Mathematics 11 (Punjab Board)
Exercise 6.2 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 6: Sequences and Series.
Question 1: Find the common difference and write the next two terms of each arithmetic sequence.
Common difference \(d\) do consecutive terms ka farq hai — phir aakhri term mein \(d\) jor kar agle terms likho.
(i) \(9, 16, 23, \dots\)
\[
d = 16 - 9 = 7 \implies 30, \; 37
\]
(ii) \(5, 5+\sqrt{2}, 5+2\sqrt{2}, \dots\)
\[
d = \sqrt{2} \implies 5+3\sqrt{2}, \; 5+4\sqrt{2}
\]
Answer: (i) \(d = 7\); next terms \(30, 37\), (ii) \(d = \sqrt{2}\); next terms \(5+3\sqrt{2}, 5+4\sqrt{2}\)
Question 2: Write the first three terms of each arithmetic sequence, with given information.
\(a_2 = a_1 + d\) aur \(a_3 = a_2 + d\) — bas \(d\) jorte jao.
(i) \(a_1 = 2, d = 13\)
\[
2, \quad 15, \quad 28
\]
(ii) \(a_1 = 12, d = -13\)
\[
12, \quad -1, \quad -14
\]
Answer: (i) \(2, 15, 28\), (ii) \(12, -1, -14\)
Question 3: Find \(a_{n+1}\) and \(a_{2n}\) if \(a_n = 4 + 3n\).
\(a_{n+1}\) ke liye \(n\) ki jagah \(n+1\), aur \(a_{2n}\) ke liye \(2n\) rakho.
\[
a_{n+1} = 4 + 3(n+1) = 3n + 7, \qquad a_{2n} = 4 + 3(2n) = 6n + 4
\]
Answer: \(a_{n+1} = 3n + 7\), \(a_{2n} = 6n + 4\)
Question 4: Find the indicated term of each of the following arithmetic sequences:
Formula \(a_n = a_1 + (n-1)d\) — pehle \(d\) nikalo (agar na diya ho), phir indicated term.
(i) \(a_1 = 3, d = 7, a_{14}\)
\[
a_{14} = 3 + 13(7) = 94
\]
(ii) \(8, 3, -2, \dots, a_{12}\)
\[
d = -5, \quad a_{12} = 8 + 11(-5) = -47
\]
Answer: (i) \(94\), (ii) \(-47\)
Question 5: The 18th and 30th terms of an arithmetic sequence are 367 and 499 respectively. How many terms of this sequence are less than 1000?
Do equations se \(d\) aur \(a\) nikalo, phir \(a_n < 1000\) ki inequality solve karo.
\[
a + 17d = 367, \quad a + 29d = 499 \implies 12d = 132 \implies d = 11, \; a = 180
\]
\[
180 + (n-1)11 < 1000 \implies n - 1 < 74.5 \implies n \le 75
\]
Answer: \(75\) terms
Question 6: Is 301 a term of the A.P. \(5, 11, 17, \dots\)?
Maan lo 301, \(n\)th term hai. Agar \(n\) poora adad (positive integer) aaye to term hai, warna nahi.
\[
5 + (n-1)6 = 301 \implies n - 1 = 49.33 \implies n = 50.33
\]
Since \(n\) is not a positive integer, 301 is not a term of the A.P.
Answer: No, 301 is not a term.
Question 7: If \(2x, x+8, 3x+1\) are in A.P., then find the value of \(x\).
A.P. ki shart: darmiyani term ka double, dono taraf wali terms ke sum ke barabar — \(2b = a + c\).
\[
2(x+8) = 2x + (3x+1) \implies 2x + 16 = 5x + 1 \implies x = 5
\]
Answer: \(x = 5\)
Question 8: Which term of the A.P. \(3, 8, 13, \dots\) is 123?
\(a_n = 123\) rakh kar \(n\) nikalo.
\[
3 + (n-1)5 = 123 \implies n - 1 = 24 \implies n = 25
\]
Answer: \(25^{th}\) term
Question 9: Which term of the A.P. \(30, 29.5, 29, \dots\) is the first negative term?
