NCERT Solutions for Class 10 Maths Chapter 5

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions समांतर श्रेढ़ियाँ

Exercise 5.1

Question 1:
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
निम्नलिखित स्थितियों में से किन स्थितियों में संबंद्ध संख्याओं की सूची A.P. है और क्यों?
(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
(i) प्रत्येक किलो मीटर के बाद का टैक्सी का किराया, जबकि प्रथम किलो मीटर के लिए किराया ₹ 15 है और प्रत्येक अतिरिक्त किलो के लिए किराया ₹ 8 है।
Answer
Taxi fare for 1 km = 15
Taxi fare for first 2 kms = 15+8 = 23
Taxi fare for first 3 kms = 23+8 = 31
Taxi fare for first 4 kms = 31+8 = 39
And so on……
Thus, 15, 23, 31, 39 … forms an A.P. because every next term is 8 more than the preceding term.
प्रथम किलोमीटर का किराया = 15 रुपये |
अतिरिक्त किलोमीटर का किराया = 8 रुपये
श्रृंखला : 15, 23, 31, 39 …..
जाँच:
a = 15
d₁ = a₂ – a₁ = 23 – 15 = 8
d₂ = a₃ – a₂ = 31 – 23 = 8
d₃ = a₄ – a₃ = 39 – 31 = 8
चूँकि सभी अंतरों का अंतर सामान है अर्थात सार्वअंतर = 8 है | इसलिए दिया गया सूची A. P है |

(ii) The amount of air present in a cylinder when a vacuum pump removes \(\frac{1}{4}\) of the air remaining in the cylinder at a time.
(ii) किसी बेलन में उपस्थित हवा की मात्रा, जबकि वायु निकालने वाला पंप प्रत्येक बार बेलन की शेष हवा का \(\frac{1}{4}\) भाग बाहर निकाल देता है।
Answer
Let the amount of air be 1
\(T_1 = 1\)
Air removes = \(\frac{1}{4}\)
\(T_2 = 1 – \frac{1}{4} = \frac{3}{4}\)
Air removes = \(\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}\)
\(T_3 = \frac{3}{4} – \frac{3}{16} = \frac{9}{16}\)
Air removes = \(\frac{9}{16} \times \frac{1}{4} = \frac{9}{64}\)
\(T_4 = \frac{9}{16} – \frac{9}{64} = \frac{27}{64}\)
Series: \(1, \frac{3}{4}, \frac{9}{16}, \frac{27}{64}\\
d_1 = \frac{3}{4} – 1 = \frac{-1}{4}\\
d_2 = \frac{9}{16} – \frac{3}{4} = \frac{-3}{16}\)
this series do not have the common difference between them. Therefore, this series is not an A.P.

माना बेलन में हवा की मात्रा 1 है |
\(T_1 = 1\)
हवा निकला = \(\frac{1}{4}\)
\(T_2 = 1 – \frac{1}{4} = \frac{3}{4}\)
हवा निकला = \(\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}\)
\(T_3 = \frac{3}{4} – \frac{3}{16} = \frac{9}{16}\)
हवा निकला = \(\frac{9}{16} \times \frac{1}{4} = \frac{9}{64}\)
\(T_4 = \frac{9}{16} – \frac{9}{64} = \frac{27}{64}\)
शृंखला: \(1, \frac{3}{4}, \frac{9}{16}, \frac{27}{64}\\
d_1 = \frac{3}{4} – 1 = \frac{-1}{4}\\
d_2 = \frac{9}{16} – \frac{3}{4} = \frac{-3}{16}\)
उपरोक्त शृंखला समांतर श्रेणी में नहीं है, क्योंकि सार्वांतर समान नहीं है।

(iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
प्रत्येक मीटर की खुदाई के बाद, एक कुआँ खोदने में आई लागत, जबकि प्रथम मीटर खुदाई की लागत ₹ 150 है और बाद में प्रत्येक मीटर खुदाई की लागत ₹ 50 बढ़ती जाती है।
Answer
Cost of digging for first metre = 150
Cost of digging for first 2 metres = 150 + 50 = 200
Cost of digging for first 3 metres = 200 + 50 = 250
Cost of digging for first 4 metres = 250 + 50 = 300
Clearly, 150, 200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term.
प्रथम मीटर का लागत = 150,
दुसरे मीटर खुदाई की लागत = 150 + 50 = 200
तीसरे मीटर खुदाई की लागत = 200 + 50 = 250
श्रृंखला : 150, 200, 250, 300 …..
जाँच:
a = 150
d₁ = a₂ – a₁ = 200 – 150 = 50
d₂ = a₃ – a₂ = 250 – 200 = 50
d₃ = a₄ – a₃ = 300 – 250 = 50
सार्व अंतर = 50
यहाँ सार्व अंतर समान है इसलिए यह श्रृंखला A.P है|

(iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8 % per annum.
खाते में प्रत्येक वर्ष का मिश्रधन, जबकि ₹ 10000 की राशि 8% वार्षिक की दर से चक्रवृद्धि ब्याज पर जमा की जाती है।
Answer
Principal for first year = 10000
Interest of second year = \(\frac{10000 \times 8 \times 1}{100} = 800\)
Principal of second year = 10800
Interest of third year = \(\frac{10800 \times 8 \times 1}{100} = 864\)
Principal of third year = 11664
Series: 10000, 10800, 11664 ……
Clearly, this series do not have the same difference between them. Therefore, this is not an A.P.

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पहले वर्ष की राशि = 10000
दूसरे वर्ष का चक्रवृद्धि ब्याज = \(\frac{10000 \times 8 \times 1}{100} = 800\)
दूसरे वर्ष की राशि = 10800
तीसरे वर्ष का चक्रवृद्धि ब्याज = \(\frac{10800 \times 8 \times 1}{100} = 864\)
तीसरे वर्ष की राशि = 11664
श्रृंखला: 10000, 10800, 11664 ……
स्पष्ट है कि इस श्रृंखला का सार्व अंतर समान नहीं है अत: A.P नहीं है |

Question 2:
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
प्रश्न 2. दी हुई A.P के प्रथम चार पद लिखिए, जबकि प्रथम पद a और सार्व अंतर d निम्नलिखित हैं :
\((i)\;\; a = 10, \; d = 10 \quad (ii)\; a = –2, \; d = 0\\
(iii)\; a = 4, \; d = – 3 \quad (iv)\; a = – 1, \; d = \frac{1}{2}\\
(v)\;\; a = – 1.25, \; d = – 0.25\)
Answer
(i) a = 10, d = 10
a₁ = a = 10
a₂ = a₁ + d = 10 + 10 = 20;
a₃ = a₂ + d = 20 + 10 = 30;
a₄ = a₃ + d = 30 + 10 = 40;
a₅ = a₄ + d = 40 + 10 = 50
Therefore, the series will be 10, 20, 30, 40, 50 …
First four terms of this A.P. will be 10, 20, 30, and 40.

(ii)  a = – 2, d = 0
a₁ = a = -2
a₂ = a₁ + d = – 2 + 0 = – 2
a₃ = a₂ + d = – 2 + 0 = – 2
a₄ = a₃ + d = – 2 + 0 = – 2
Therefore, the series will be – 2, – 2, – 2, – 2 …
First four terms of this A.P. will be – 2, – 2, – 2 and – 2.

(iii) a = 4, d = – 3
a₁ = a = 4
a₂ = a₁ + d = 4 – 3 = 1
a₃ = a₂ + d = 1 – 3 = – 2
a₄ = a₃ + d = – 2 – 3 = – 5
Therefore, the series will be 4, 1, – 2 – 5 …
First four terms of this A.P. will be 4, 1, – 2 and – 5.

(iv) a = – 1, d = 1/2
a₁ = a = -1
a₂ = a₁ + d = -1 + 1/2 = -1/2
a₃ = a₂ + d = -1/2 + 1/2 = 0
a₄ = a₃ + d = 0 + 1/2 = 1/2
Clearly, the series will be-1, -1/2, 0, 1/2
First four terms of this A.P. will be -1, -1/2, 0 and 1/2.

