i) the payment in the \(24^{th}\) month (ii) the total amount paid over 24 months (iii) if the company can only afford Rs. 400,000 in total, can they complete the lease? (iv) the maximum months \(n\) with \(S_n \le 400000\).
A.P. hai (\(a_1 = 15000, d = 500\)). \(a_{24}\) aur \(S_{24}\) nikalo, phir inequality solve karo.
(i) 24th month payment:
\[
a_{24} = 15000 + 23(500) = 26500
\]
(ii) Total over 24 months:
\[
S_{24} = \frac{24}{2}(15000 + 26500) = 498000
\]
(iii) \(498000 > 400000\) — no, they cannot complete it.
(iv) Maximum \(n\):
\[
\frac{n}{2}\big[30000 + 500(n-1)\big] \le 400000 \implies n^2 + 59n - 1600 \le 0 \implies n = 20
\]
Answer: (i) Rs. \(26500\), (ii) Rs. \(498000\), (iii) No, (iv) \(20\) months