Sab se pehla negative term woh hai jahan \(a_n < 0\) pehli dafa ho — inequality solve karo.
\[
30 + (n-1)(-0.5) < 0 \implies n - 1 > 60 \implies n > 61
\]
Answer: \(62^{nd}\) term
Question 10: The 7th and 21st terms of an A.P. are 37 and 107 respectively. Find the A.P. and its 100th term.
Do equations se \(a\) aur \(d\) nikalo, phir A.P. likho aur \(a_{100}\) nikalo.
\[
a + 6d = 37, \quad a + 20d = 107 \implies d = 5, \; a = 7
\]
\[
\text{A.P.: } 7, 12, 17, \dots \qquad a_{100} = 7 + 99(5) = 502
\]
Answer: A.P. is \(7, 12, 17, \dots\); \(a_{100} = 502\)
Question 11: If \(\frac{1}{a-c}, \frac{1}{b-c}, \frac{1}{b-a}\) are in A.P., show that \(a-b = b-c\).
A.P. ki shart \(2T_2 = T_1 + T_3\) lagao, simplify karo — perfect squares banenge.
\[
\frac{2}{b-c} = \frac{1}{a-c} + \frac{1}{b-a} = \frac{b-c}{(a-c)(b-a)}
\]
\[
2(a-c)(b-a) = (b-c)^2 \implies (a-b)^2 + (a-c)^2 = 0
\]
\[
\implies a = b = c \implies a - b = b - c = 0
\]
Answer: Proved.
Question 12: How many numbers of three digits are divisible by 7?
7 se divisible pehla 3-digit number 105 aur aakhri 994 hai — yeh A.P. hai (\(d = 7\)), \(n\) nikalo.
\[
105 + (n-1)7 = 994 \implies n - 1 = 127 \implies n = 128
\]
Answer: \(128\)
Question 13: Find the 8th term from the end of the A.P. \(8, 11, 14, \dots, 185\).
A.P. ko ulta kar lo: pehli term 185, \(d = -3\) — phir 8th term nikalo.
\[
a_8 = 185 + 7(-3) = 185 - 21 = 164
\]
Answer: \(164\)
Question 14: Find the \(n^{th}\) term of the progression \(\left(\frac{3}{7}\right)^{10}, \left(\frac{10}{7}\right)^{10}, \left(\frac{17}{7}\right)^{10}, \dots\). Is the progression an A.P.?
Base sequence \(\frac{3}{7}, \frac{10}{7}, \frac{17}{7}, \dots\) A.P. hai (\(d = 1\)) — uska \(n\)th term nikaal kar power 10 lagao.
\[
\text{Base: } a_n = \frac{3}{7} + (n-1) = \frac{7n-4}{7} \implies T_n = \left(\frac{7n-4}{7}\right)^{10}
\]
Since the terms are raised to the power 10, consecutive differences are not constant — it is not an A.P.
Answer: \(T_n = \left(\frac{7n-4}{7}\right)^{10}\); not an A.P.
Question 15: If the arithmetic progressions \(3, 10, 17, \dots\) and \(63, 65, 67, \dots\) are such that their \(n^{th}\) terms are equal, then find the value of \(n\).
Dono ke \(n\)th terms barabar rakh kar \(n\) solve karo.
\[
7n - 4 = 2n + 61 \implies 5n = 65 \implies n = 13
\]
Answer: \(n = 13\)
Question 16: If the \(p^{th}\) term of an A.P. is \(q\) and the \(q^{th}\) term is \(p\), prove that its \(n^{th}\) term is \((p+q-n)\).
Do equations banao, subtract karke \(d = -1\) nikalo, phir \(a\) aur \(a_n\).
\[
a + (p-1)d = q, \quad a + (q-1)d = p \implies (p-q)d = q-p \implies d = -1
\]
\[
a = p + q - 1 \implies a_n = (p+q-1) + (n-1)(-1) = p+q-n
\]
Answer: Proved.
Question 17: If \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in A.P., show that \(b = \frac{2ac}{a+c}\).