(v) a = – 1.25, d = – 0.25
a₁ = a = – 1.25
a₂ = a₁ + d = – 1.25 – 0.25 = – 1.50
a₃ = a₂ + d = – 1.50 – 0.25 = – 1.75
a₄ = a₃ + d = – 1.75 – 0.25 = – 2.00
Clearly, the series will be 1.25, – 1.50, – 1.75, – 2.00 ……..
First four terms of this A.P. will be – 1.25, – 1.50, – 1.75 and – 2.00.

Question 3:
For the following APs, write the first term and the common difference:
प्रश्न 3. निम्नलिखित में से प्रत्येक समान्तर श्रेणी के लिए प्रथम पद और सार्वअंतर लिखिए |
\((i)\;\; 3, 1, – 1, – 3, . . . \quad (ii) – 5, – 1, 3, 7, . . .\\
(iii)\; \frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, . . .\; \;(iv)\; 0.6, 1.7, 2.8, 3.9, . . .\)
Answer
(i) 3, 1, – 1, – 3 …
Here, first term, a = 3
Common difference, d = Second term – First term = 1 – 3 = – 2

(ii) – 5, – 1, 3, 7 …
Here, first term, a = – 5
Common difference, d = Second term – First term = ( – 1) – ( – 5) = – 1 + 5 = 4

(iii) 1/3, 5/3, 9/3, 13/3 ….
Here, first term, a = 1/3
Common difference, d = Second term – First term = 5/3 – 1/3 = 4/3

(iv) 0.6, 1.7, 2.8, 3.9 …
Here, first term, a = 0.6
Common difference, d = Second term – First term = 1.7 – 0.6 = 1.1

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Question 4:
Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
प्रश्न 4. निम्नलिखित में से कौन-कौन A.P हैं? यदि कोई A.P है, तो इसका सार्व अंतर ज्ञात कीजिए और इनके तीन पद लिखिए |
\((i)\; 2, 4, 8, 16, . . . \quad\quad (ii) 2, \frac{5}{2}, 3, \frac{7}{2}, . . .\quad\quad\quad \quad \quad (iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . \\ (iv) – 10, – 6, – 2, 2, . . . (v) 3, 3  + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}, . . . (vi) 0.2, 0.22, 0.222, 0.2222, . . \\(vii) 0, – 4, – 8, –12, . . . \;(viii) \frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2},\frac{-1}{2},…\; (ix) 1, 3, 9, 27, . . .\\ (x) a, 2a, 3a, 4a, . . . \; (xi) a, a^2, a^3, a^4, …\; (xii) \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32},…\\ (xiii) \sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, …\;(xiv) 1^2, 3^2, 5^2,7^2,…\;(xv) 1^2, 5^2, 7^2, 73,…\)
Answer
(i) 2, 4, 8, 16 …
a₂ – a₁ = 4 – 2 = 2
a₃ – a₂ = 8 – 4 = 4
a₄ – a₃ = 16 – 8 = 8
⇒ a_n+1 – a_n is not the same every time.
Therefore, the given numbers are forming an A.P.

(ii) 2, 5/2, 3, 7/2 ….
a₂ – a₁ = 5/2 – 2 = 1/2
a₃ – a₂ = 3 – 5/2 = 1/2
a₄ – a₃ = 7/2 – 3 = 1/2
⇒ a_n+1 – a_n is same every time.
Therefore, d = 1/2 and the given numbers are in A.P.
Three more terms are
a₅ = 7/2 + 1/2 = 4
a₆ = 4 + 1/2 = 9/2
a₇ = 9/2 + 1/2 = 5

(iii) -1.2, – 3.2, -5.2, -7.2 …
a₂ – a₁ = ( -3.2) – ( -1.2) = -2
a₃ – a₂ = ( -5.2) – ( -3.2) = -2
a₄ – a₃ = ( -7.2) – ( -5.2) = -2
⇒ a_n+1 – a_n is same every time.
Therefore, d = -2 and the given numbers are in A.P.
Three more terms are
a₅ = – 7.2 – 2 = – 9.2
a₆ = – 9.2 – 2 = – 11.2
a₇ = – 11.2 – 2 = – 13.2

(iv) -10, – 6, – 2, 2 …
a₂ – a₁ = (-6) – (-10) = 4
a₃ – a₂ = (-2) – (-6) = 4
a₄ – a₃ = (2) – (-2) = 4
⇒ a_n+1 – a_n is same every time.
Therefore, d = 4 and the given numbers are in A.P.
Three more terms are
a₅ = 2 + 4 = 6
a₆ = 6 + 4 = 10
a₇ = 10 + 4 = 14

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
a₂ – a₁ = 3 + √2 – 3 = √2
a₃ – a₂ = (3 + 2√2) – (3 + √2) = √2
a₄ – a₃ = (3 + 3√2) – (3 + 2√2) = √2
⇒ a_n+1 – a_n is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more terms are
a₅ = (3 + √2) + √2 = 3 + 4√2
a₆ = (3 + 4√2) + √2 = 3 + 5√2
a₇ = (3 + 5√2) + √2 = 3 + 6√2

(vi) 0.2, 0.22, 0.222, 0.2222 ….
a₂ – a₁ = 0.22 – 0.2 = 0.02
a₃ – a₂ = 0.222 – 0.22 = 0.002
a₄ – a₃ = 0.2222 – 0.222 = 0.0002
⇒ a_n+1 – a_n is not the same every time.
Therefore, the given numbers are forming an A.P.

(vii) 0, -4, -8, -12 …
a₂ – a₁ = (-4) – 0 = -4
a₃ – a₂ = (-8) – (-4) = -4
a₄ – a₃ = (-12) – (-8) = -4
⇒ a_n+1 – a_n is same every time.
Therefore, d = -4 and the given numbers are in A.P.
Three more terms are
a₅ = -12 – 4 = -16
a₆ = -16 – 4 = -20
a₇ = -20 – 4 = -24

(viii) -1/2, -1/2, -1/2, -1/2 ….
a₂ – a₁ = (-1/2) – (-1/2) = 0
a₃ – a₂ = (-1/2) – (-1/2) = 0
a₄ – a₃ = (-1/2) – (-1/2) = 0
⇒ a_n+1 – a_n is same every time.
Therefore, d = 0 and the given numbers are in A.P.
Three more terms are
a₅ = (-1/2) – 0 = -1/2
a₆ = (-1/2) – 0 = -1/2
a₇ = (-1/2) – 0 = -1/2

(ix) 1, 3, 9, 27 …
a₂ – a₁ = 3 – 1 = 2
a₃ – a₂ = 9 – 3 = 6
a₄ – a₃ = 27 – 9 = 18
⇒ a_n+1 – a_n is not the same every time.
Therefore, the given numbers are forming an A.P.

(x) a, 2a, 3a, 4a …
a₂ – a₁ = 2a – a = a
a₃ – a₂ = 3a – 2a = a
a₄ – a₃ = 4a – 3a = a
⇒ a_n+1 – a_n is same every time.
Therefore, d = a and the given numbers are in A.P.
Three more terms are
a₅ = 4a + a = 5a
a₆ = 5a + a = 6a
a₇ = 6a + a = 7a

(xi) a, a², a³, a⁴ …
a₂ – a₁ = a² – a = (a – 1)
a₃ – a₂ = a³ –a² = a² (a – 1)
a₄ – a₃ = a⁴ – a³ = a³(a – 1)
⇒ a_n+1 – a_n is not the same every time.
Therefore, the given numbers are forming an A.P.

(xii) √2, √8, √18, √32 …
a₂ – a₁ = √8 – √2  = 2√2 – √2 = √2
a₃ – a₂ = √18 – √8 = 3√2 – 2√2 = √2
a₄ – a₃ = 4√2 – 3√2 = √2
⇒ a_n+1 – a_n is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more terms are
a₅ = √32  + √2 = 4√2 + √2 = 5√2 = √50
a₆ = 5√2 +√2 = 6√2 = √72
a₇ = 6√2 + √2 = 7√2 = √98

(xiii) √3, √6, √9, √12 …
a₂ – a₁ = √6 – √3 = √3 × 2 -√3 = √3(√2 – 1)
a₃ – a₂ = √9 – √6 = 3 – √6 = √3(√3 – √2)
a₄ – a₃ = √12 – √9 = 2√3 – √3 × 3 = √3(2 – √3)
⇒ a_n+1 – a_n is not the same every time.
Therefore, the given numbers are forming an A.P.

