Mathematics 11 (Punjab Board)
Exercise 6.4 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 6: Sequences and Series.
Question 1: Sum the series:
A.P. ka sum formula \(S_n = \frac{n}{2}[2a_1 + (n-1)d]\) lagao.
(i) \(3 + 6 + 9 + \dots + a_{20}\)
\[
a_1 = 3, \; d = 3 \implies S_{20} = \frac{20}{2}(3 + 60) = 630
\]
(ii) \(\frac{4}{\sqrt{5}} + \sqrt{5} + \frac{6}{\sqrt{5}} + \dots + a_n\)
\[
d = \frac{1}{\sqrt{5}} \implies S_n = \frac{n}{2}\left[\frac{8}{\sqrt{5}} + \frac{n-1}{\sqrt{5}}\right] = \frac{n(n+7)}{2\sqrt{5}}
\]
Answer: (i) \(630\), (ii) \(\frac{n(n+7)}{2\sqrt{5}}\)
Question 2: Find \(S_n\) for each arithmetic series:
Jo data diya ho uske mutabiq formula chuno — aakhri term maloom ho to \(\frac{n}{2}(a_1 + a_n)\).
(i) \(a_1 = 4, n = 25, a_n = 100\)
\[
S_{25} = \frac{25}{2}(4 + 100) = 1300
\]
(ii) \(a_1 = 40, n = 20, d = -3\)
\[
S_{20} = \frac{20}{2}\big[80 + 19(-3)\big] = 10(23) = 230
\]
(iii) \(a_n = 52, n = 21, d = -4\)
\[
a_1 = 52 + 80 = 132, \qquad S_{21} = \frac{21}{2}(132 + 52) = 1932
\]
Answer: (i) \(1300\), (ii) \(230\), (iii) \(1932\)
Question 3: Find \(a_1\) for the arithmetic series: \(d = 8, n = 19, S_n = 1786\).
\(S_n\) ke formula mein values rakh kar \(a_1\) solve karo.
\[
1786 = \frac{19}{2}\big[2a_1 + 18(8)\big] \implies 2a_1 + 144 = 188 \implies a_1 = 22
\]
Answer: \(a_1 = 22\)
Question 4: How many terms of the series \(96 + 93 + 90 + \dots\) amount to 1071?
\(S_n = 1071\) rakh kar quadratic banao — dono roots check karo, jo valid ho woh lo.
\[
\frac{n}{2}\big[192 - 3(n-1)\big] = 1071 \implies n^2 - 65n + 714 = 0
\]
\[
(n-51)(n-14) = 0 \implies n = 51 \text{ or } n = 14
\]
Answer: \(14\) or \(51\) terms
Question 5: If the three sides of a right-angled triangle having perimeter 36 cm are in A.P., find them.
Sides \(a-d, a, a+d\) lo. Perimeter se \(a\) nikalo, phir Pythagoras se \(d\).
\[
3a = 36 \implies a = 12
\]
\[
(12+d)^2 = 12^2 + (12-d)^2 \implies 48d = 144 \implies d = 3
\]
\[
\text{Sides: } 9, 12, 15 \text{ cm}
\]
Answer: \(9\) cm, \(12\) cm, \(15\) cm
Question 6: Sum the series to \(3n\) terms:
Teen-teen terms ke groups banao — har group ka sum aik A.P. banayega, phir us A.P. ka sum nikalo.
(i) \(3 + 5 - 7 + 9 + 11 - 13 + \dots\)
\[
(3+5-7) + (9+11-13) + \cdots = 1 + 7 + 13 + \cdots
\]
\[
S = \frac{n}{2}\big[2 + 6(n-1)\big] = n(3n-2)
\]
(ii) \(1 + 4 - 7 + 10 + 13 - 16 + \dots\)
\[
(1+4-7) + (10+13-16) + \cdots = -2 + 7 + 16 + \cdots
\]
\[
S = \frac{n}{2}\big[-4 + 9(n-1)\big] = \frac{n(9n-13)}{2}
\]
Answer: (i) \(n(3n-2)\), (ii) \(\frac{n(9n-13)}{2}\)
Question 7: Find the sum of 20 terms of the series whose \(n^{th}\) term is \(3n + 1\).
Yeh A.P. hai (\(d = 3\)): \(a_1 = 4\), phir \(S_{20}\) nikalo.
\[
S_{20} = \frac{20}{2}\big[8 + 19(3)\big] = 10 \times 65 = 650
\]
Answer: \(650\)
Question 8: The 5th and 9th terms of an A.P. are 11 and 17 respectively. Find the sum of 20 terms.
Pehle \(d\) aur \(a_1\) nikalo, phir \(S_{20}\).
\[
d = \frac{17-11}{4} = 1.5, \quad a_1 = 11 - 4(1.5) = 5
\]
\[
S_{20} = \frac{20}{2}\big[10 + 19(1.5)\big] = 10 \times 38.5 = 385
\]
Answer: \(385\)
Question 9: Obtain the sum of all integers in the first 1000 positive integers which are neither divisible by 5 nor by 2.
Total sum mein se 2 ya 5 se divisible numbers ka sum minus karo (inclusion-exclusion).
\[
\text{Total} = \frac{1000 \cdot 1001}{2} = 500500
\]
\[
\sum(\text{div. by } 2) = 2 + 4 + \cdots + 1000 = 250500
\]
\[
\sum(\text{div. by } 5) = 5 + 10 + \cdots + 1000 = 100500
\]
\[
\sum(\text{div. by } 10) = 10 + 20 + \cdots + 1000 = 50500
\]
\[
500500 - 250500 - 100500 + 50500 = 200000
\]
Answer: \(200000\)
Question 10: The sum of 9 terms of an A.P. is 171 and its eighth term is 31. Find the series.