A.P. ki shart \(\frac{2}{b} = \frac{1}{a} + \frac{1}{c}\) — simplify karke \(b\) nikalo.
\[
\frac{2}{b} = \frac{1}{a} + \frac{1}{c} = \frac{a+c}{ac} \implies b = \frac{2ac}{a+c}
\]
Answer: Proved.
Question 18: If \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in A.P., show that the common difference is \(\frac{a-c}{2ac}\).
\(d = \frac{1}{b} - \frac{1}{a}\) — Q17 se \(\frac{1}{b} = \frac{a+c}{2ac}\) rakho.
\[
d = \frac{a+c}{2ac} - \frac{1}{a} = \frac{a+c-2c}{2ac} = \frac{a-c}{2ac}
\]
Answer: Proved.
Question 19: If \(a_k\) and \(a_m\) denote two different terms of an A.P., show that its \(n^{th}\) term is \(a_k + (n-k)\left(\frac{a_k - a_m}{k-m}\right)\).
\(a_k - a_m = (k-m)d\) se \(d\) nikalo, phir \(a_n = a_k + (n-k)d\) mein rakho.
\[
d = \frac{a_k - a_m}{k-m} \implies a_n = a_k + (n-k)\left(\frac{a_k - a_m}{k-m}\right)
\]
Answer: Proved.
Question 20: If \(a_1, a_2, a_3, \dots, a_n\) are positive and in A.P., prove that \(\frac{1}{\sqrt{a_1}+\sqrt{a_2}} + \frac{1}{\sqrt{a_2}+\sqrt{a_3}} + \cdots + \frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}} = \frac{n-1}{\sqrt{a_n}+\sqrt{a_1}}\).
Har term ko rationalize karo — conjugate se multiply karne par telescoping sum banegi.
\[
\frac{1}{\sqrt{a_r}+\sqrt{a_{r+1}}} = \frac{\sqrt{a_{r+1}}-\sqrt{a_r}}{a_{r+1}-a_r} = \frac{\sqrt{a_{r+1}}-\sqrt{a_r}}{d}
\]
\[
\text{Sum} = \frac{1}{d}\big[(\sqrt{a_2}-\sqrt{a_1}) + \cdots + (\sqrt{a_n}-\sqrt{a_{n-1}})\big] = \frac{\sqrt{a_n}-\sqrt{a_1}}{d}
\]
\[
d = \frac{a_n-a_1}{n-1} \implies \text{Sum} = \frac{(n-1)(\sqrt{a_n}-\sqrt{a_1})}{a_n-a_1} = \frac{n-1}{\sqrt{a_n}+\sqrt{a_1}}
\]
Answer: Proved.
Question 21: If the roots of the equation \((b-c)x^2 + (c-a)x + (a-b) = 0\) are equal, show that \(a, b, c\) are in A.P.
Equal roots ka matlab discriminant zero — expand karke perfect square banao.
\[
D = (c-a)^2 - 4(b-c)(a-b) = 0
\]
\[
a^2 + 4b^2 + c^2 + 2ac - 4ab - 4bc = 0 \implies (a + c - 2b)^2 = 0
\]
\[
a + c = 2b \implies a, b, c \text{ are in A.P.}
\]
Answer: Proved.
Question 22: If the sides of a right-angled triangle are in A.P., find the ratio of its sides.
Sides \(a-d, a, a+d\) lo (hypotenuse sab se bari). Pythagoras lagao.
\[
(a+d)^2 = a^2 + (a-d)^2 \implies 4ad = a^2 \implies a = 4d
\]
\[
\text{Sides: } 3d, 4d, 5d \implies \text{ratio } 3:4:5
\]
Answer: \(3:4:5\)
Question 23: If the \(n^{th}\) term of a progression is a linear expression in \(n\), then prove that this progression is an A.P.
\(a_n = pn + q\) rakho aur \(a_n - a_{n-1}\) nikalo — agar \(n\) se azad constant aaye to A.P. hai.
\[
a_n - a_{n-1} = (pn+q) - \big(p(n-1)+q\big) = p \quad (\text{constant})
\]
Since the difference is independent of \(n\), the progression is an A.P.
Answer: Proved.