(xiv) 1², 3², 5², 7² …Or, 1, 9, 25, 49 …..
a₂ − a₁ = 9 − 1 = 8
a₃ − a₂ = 25 − 9 = 16
a₄ − a₃ = 49 − 25 = 24
⇒ a_n+1 – a_n is not the same every time.
Therefore, the given numbers are forming an A.P.

(xv) 1², 5², 7², 73 …
Or 1, 25, 49, 73 …
a₂ − a₁ = 25 − 1 = 24
a₃ − a₂ = 49 − 25 = 24
a₄ − a₃ = 73 − 49 = 24
i.e., a_k+1 − a_k is same every time.
⇒ a_n+1 – a_n is same every time.
Therefore, d = 24 and the given numbers are in A.P.
Three more terms are
a₅ = 73+ 24 = 97
a₆ = 97 + 24 = 121
a₇ = 121 + 24 = 145

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NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions समांतर श्रेढ़ियाँ

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Exercise 5.2

Question 1:
Fill in the blanks in the following table, given that a is the first term, d the common difference and \(a_n\) the nth term of the AP:
प्रश्न 1. निम्नलिखित सारणी में, रिक्त स्थानों को भरिए, जहाँ A.P का प्रथम पद a, सार्व अंतर d और nवाँ पद a_n है:

Answer
(i) a = 7, d = 3, n = 8, a_n = ?
a_n = a + (n − 1) d
= 7 + (8 − 1) 3
= 7 + (7) 3
= 7 + 21 = 28
Hence, a_n = 28

(ii) a = −18, n = 10, a_n = 0, d = ?
a_n = a + (n − 1) d
\(\Rightarrow\) 0 = − 18 + (10 − 1) d
\(\Rightarrow\) 18 = 9d
d = 18/9 = 2
Hence, common difference, d = 2

(iii) d = −3, n = 18, a_n = −5
a_n = a + (n − 1) d
\(\Rightarrow\) −5 = a + (18 − 1) (−3)
\(\Rightarrow\) −5 = a + (17) (−3)
\(\Rightarrow\) −5 = a − 51
a = 51 − 5 = 46
Hence, a = 46

(iv) a = −18.9, d = 2.5, a_n = 3.6, n = ?
a_n = a + (n − 1) d
\(\Rightarrow\) 3.6 = − 18.9 + (n − 1) 2.5
\(\Rightarrow\) 3.6 + 18.9 = (n − 1) 2.5
\(\Rightarrow\) 22.5 = (n − 1) 2.5
\(\Rightarrow\) (n – 1) = 22.5/2.5
\(\Rightarrow\) n – 1 = 9
n = 10
Hence, n = 10

(v) a = 3.5, d = 0, n = 105, a_n = ?
a_n = a + (n − 1) d
\(\Rightarrow\) a_n = 3.5 + (105 − 1) 0
\(\Rightarrow\) a_n = 3.5 + 104 × 0
a_n = 3.5
Hence, a_n = 3.5

Question 2:
Choose the correct choice in the following and justify :
प्रश्न 2. निम्नलिखित में सही उत्तर चुनिए और उसका औचित्य दीजिए:
(i) 30th term of the AP: 10, 7, 4, . . . , is
(i) A.P. : 10, 7, 4, …. का 30वाँ पद है:
(a) 97 (b) 77 (c) –77 (d) – 87
Answer
A.P. 10, 7, 4, …
First term, a = 10
Common difference, d = a₂ − a₁ = 7 − 10 = −3
a_n = a + (n − 1) d
a₃₀ = 10 + (30 − 1) (−3)
a₃₀ = 10 + (29) (−3)
a₃₀ = 10 − 87 = −77 (C)

(ii) 11th term of the AP: –A.P. : -3, -1/2, 2 …., is
(ii) A.P. : -3, -1/2, 2 …., का 11वाँ पद है:
(a) 28 (b) 22 (c) -38 (d) \(-48\frac{1}{2}\)
Answer
A.P. is -3, -1/2, ,2 …
First term a = – 3
Common difference, d = a₂ − a₁ = (-1/2) – (-3)
= (-1/2) + 3 = 5/2
a_n = a + (n − 1) d
a₁₁ = 3 + (11 -1)(5/2)
a₁₁ = 3 + (10)(5/2)
a₁₁ = -3 + 25
a₁₁ = 22 (B)

Question 3:
In the following APs, find the missing terms in the boxes :
प्रश्न 3. निम्नलिखित समांतर श्रेढ़ी में, रिक्त खानों (boxes) के पदों को ज्ञात कीजिए | 

Answer
(i) a = 2
and, a₃ = 26
 a_n = a + (n − 1) d
\(\Rightarrow\) a₃ = 2 + (3 – 1) d
\(\Rightarrow\) 26 = 2 + 2d
\(\Rightarrow\) 24 = 2d
d = 12
a₂ = 2 + (2 – 1) 12
= 14
Therefore, 14 is the missing term.

(ii) a₂ = 13
and a₄ = 3
a_n = a + (n − 1) d
a₂ = a + (2 – 1) d
\(\Rightarrow\) 13 = a + d … (i)
and, a₄ = a + (4 – 1) d
\(\Rightarrow\) 3 = a + 3d … (ii)
On subtracting (i) from (ii), we get
– 10 = 2d
d = – 5
From equation (i), we get
13 = a + (-5)
a = 18
a₃ = 18 + (3 – 1) (-5)
= 18 + 2 (-5) = 18 – 10 = 8
Therefore, the missing terms are 18 and 8 respectively.

(iii) a = 5
and a₄ = 19/2
a_n = a + (n − 1) d
a₄ = a + (4 – 1) d
\(\Rightarrow\) 19/2 = 5 + 3d
\(\Rightarrow\) 19/2 – 5 = 3d3d = 9/2
d = 3/2
and, a₂ = a + (2 – 1) d
\(\Rightarrow\) a₂ = 5 + 3/2
a₂ = 13/2
and, a₃ = a + (3 – 1) d
\(\Rightarrow\) a₃ = 5 + 2×3/2
a₃ = 8Therefore, the missing terms are 13/2 and 8 respectively.

(iv) a = −4
and a₆ = 6
a_n = a + (n − 1) d
\(\Rightarrow\) a₆ = a + (6 − 1) d
\(\Rightarrow\) 6 = − 4 + 5d
\(\Rightarrow\) 10 = 5d
d = 2
a₂ = a + d = − 4 + 2 = −2
a₃ = a + 2d = − 4 + 2 (2) = 0
a₄ = a + 3d = − 4 + 3 (2) = 2
a₅ = a + 4d = − 4 + 4 (2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.

(v) a₂ = 38
a₆ = −22
a_n = a + (n − 1) d
\(\Rightarrow\) a₂ = a + (2 − 1) d
\(\Rightarrow\) 38 = a + d … (i)
and a₆ = a + (6 − 1) d
\(\Rightarrow\) −22 = a + 5d … (ii)
On subtracting equation (i) from (ii), we get
− 22 − 38 = 4d
−60 = 4d
d = −15
a = a₂ − d = 38 − (−15) = 53
a₃ = a + 2d = 53 + 2 (−15) = 23
a₄ = a + 3d = 53 + 3 (−15) = 8
a₅ = a + 4d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7 respectively.