\(S_9 = \frac{9}{2}(a_1 + a_9) = 171\) aur \(a_8 = 31\) — dono se \(a_1, d\) nikalo.
\[
a_1 + a_9 = 38, \quad a_1 + 7d = 31 \implies 2a_1 + 8d = 38
\]
\[
2a_1 + 14d = 62 \implies 6d = 24 \implies d = 4, \; a_1 = 3
\]
Answer: \(3, 7, 11, 15, \dots\)
Question 11: The 5th term of an arithmetic progression is 21 and the sum of first six terms is 90. Find the 18th term.
\(a + 4d = 21\) aur \(\frac{6}{2}(2a + 5d) = 90\) — solve karke \(a_{18}\) nikalo.
\[
a + 4d = 21, \quad 2a + 5d = 30 \implies d = 4, \; a = 5
\]
\[
a_{18} = 5 + 17(4) = 73
\]
Answer: \(73\)
Question 12: The sum of three numbers in an A.P. is 24 and their product is 440. Find the numbers.
Numbers \(a-d, a, a+d\) lo — sum se \(a = 8\), phir product se \(d\).
\[
3a = 24 \implies a = 8, \qquad (8-d)(8)(8+d) = 440 \implies 64 - d^2 = 55
\]
\[
d^2 = 9 \implies d = \pm 3 \implies \text{numbers: } 5, 8, 11
\]
Answer: \(5, 8, 11\)
Question 13: The first four terms of an A.P. are 2, 6, 10 and 14. Find the least number of terms needed so that the sum of the terms is greater than 2000.
\(S_n > 2000\) ki inequality banao aur sab se chhota poora \(n\) nikalo.
\[
\frac{n}{2}\big[4 + 4(n-1)\big] > 2000 \implies 2n^2 > 2000 \implies n > 31.6
\]
Answer: \(32\) terms
Question 14: Find four numbers in A.P. whose sum is 32 and the sum of whose squares is 276.
Chaar numbers hon to \(a-3d, a-d, a+d, a+3d\) lo (common difference \(2d\)) — sum se \(a\), squares se \(d\).
\[
4a = 32 \implies a = 8
\]
\[
2(a^2 + 9d^2) + 2(a^2 + d^2) = 276 \implies 4a^2 + 20d^2 = 276
\]
\[
256 + 20d^2 = 276 \implies d^2 = 1 \implies d = \pm 1
\]
\[
\text{Numbers: } 5, 7, 9, 11
\]
Answer: \(5, 7, 9, 11\)
Question 15: Find the five numbers in A.P. whose sum is 25 and the sum of whose squares is 135.
Numbers \(a-2d, a-d, a, a+d, a+2d\) — sum se \(a = 5\), squares se \(d\).
\[
5a = 25 \implies a = 5, \qquad 5a^2 + 10d^2 = 135 \implies 125 + 10d^2 = 135
\]
\[
d^2 = 1 \implies d = \pm 1 \implies \text{numbers: } 3, 4, 5, 6, 7
\]
Answer: \(3, 4, 5, 6, 7\)
Question 16: If \(\frac{1}{a+b}, \frac{1}{c+a}, \frac{1}{b+c}\) are in A.P. then show that \(a^2, b^2, c^2\) are in A.P.
A.P. ki shart lagao, simplify karo — \(a^2 + c^2 = 2b^2\) nikalna hai.
\[
\frac{2}{c+a} = \frac{1}{a+b} + \frac{1}{b+c} = \frac{2b + a + c}{(a+b)(b+c)}
\]
\[
2(a+b)(b+c) = (c+a)(a+c+2b)
\]
\[
2b^2 = a^2 + c^2 \implies a^2, b^2, c^2 \text{ are in A.P.}
\]
Answer: Proved.
Question 17: The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, find the number of terms.
Pehle 4 terms se \(d\) nikalo, phir aakhri 4 terms ka sum \(n\) ke terms mein likh kar solve karo.
\[
\frac{4}{2}\big[22 + 3d\big] = 56 \implies 22 + 3d = 28 \implies d = 2
\]
\[
\text{Last four: } a_{n-3} + a_n = \frac{112}{2} = 56 \implies \big[11 + (n-4)2\big] + \big[11 + (n-1)2\big] = 56
\]
\[
4n + 12 = 56 \implies n = 11
\]
Answer: \(11\) terms
Question 18: The first, second and last terms of an A.P. are \(a, b\) and \(c\) respectively. Show that the sum of the A.P. is \(\frac{(b+c-2a)(c+a)}{2(b-a)}\).
\(d = b - a\) aur \(a_n = c\) se \(n\) nikalo, phir \(S_n = \frac{n}{2}(a_1 + a_n)\) mein rakho.
\[
n = \frac{c-a}{b-a} + 1 = \frac{b+c-2a}{b-a}
\]
\[
S_n = \frac{n}{2}(a + c) = \frac{(b+c-2a)(c+a)}{2(b-a)}
\]
Answer: Proved.
Question 19: Show that the sum of \(n\) A.Ms. between \(a\) and \(b\) is \(n\) times the single A.M. between them.
\(n\) means \(a+d, a+2d, \dots, a+nd\) hain — inka sum A.P. sum formula se nikalo.
\[
\text{Sum} = \frac{n}{2}\big[(a+d) + (a+nd)\big] = \frac{n}{2}\big[2a + (n+1)d\big]
\]
\[
d = \frac{b-a}{n+1} \implies \text{Sum} = \frac{n}{2}(a + b) = n \cdot \frac{a+b}{2}
\]
Answer: Proved.