Question 4:
Which term of the AP : 3, 8, 13, 18, . . . ,is 78?
प्रश्न 4. A.P. : 3, 8, 13, 18, . . . का कौन सा पद 78 है ?
Answer
3, 8, 13, 18, . . . ,is 78
a = 3, d = 8 – 3 = 5
a_n = a + (n – 1)d = 78
\(\Rightarrow\) 3 + (n – 1) 5 = 78
\(\Rightarrow\) (n – 1) 5 = 78 – 3 = 75
\(\Rightarrow\) n – 1 = \(\frac{75}{5}\) = 15
\(\Rightarrow\) n = 15 + 1 = 16

Question 5:
Find the number of terms in each of the following APs :
प्रश्न 5. निम्नलिखित समांतर श्रेढियों में से प्रत्येक श्रेढ़ी में कितने पद हैं ?
(i) 7, 13, 19, …., 205
(ii) 17, 15½, 13, …., -47
Answer
(i) 7, 13, 19, …., 205
a = 7, d = 13 – 7 = 6
a_n = a + (n – 1)d = 205
\(\Rightarrow\) 7 + (n – 1) 6 = 205
\(\Rightarrow\) (n – 1) 6 = 205 – 7 = 198
\(\Rightarrow\) n – 1 = \(\frac{198}{6}\) = 33
\(\Rightarrow\) n = 33 + 1 = 34

(ii) 18, 15½, 13, …., -47
a = 18, d = 15½ – 18 = \(\frac{-5}{2}\)
a_n = a + (n – 1)d = -47
\(\Rightarrow\) 18 + (n – 1) \(\frac{-3}{2}\) = -47
\(\Rightarrow\) (n – 1)\(\frac{-3}{2}\) = -47 – 18 = -65
\(\Rightarrow\) n – 1 = -65 \(\times \frac{2}{-5}\) = 26
\(\Rightarrow\) n = 26 + 1 = 27

Question 6:
Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . .
क्या A.P. , 11, 8, 5, 2, … का एक पद – 150 है? क्यों?
Answer
11, 8, 5, 2, … 
a = 11, d = 8−11 = −3
n ^th term = – 150
a _n = a +( n −1) d =150
\(\Rightarrow\) 11 + (n -1 ) (-3) = -150
\(\Rightarrow\)  n – 1 = \(\frac{-161}{-3}\)
\(\Rightarrow\) n = \(\frac{164}{3}\) = \(54\frac{2}{3}\)
Clearly, n is not an integer but a fraction.
Therefore, – 150 is not a term of this A.P.

Question 7:
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
उस A.P. का 31 वाँ पद ज्ञात कीजिए, जिसका 11 वाँ पद 38 है और 16 वाँ पद 73 है।
Answer
11 ^th term, a ₁₁ = 38 and 16 ^th term, a ₁₆ = 73
We know that, a _n = a+(n−1)d
a ₁₁ = a+(11−1)d =38
\(\Rightarrow\) a+10d = 38… (i) 
and, a ₁₆ = a +(16−1) d = 73
\(\Rightarrow\) a +15 d = 73 … (ii)
On subtracting equation (i) from (ii) ,
\(\quad\) 5d = 35 \(\Rightarrow\) d = 7
Putting the value of d in equation (i) ,
\(\;\) a +10×(7) = 38 \(\Rightarrow\) a = 38 − 70 \(\Rightarrow\) a = −32
Now, a ₃₁ = a +(31−1) d = − 32 + 30 (7) = − 32 + 210 = 178
Hence, 31 ^st term is 178.

Question 8:
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
एक A.P. में 50 पद हैं, जिसका तीसरा पद 12 है और अंतिम पद 106 है। इसका 29 वाँ पद ज्ञात कीजिए।
Answer
3 ^rd term, a ₃ = 12 and n ^th term, a _n = 106 and n = 50
We know that, a _n = a +( n −1) d 
\(\;\) a ₃ = a + (3 − 1)d = 12
\(\Rightarrow\) a + 2d = 12 … (i) 
and, a ₅₀ = a + (50 − 1)d =  106
\(\Rightarrow\) a +49d  = 106 … (ii) 
On subtracting equation (i) from (ii),
we get, 47d = 94 \(\Rightarrow\) d = 2
Putting the value of d in equation (i) ,
\(\;\) a  + 2(2) = 12 \(\Rightarrow\) a = 8 
Now, a ₂₉ = a  + (29−1) d\(\Rightarrow\) a ₂₉ = 8 + (28) 2 \(\Rightarrow\) a ₂₉ = 8 + 56 = 64
Therefore, 29 ^th term is 64.

Question 9:
If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?
यदि किसी A.P. के दिसरे और नौवें क्रमशः 4 और -8 हैं, तो इसका कौन-सा पद शून्य होगा?
Answer
3 ^rd term, a ₃ = 4 and 9 ^th term, a ₉ = −8
We know that, a _n = a +( n −1) d 
Therefore, a ₃ = a  + (3−1) d  = 4
\(\Rightarrow\) a  + 2d  = 4 …. (i) 
and, a ₉ = a +(9−1) d  = −8
\(\Rightarrow\) a +8 d  = – 8 … (ii) 
On subtracting equation (i) from (ii) ,
we will get, 6 d  = -12 \(\Rightarrow\) d = −2
Putting the value of d in equation (i) ,
\(\;\) a  + 2(−2) = 4 \(\Rightarrow\) a  = 8
Let n ^th term of this A.P. be zero. 
\(\;\) a _n = a +( n −1) d = 0
\(\Rightarrow\) 8 + ( n  − 1)(−2) = 0 \(\Rightarrow\) n – 1 = 4 \(\Rightarrow\) n = 5
Hence, 5 ^th term of this A.P. is 0.

Question 10:
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
किसी A.P. का 17 वाँ पद उसके 10 वें पद से 7 अधिक है। इसका सार्व अंतर ज्ञात कीजिए।
Answer
\(\;\) a ₁₇ = a ₁₀ + 7 [a _n = a +( n −1) d ]
\(\Rightarrow\) a ₁₇ – a ₁₀ = 7
\(\Rightarrow\) [a + (17 – 1)d] – [a + (10 – 1)d] = 7
\(\Rightarrow\)  a +16 d  − a  – 9d  = 7
\(\Rightarrow\) 7d = 7 \(\Rightarrow\) d = 1
Therefore, the common difference is 1.

Question 11:
Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?
A.P. : 3, 15, 27, 39, … का कौन-सा पद उसके 54 वें पद से 132 अधिक होगा?
Answer
3, 15, 27, 39, … 
a = 3, d = a ₂ − a ₁ = 15 − 3 = 12
We know that, a _n = a +( n −1) d 
Therefore, a ₅₄ = a + (54−1) d  = 3 + (53)(12) = 3 + 636 = 639 
a ₅₄ = 639 + 132 = 771
We have to find the term of this A.P. which is 132 more than a _54, i.e.771.
Let n ^th term be 771. 
\(\;\) a _n = a  + ( n −1) d 
\(\Rightarrow\) 771 = 3 + ( n  − 1)12
\(\Rightarrow\) 768 = ( n −1)12
\(\Rightarrow\) ( n −1) = 64 \(\Rightarrow\) n = 65
Therefore, 65 ^th term was 132 more than 54 ^th term.

Question 12:
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
दो समांतर श्रेढ़ियों का सार्व अंतर समान है। यदि इनके 100 वें पदों का अंतर 100 है, तो इनके 1000 वें पदों का अंतर क्या होगा?
Answer
Let, the first term of two APs be a ₁ and a ₂ respectively
And the common difference of these APs be d .
We know, a _n = a +( n −1) d 
Given that, difference between 100 ^th term of the two APs = 100
Therefore, ( a ₁ +99 d ) − ( a ₂ +99 d ) = 100 \(\Rightarrow\) a ₁ − a ₂ = 100 … (i) 
Now,
Difference between 1000 ^th terms of the two APs = ( a ₁ +999 d ) − ( a ₂ +999 d ) = a ₁ − a ₂  = 100 [From (i)]
Hence, the difference between 1000 ^th terms of the two A.P. will be 100.

Question 13:
How many three-digit numbers are divisible by 7?
तीन अंकों वाली कितनी संखयाएं 7 से विभाज्य हैं?
Answer
3-digits numbers which are divisible by 7,
105, 112, 119, ……., 994
a = 105, d = 112 – 105 = 7
and a_n = 994
\(\Rightarrow\) a + (n – 1)d = 994
\(\Rightarrow\) 105 + (n – 1) 7 = 994
\(\Rightarrow\) n – 1 = \(\frac{889}{7}\) = 127
\(\Rightarrow\) n = 128
Therefore, 128 three-digit numbers are divisible by 7.

Question 14:
How many multiples of 4 lie between 10 and 250?
10 और 250 के बीच में 4 के कितने गुणज है?
Answer
Multiples of 4 lie between 10 and 250 are: 12, 16, 20, …, 248
a = 12, d = 16 – 12 = 4
and, a_n = 248
\(\Rightarrow\) a + (n – 1) d = 248
\(\Rightarrow\) 12 + (n – 1) 4 = 248
\(\Rightarrow\) n – 1 = \(\frac{236}{4}\) = 59
\(\Rightarrow\) n = 60
Therefore, there are 60 multiples of 4 between 10 and 250.

Question 15:
For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?
n के किस मान के लिए, दोनों समांतर श्रेढ़ियों 63, 65,67, … और 3, 10, 17, … के nवें पद बराबर होंगे?
Answer
Given two APs as; 63, 65, 67,… and 3, 10, 17,….
Taking first AP,
63, 65, 67, …
a = 63, d = a ₂ −a ₁ = 65 − 63 = 2
We know, n ^th term of this A.P. = a _n = a+(n−1)d 
\(\Rightarrow\) a _n = 63 + ( n  − 1)2 = 63 + 2n −2 = 61 + 2n … (i)
Taking second AP,
3, 10, 17, …
a = 3, d = a ₂ − a ₁ = 10 − 3 = 7
\(\Rightarrow\) a _n = 3 + (n  − 1)7 = 3 + 7n −7 = -4 + 7n … (ii)
Given, n ^th term of these A.P.s are equal to each other.
we get, 61 + 2n  = −4 + 7n 
\(\Rightarrow\) 61 + 4 = 7n – 2n \(\Rightarrow\) 5n = 65 \(\Rightarrow\) n = 13
Therefore, 13 ^th terms of both these A.P.s are equal to each other.

Question 16;
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
वह A.P. ज्ञात कीजिए जिसका तीसरा पद 16 है और 7 वाँ पद 5 वें पद से 12 अधिक है।
Answer
Third term, a ₃ = 16 \(\Rightarrow\) a  + (3 − 1) d = 16 \(\Rightarrow\) a +2 d = 16 … (i)
It is given that, 7 ^th term exceeds the 5 ^th term by 12. 
\(\therefore\) a ₇  = 12 + a ₅
\(\therefore\) a ₇  – a ₅ = 12 \(\Rightarrow\) [a +(7−1) d ] − [a  + (5 − 1) d ] = 12
\(\Rightarrow\)   a  + 6 d  − a  – 4 d  = 12 \(\Rightarrow\) 2 d = 12 \(\Rightarrow\) d = 6
Putting the value of d in equation (i), we get, a  + 2(6) = 16 \(\Rightarrow\) a  + 12 = 16 \(\Rightarrow\) a = 4
Therefore, A.P. will be 4, 10, 16, 22, …

Question 17:
Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.
A.P. : 3, 8, 13, …, 253 में अंतिम पद से 20 वाँ पद ज्ञात कीजिए।
Answer
A. P. 3, 8, 13, …, 253
Common difference, d= 5.
Therefore, we can write the given AP in reverse order as; 253, 248, 243, …, 13, 8, 5
Now for the new AP, first term, a = 253 and common difference, d = 248 − 253 = −5 n = 20
Therefore, a ₂₀ = a  + (20 − 1) d  \(\Rightarrow\) a ₂₀ = 253 + (19)(−5) \(\Rightarrow\) a ₂₀ = 253 − 95 \(\Rightarrow\) a = 158
Therefore, 20 ^th term from the last term of the AP 3, 8, 13, …, 253 . is 158.

Question 18:
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
किसी A.P. के चौथे और 8 वें पदों का योग 24 है तथा छठे और 10 वें पदों का योग 44 है। इस A.P. के प्रथम तीन पद ज्ञात कीजिए।
Answer
It is given that a ₄ + a ₈ = 24 \(\Rightarrow\) a + 3d + a + 7d = 24 \(\Rightarrow\) 2a + 10d = 24
\(\Rightarrow\) a + 5 d = 12 …(1)
and a ₆ + a ₁₀ = 44 \(\Rightarrow\) a + 5d + a + 9d = 44 \(\Rightarrow\) 2a + 14d = 44
\(\Rightarrow\) a + 7d = 22 … (2)
On subtracting equation (1) from (2) ,
we get, 2d = 10 \(\Rightarrow\) d = 5
Putting the value of d in equation (i) , we get, a  + 5(5) = 12 \(\Rightarrow\) a +25 = 12 \(\Rightarrow\) a = −13 
Now a ₁ = – 13, a ₂ = a + d = − 13+5 = −8;  a ₃ = a ₂ + d = − 8+5 = −3
Therefore, the first three terms of this A.P. are −13, −8, and −3.

Question 19:
Subba Rao started work in 1995 at an annual salary of 5000 and received an increment of 200 each year. In which year did his income reach ₹ 7000?
सुब्बा राव ने 1995 में 5000 के मासिक वेतन पद कार्य आरंभ किया और प्रत्येक वर्ष 200 की वेतन वृद्धि प्राप्त की। किस वर्ष में उसका वेतन ₹ 7000 हो गया?
Answer
The incomes of Subba Rao increases every year by Rs.200 and hence, forms an AP.
A.P. 5000, 5200, 5400, … 7000
Here, a = 5000 and d = 200
 a _n =  7000 \(\Rightarrow\) a  + ( n  − 1) d  = 7000 \(\Rightarrow\) 5000 + ( n  − 1)200 = 7000
\(\Rightarrow\) 200( n −1)= 2000 \(\Rightarrow\)  n −1 = 10 \(\Rightarrow\) n = 11
Therefore, in 11th year, his salary will be Rs 7000.

Question 20:
Ramkali saved 5 in the first week of a year and then increaased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.
रामकली ने किसी वर्ष के प्रथम सप्ताह में 50 की बचत की और फिर अपनी साप्ताहिक बचत ₹ 17.5 बढ़ाती गई। यदि nवें सप्ताह में उसकी साप्ताहिक बचत ₹ 207.50 हो जाती है, तो n ज्ञात कीजिए।
Answer
Ramkali saved Rs.5 in first week and then started saving each week by Rs.1.75.
Hence, AP is 5, 6.75, 8.50, ….., 20.75
First term, a = 5 and common difference, d = 1.75
\(\quad\) a _n = 20.75
\(\Rightarrow\) 20.75 = 5 + ( n  – 1) × 1.75 \(\Rightarrow\) 15.75 = ( n  – 1) × 1.75
\(\Rightarrow\) ( n -1) = \(\frac{15.75}{1.75}\) = 9
Hence, n is 10.

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NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions समांतर श्रेढ़ियाँ

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Exercise 5.3

Question 1:
Find the sum of the following AP’s.
(i) 2, 7, 12… to 10 terms
(ii) –37, –33, –29… to 12 terms
(iii) 0.6, 1.7, 2.8… to 100 terms
(iv) \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10}\), . . . to 11 terms
Answer
(i) 2, 7, 12… to 10 terms
Here, a = 2 , d = 7 – 2 = 5 and n = 10
\(\quad\) S_n = \(\frac{n}{2}\) [2a + (n – 1 )d]
\(\Rightarrow\) S₁₀ = \(\frac{10}{2}\) [2(2) + (10 – 1 )5]
\(\Rightarrow\) S₁₀ = 5 [4 + 9 \(\times\)5]
\(\Rightarrow\) S₁₀ = 5 [4 + 45]
\(\Rightarrow\) S₁₀ = 5 [49] = 245

(ii) –37, –33, –29… to 12 terms
Here, a = -37 , d = -33 – (-37) = 4 and n = 12
\(\quad\) S_n = \(\frac{n}{2}\) [2a + (n – 1 )d]
\(\Rightarrow\) S₁₂ = \(\frac{12}{2}\) [2(-37) + (12 – 1 )4]
\(\Rightarrow\) S₁₂ = 6 [-74 + 11 \(\times\)4]
\(\Rightarrow\) S₁₂ = 6 [-74 + 44]
\(\Rightarrow\) S₁₂ = 6 [-30] = -180

(iii) 0.6, 1.7, 2.8… to 100 terms
Here, a = 0.6 , d = 1.7 – 0.6 = 1.1 and n = 100
\(\quad\) S_n = \(\frac{n}{2}\) [2a + (n – 1 )d]
\(\Rightarrow\) S₁₀₀ = \(\frac{100}{2}\) [2(0.6) + (100 – 1 )1.1]
\(\Rightarrow\) S₁₀₀ = 50 [1.2 + 99 \(\times\)1.1]
\(\Rightarrow\) S₁₀₀ = 50 [1.2 + 108.9]
\(\Rightarrow\) S₁₀₀ = 50 [110.1] = 5505

(iv) \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10}\), . . . to 11 terms
Here, a = \(\frac{1}{15}\), d = \(\frac{1}{12} -\frac{1}{15}\) = \(\frac{1}{60}\) and n = 11
\(\quad\) S_n = \(\frac{n}{2}\) [2a + (n – 1 )d]
\(\Rightarrow\) S₁₁ = \(\frac{11}{2}\) [\(2\times\frac{1}{15}\) + (11 – 1 )\(\frac{1}{60}\)]
\(\Rightarrow\) S₁₁ = \(\frac{11}{2}\) \( [\frac{2}{15}\) + \(\frac{10}{60} ] \)
\(\Rightarrow\) S₁₁ = \(\frac{11}{2}\) [\(\frac{18}{60}\)] = \(\frac{33}{20}\)

Question 2:
 Find the sums given below
(i) 7 + 10 \(\frac{1}{2}\) + 14 + ……………… +84
(ii) 34 + 32 + 30 + ……….. + 10
(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)
Answer
(i) 7 + 10 \(\frac{1}{2}\) + 14 + ……………… +84
Here, a = 7, d = 10 \(\frac{1}{2}\) – 7 = \(\frac{21}{2}\) – 7 = \(\frac{7}{2}\)
\(\quad\) a_n = 84 \(\Rightarrow\) a + (n – 1) = 84
\(\Rightarrow\) 7 + (n – 1 ) \(\frac{7}{2}\) = 84 \(\Rightarrow\) (n – 1) \(\frac{7}{2}\) = 77
\(\Rightarrow\) n – 1 = 22 \(\Rightarrow\) n = 23
Now, S₂₃ = \(\frac{23}{2}\) [2a + (23 – 1)d ] = \(\frac{23}{2}\)\(\left[2 \times 7 + 22 \times \frac{7}{2}\right]\)
\(\quad\)\(\quad\) \(\quad\) = \(\frac{23}{2}\)[14 + 77] = \(\frac{23}{2}\times\)91 = \(\frac{2093}{2}\) = 1046\(\frac{1}{2}\)

(ii) 34 + 32 + 30 + ……….. + 10
Here, a = 34, d = 32 – 34 = – 2
\(\quad\) a_n = 10 \(\Rightarrow\) a + (n – 1) = 10
\(\Rightarrow\) 34 + (n – 1 ) (-2) = 10 \(\Rightarrow\) (n – 1) (-2) = – 24
\(\Rightarrow\) n – 1 = 12 \(\Rightarrow\) n = 13
Now, S₁₃ = \(\frac{13}{2}\) [2a + (13 – 1)d ] = \(\frac{13}{2}\) [2 \(\times\) 34 + 12 \(\times\) (-2)]
\(\quad\)\(\quad\) \(\quad\) = \(\frac{13}{2}\)[68 – 24] = \(\frac{13}{2}\times\)44 = 13\(\times\)22 = 286

(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)
Here, a = -5, d = -8 – (-5) = – 3
\(\quad\) a_n = -230 \(\Rightarrow\) a + (n – 1) = -230
\(\Rightarrow\) -5 + (n – 1 ) (-3) = -230 \(\Rightarrow\) (n – 1) (-3) = – 230 + 5 \(\Rightarrow\) (n – 1) (-3) = -225
\(\Rightarrow\) n – 1 = \(\frac{-225}{-3}\) \(\Rightarrow\) n – 1 = 75 \(\Rightarrow\) n = 76
Now, S₇₆ = \(\frac{76}{2}\) [2a + (76 – 1)d ] = \(\frac{76}{2}\) [2 \(\times\) (-5) + 75 \(\times\) (-3)]
\(\quad\)\(\quad\) \(\quad\) = 38 \(\times\)[-10 – 225] = 38 \(\times\) (-235) = -8930

Question 3:
In an AP
(i) Given a = 5, d = 3, a_n = 50, find n and S_n.
(ii) Given a = 7, a₁₃ = 35, find d and S₁₃.
(iii) Given a12 = 37, d = 3, find a and S₁₂.
(iv) Given a3 = 15, S₁₀ = 125, find d and a₁₀.
(v) Given d = 5, S₉ = 75, find a and a₉.
(vi) Given a = 2, d = 8, S_n = 90, find n and a_n.
(vii) Given a = 8, a_n = 62, S_n = 210, find n and d.
(viii) Given a_n = 4, d = 2, S_n = − 14, find n and a.
(ix) Given a = 3, n = 8, S = 192, find d.
(x) Given l = 28, S = 144 and there are total 9 terms. Find a.
Answer
(i) Given a = 5, d = 3, a_n = 50
\(\quad\) a_n = 50 \(\Rightarrow\) a + (n – 1) d = 50
\(\Rightarrow\) 5 + (n – 1) 3 = 50 \(\Rightarrow\) n – 1 = \(\frac{45}{3}\) = 15
\(\Rightarrow\) n = 15 + 1 = 16
Now, S_n= \(\frac{n}{2}\) [2a + (n – 1) d]
\(\quad\) S₁₆ = \(\frac{16}{2}\) [2 \(\times\)5 + (16 – 1) 3]
\(\quad\) \(\quad\) = 8 [10 + 45] = 8 \(\times\) 55 = 440

(ii) Given a = 7, a₁₃ = 35, find d and S₁₃.
\(\quad\) a₁₃ = 35 \(\Rightarrow\) 7 + (13 – 1) d = 35
\(\Rightarrow\) 12d = 35 – 7 \(\Rightarrow\) d = \(\frac{28}{12}\) = \(\frac{7}{3}\)
Now, S_n= \(\frac{n}{2}\) [2a + (n – 1) d]
\(\quad\) S₁₃ = \(\frac{13}{2}\) [2 \(\times\)7 + (13 – 1)\(\frac{7}{3}\)]
\(\quad\) \(\quad\) = \(\frac{13}{2}\) [14 + 28] = \(\frac{13}{2}\) \(\times\) 42 = 273

(iii) Given a₁₂ = 37, d = 3, find a and S₁₂.
\(\quad\)a₁₂ = 37 \(\Rightarrow\) a + (12 -1 )d = 37 \(\Rightarrow\) a + 11(3) = 37 \(\Rightarrow\) a = 37 – 33
\(\Rightarrow\) a = 4
Now, S₁₂ = \(\frac{12}{2}\)[2a + (12 -1)d]
\(\quad\) = 6[2(4) + 11(3)] = 6 \(\times\) 41 = 246

(iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀.
\(\quad\) a₃ = 15 \(\Rightarrow\) a + 2d = 15 … (1)
and, S₁₀ = 125 \(\Rightarrow \frac{10}{2}\) [2a + 9d] = 125
\(\Rightarrow\) 2a + 9d = 25 …. (2)
Solving (1) and (2), 5d = – 5 \(\Rightarrow\) d = -1

(v) Given d = 5, S₉ = 75, find a and a₉.
\(\quad\)
(vi) Given a = 2, d = 8, S_n = 90, find n and a_n.
\(\quad\)
(vii) Given a = 8, a_n = 62, S_n = 210, find n and d.
\(\quad\)
(viii) Given a_n = 4, d = 2, S_n = − 14, find n and a.
\(\quad\)
(ix) Given a = 3, n = 8, S = 192, find d.
\(\quad\) S = 192 \(\Rightarrow\) \(\frac{n}{2}\)[a + a_n] = 192
\(\Rightarrow\) \(\frac{8}{2}\)[3 + a_n] = 192 \(\Rightarrow\) 3 + a_n = 192 \(\times \frac{2}{8}\) = 48
\(\Rightarrow\) a_n = 45 \(\Rightarrow\) a + (n – 1)d = 45 \(\Rightarrow\) 3 + (8 – 1) d = 45
\(\Rightarrow\) 7d = 42 \(\Rightarrow\) d = 6

(x) Given l = 28, S = 144 and there are total 9 terms. Find a.
\(\quad\) l = a_n = 28 \(\Rightarrow\) a + (9 – 1)d = 28 \(\Rightarrow\) a + 8d = 28 … (1)
\(\quad\) S= 144 \(\Rightarrow\) \(\frac{n}{2}\)[a + a_n] = 28 \(\Rightarrow\) \(\frac{9}{2}\)[a + 28] = 144
\(\Rightarrow\) a + 28 = 144 \(\times \frac{2}{9}\) = 32 \(\Rightarrow\) a = 32 – 28 = 4
Putting the value of a in equation (1), 4 + 8d = 28 \(\Rightarrow\) 8d = 24 \(\Rightarrow\) d = 3

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NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions समांतर श्रेढ़ियाँ

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Question 4:
How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?
Answer
A. P. 9, 17, 25, . . .
Here, a = 9, d = 17 – 9 = 8
\(\quad\) S_n = 636
\(\Rightarrow\) \(\frac{n}{2}\)[2a + (n – 1)d] = 636
\(\Rightarrow\) \(\frac{n}{2}\)[2(9) + (n – 1)(8)] = 636
\(\Rightarrow\) n[18 + 8n – 8] = 636 \(\times\) 2
\(\Rightarrow\) n(8n + 10) = 1272
\(\Rightarrow\) 8n² + 10 n – 1272 = 0
\(\Rightarrow\) 2(4n² + 5n – 636) = 0
\(\Rightarrow\) 4n² + 53n – 48n – 636 = 0
\(\Rightarrow\) n(4n + 53) – 12(4n + 53) = 0
\(\Rightarrow\) (4n + 53) (n – 12) = 0
\(\Rightarrow\) n = 12 or \(\frac{-53}{4}\)
But, n \(\neq \frac{-53}{4}\) \(\therefore\) n = 12

Question 5:
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Answer
Given: a = 5, a_n = 45 and S = 400
\(\quad\) a_n = 45 \(\Rightarrow\) a + (n -1) = 45 \(\Rightarrow\) 5 + (n – 1)d = 45 \(\Rightarrow\) (n – 1)d = 40 … (1)
And, S = 400 \(\Rightarrow\) \(\frac{n}{2}\)[a + a_n] = 400 \(\Rightarrow\) n [5 + 45] = 800 \(\Rightarrow\) n = 16
Putting the value of n in equation (1), (16 – 1)d = 40 \(\Rightarrow\) d = \(\frac{40}{15}\) \(\Rightarrow\) d = \(\frac{8}{3}\)

Question 6:
The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Answer
Given: a = 17 and last term, an = 350 and d = 9, n ? and S = ?
last term, a_n = 350 \(\Rightarrow\) a + (n – 1)d = 350 \(\Rightarrow\) 17 + (n – 1)9 = 350
\(\Rightarrow\) (n – 1)9 = 350 – 17 \(\Rightarrow\) n – 1 = \(\frac{333}{9}\) \(\Rightarrow\) n = 37 + 1 = 38
And, S = \(\frac{n}{2}\)[a + an] = \(\frac{38}{2}\)[17 + 350] = 19 (367) = 6978

Question 7:
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Answer
Given d = 7 and 22nd term = 149
\(\quad\)a₂₂ = 149 \(\Rightarrow\) a + (22 – 1)d = 149 \(\Rightarrow\) a + 21(7) = 149 \(\Rightarrow\) a = 149 – 147
\(\Rightarrow\) a = 2
Now, S₂₂ = \(\frac{22}{2}\)[a + a₂₂] = 11[2 + 149] = 11(151) = 1661

Question 8:
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Answer
Given, a₂ = 14 and a₃ = 18
\(\quad)a₂ = 14 \(\Rightarrow\) a + (2 – 1)d = 14 \(\Rightarrow\) a + d = 14 … (1)
and, a₃ = 18 \(\Rightarrow\) a + (3 – 1)d = 18 \(\Rightarrow\) a + 2d = 18 … (2)
Subtracting (2) from (1), d = 4
Putting the value of d in equation (1), a = 10
Now, S₅₁ = \(\frac{51}{2}\)[2a + (51 – 1)d] = \(\frac{51}{2}\)[2(10) + 50(4)] = \(\frac{51}{2}\) \(\times\) 220 = 5610

Question 9:
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Answer
Given S₇ = 49 and S₁₄ = 289
\(\quad\) S₇ = 49 \(\Rightarrow\) \(\frac{7}{2}\)[2a + (7 – 1)d] = 49 \(\Rightarrow\) 2a + 6d = 49 \(\times\)\(\frac{2}{7}\)
\(\Rightarrow\) 2a + 6d = 14 … (1)
and, S₁₇ = 289 \(\Rightarrow\) \(\frac{17}{2}\)[2a + (17 – 1)d] = 289 \(\Rightarrow\) 2a + 16d = 289 \(\times\)\(\frac{2}{17}\)
\(\Rightarrow\) 2a + 16d = 34 … (2)
Subtracting (2) from (1), 10d = 20 \(\Rightarrow\) d = 2
Putting the value of d in equation (1), 2a + 12 = 14 \(\Rightarrow\) a = 1
Now, S = \(\frac{n}{2}\)[2a + (n – 1)d ] = \(\frac{n}{2}\)[2(1) + (n – 1)2] ] = \(\frac{n}{2}\) [2 + 2n – 2] = n².

Question 10:
Show that a₁, , a₂, , . . ., an form an AP where an is defined as below :
(i) a_n = 3 + 4n (ii) a_n = 9 – 5n
Also find the sum of the first 15 terms in each case.
Answer
(i) a_n = 3 + 4n
for n = 1, a₁ = 3 + 4(1) = 7;
for n = 2, a₂ = 3 + 4(2) = 11;
for n = 3, a₃ = 3 + 4(3) = 15;
for n = 4, a₄ = 3 + 4(4) = 19;
It can be observed that
a₂ – a₁ = 11 – 7 = 4
a₃ – a₂ = 11 – 7 = 4
a₄ – a₃ = 11 – 7 = 4
i.e., a_n+1 – a_n is same every time.
Therefore, this is an AP with common differnece as 4 and first term as 7.
Now, S₁₅ = \(\frac{15}{2}\)[2a + (15 – 1)d] = \(\frac{15}{2}\) [2(7) + 14(4)] = \(\frac{15}{2}\) \(\times\) 70 = 525

(ii) a_n = 9 – 5n
for n = 1, a₁ = 9 – 5(1) = 4;
for n = 2, a₂ = 9 – 5(2) = -1;
for n = 3, a₃ = 9 – 5(3) = -6;
for n = 4, a₄ = 9 – 5(4) = -11;
It can be observed that
a₂ – a₁ = -1 – 4 = -5
a₃ – a₂ = -6 – (-1) = -5
a₄ – a₃ = -11 – (-6) = -5
i.e., a_n+1 – a_n is same every time.
Therefore, this is an AP with common differnece as -5 and first term as 4.
Now, S₁₅ = \(\frac{15}{2}\)[2a + (15 – 1)d] = \(\frac{15}{2}\) [2(4) + 14(-5)] = \(\frac{15}{2}\) \(\times\) (-62) = -465

Question 11:
If the sum of the first n terms of an AP is 4n – n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Answer
Given: S_n = 4n – n²
for n = 1, S₁ = 4(1) – (1)² = 3
for n = 2, S₂ = 4(2) – (2)² = 4
for n = 3, S₃ = 4(3) – (3)² = 3
First term, a = S₁ = 3
The sume of first two terms = S₂ = 4
Second term = a₂ = S₂ – S₁ = 4 – 3 = 1
And, 3rd term = a₃ = S₃ – S₂ = 3 – 4 = -1
Also, 10th term = a₁₀ = S₁₀ – S₉ = [4(10) – (10)²] – [4(9) – (9)²] = 40 – 100 – 36 + 81 = 4 – 19 = – 15
And, nth term = a_n = S_n – S_{n – 1} = [4n – n²] – [4(n -1) – (n – 1)²]
\(\quad\)\(\quad\)\(\quad\)\(\quad\)\(\quad\)\(\quad\)\(\quad\) = 4n – n² – (4n – 4 – n² – 1 + 2n) = 4n – n² – 4n + 4 + n² + 1 – 2n = 5 – 2n

Question 12:
Find the sum of the first 40 positive integers divisible by 6.
Answer
The first positive integers that are divisible by 6 are 6, 12, 18, 24 … 40 terms.
Here, first term = a = 6 and Common difference = d = 12 – 6 = 6, n = 40
\(\quad\) S₄₀ = \(\frac{40}{2}\)[2a + (40 – 1) d] = 20 [2(6) + 39\(\times\)6] = 20[12 + 234] = 4920

Question 13:
Find the sum of the first 15 multiples of 8.
Answer
The first 15 multiples of 8 are 8,16, 24, 32 … 15 times
First term = a = 8 and Common difference = d = 16 – 8 = 8, n = 15
\(\quad\) S₁₅ = \(\frac{15}{2}\)[2a + (15 – 1) d] = \(\frac{15}{2}\)[2(8) + 14\(\times\)8] = \(\frac{15}{2}\)[16 + 112]
\(\quad\) \(\quad\) \(\quad\) = \(\frac{15}{2}\)[128]= 15 \(\times\)64 = 960

Question 14:
Find the sum of the odd numbers between 0 and 50.
Answer
The odd numbers between 0 and 50 are 1, 3, 5, 7 … 49
First term = a = 1, Common difference = 3 – 1 = 2, Last term = l = 49
\(\quad\) a_n = 49 ⇒ 49 = 1 + (n − 1) 2 ⇒ 49 = 1 + 2n − 2 ⇒ 50 = 2n ⇒ n = 25
\(\quad\) S_n = \(\frac{n}{2}\) [a + a_n] = \(\frac{25}{2}\)[1 + 49] = 625

Question 15:
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows:
200 for the first day, 250 for the second day, 300 for the third day, etc., the penalty for each succeeding day being 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Answer
Penalty for first day = Rs 200, Penalty for second day = Rs 250 Penalty for third day = Rs 300
It is given that penalty for each succeeding day is Rs 50 more than the preceding day.
So, we have an AP of the form 200, 250, 300, 350 … 30 terms
First term = a = 200, Common difference = d = 50, n = 30
\(\quad\) S₃₀ = \(\frac{30}{2}\)[2a + (30 – 1)d] = 15[2(200) + 29(50)] = 15 [400 + 1450] = 15 \(\times\) 1850 = 27750

Question 16:
A sum of 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is 20 less than its preceding prize, find the value of each of the prizes.
Answer
It is given that sum of seven cash prizes is equal to Rs 700.
And, each prize is R.s 20 less than its preceding term.
Let value of first prize = Rs. a
Let value of second prize =Rs (a − 20)
Let value of third prize = Rs (a − 40)
So, we have sequence of the form:
a, a − 20, a − 40, a – 60 …
First term = a, Common difference = d = (a – 20) – a = –20
n = 7 (Because there are total of seven prizes)
\(\quad\) S = \(\frac{n}{2}\)[2a + (n – 1)d] = 700
\(\quad\) S₇ = \(\frac{7}{2}\)[2a + (7 – 1)(-20)] = 700
\(\Rightarrow\) 2a – 120 = 700 \(\times\) \(\frac{2}{7}\) = 200
\(\Rightarrow\) 2a = 320 \(\Rightarrow\) a = 160
Therefore, value of first prize = Rs 160
Value of second prize = 160 – 20 = Rs 140
Value of third prize = 140 – 20 = Rs 120
Value of fourth prize = 120 – 20 = Rs 100
Value of fifth prize = 100 – 20 = Rs 80
Value of sixth prize = 80 – 20 = Rs 60
Value of seventh prize = 60 – 20 = Rs 40

Question 17:
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Answer
The number of trees planted by class I = number of sections
The number of trees planted by class II = number of sections
The number of trees planted by class III = number of sections
Therefore, we have sequence of the form 3, 6, 9 … 12 terms
To find total number of trees planted by all the students, we need to find sum of the sequence 3, 6, 9, 12 … 12 terms.
First term = a = 3, Common difference = d= 6 – 3 = 3 and n = 12
\(\quad\) S₁₂ = \(\frac{12}{2}\)[2a + (12 – 1)d] = 6[2(3) + 11(3)] = 6 (39) = 234

Question 18:
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . . as shown in Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take \(\pi\) = \(\frac{22}{7}\))

[Hint : Length of successive semicircles is l1, l2, l3, l4, . . . with centres at A, B, A, B, . . ., respectively.]
Answer
We know, Perimeter of a semi-circle = π r 
Therefore, p₁ = π(0.5) = \(\frac{\pi}{2}\) cm; p₂ = π(1) = π cm; p₃ = π(1.5) = \(\frac{3\pi}{2}\) cm
Where, P _1, P ₂ , P ₃ are the lengths of the semi-circles.
Hence we got a series here, as, \(\frac{\pi}{2}\), π, \(3\frac{\pi}{2}\), 2π, ….
P ₁ = π/2 cm P ₂ = π cm 
Common difference, d = P ₂ – P ₁ = π – \(\frac{\pi}{2}\) = \(\frac{\pi}{2}\)
First term = P ₁ = a = \(\frac{\pi}{2}\) cm
By the sum of n term formula, we know, S _n = \(\frac{n}{2}\)[2 a + ( n – 1) d ]
Therefore, the sum of the length of 13 consecutive circles is:
S ₁₃ = \(\frac{13}{2}\) [2(\(\frac{\pi}{2}\)) + (13 – 1)\(\frac{\pi}{2}\)] = \(\frac{13}{2}\) [π + 6π] = \(\frac{13}{2}\) (7π)
= \(\frac{13}{2}\) × 7 × \(\frac{22}{7}\) = 143 cm

Question 19:
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Fig. 5.5). In how many rows are the 200 logs placed and how many logs are in the top row?

Answer
We can see that the numbers of logs in rows are in the form of an A.P. 20, 19, 18,…, 200
For the given A.P., a = 20 and d = a ₂ − a ₁ = 19−20 = −1
Let a total of 200 logs be placed in n rows.
Thus, S _n = 200
\(\quad\) S _n = \(\frac{n}{2}\)[2 a +( n -1) d ] 
\(\quad\) S ₁₂ = \(\frac{12}{2}\)[2(20) + ( n  – 1)(-1)]
\(\Rightarrow\) 200 = \(\frac{12}{2}\) (40− n +1) \(\Rightarrow\) 400 = n (41- n )
\(\Rightarrow\) 400 = 41 n − n ² \(\Rightarrow\) n ² −41 n + 400 = 0 \(\Rightarrow\) n ² −16 n −25 n +400 = 0 
\(\Rightarrow\) n ( n −16)−25( n −16) = 0 \(\Rightarrow\) ( n −16)( n −25) = 0
Either ( n −16) = 0 or n −25 = 0 n = 16 or n = 25
By the nth term formula, a _n = a  + ( n −1) d \(\Rightarrow\) a ₁₆ = 20 + (16−1)(−1) \(\Rightarrow\) a ₁₆ = 20−15 \(\Rightarrow\) a ₁₆ = 5
Similarly, the 25 ^th term could be written as; a ₂₅ = 20 + (25−1)(−1) \(\Rightarrow\) a ₂₅ = 20−24 = − 4
It can be seen, the number of logs in the 16 ^th row is 5 as the numbers cannot be negative.
Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16 ^th row is 5.

Question 20:
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig. 5.6).

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint : To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]
Answer
The distance of the potatoes from the bucket are 5, 8, 11, 14…, which is in the form of AP.
Given, the distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept.
Therefore, the distance to be run w.r.t distances of potatoes, could be written as; 10, 16, 22, 28, 34,………. 
Hence, the first term, a = 10 and d = 16−10 = 6 S ₁₀ =?
We know, S ₁₀ = \(\frac{10}{2}\)[2(10) + (10 – 1)(6)] = 5[20 + 54] = 5(74) = 370
Therefore, the competitor will run a total distance of 370 m.

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EXERCISE 5.4 (Optional)*

Question 1:
Which term of the AP : 121, 117, 113, . . ., is its first negative term?
[Hint : Find n for a_n< 0]

Question 2:
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP

Question 3:
A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2\(\frac{1}{2}\) m apart, what is the length of the wood required for the rungs? [Hint : Number of rungs = \(\frac{250}{25}\) + 1 ]

Question 4:
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
[Hint : S_{x – 1} = S₄₉ – S_x ]

Question 5:
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete.
Each step has a rise of \(\frac{1}{4}\) m and a tread of \(\frac{1}{2}\) m. (see Fig. 5.8). Calculate the total volume
of concrete required to build the terrace.
[Hint : Volume of concrete required to build the first step = \(\frac{1}{4}\) \(\times\) \(\frac{1}{2}\) \(\times\) 50 m³ ]

Answer

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NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions समांतर श्रेढ़ियाँ

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