
01. Real Numbers
10
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Lecture1.2

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Lecture1.6

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02. Polynomials
12
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03. Linear Equation
12
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04. Quadratic Equation
10
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05. Arithmetic Progressions
11
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06. Some Applications of Trigonometry
7
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07. Coordinate Geometry
17
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08. Triangles
18
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09. Circles
9
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10. Areas Related to Circles
12
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11. Introduction to Trigonometry
9
Lecture11.1

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12. Surface Areas and Volumes
9
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Lecture12.7

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13. Statistics
14
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Lecture13.2

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Lecture13.7

Lecture13.8

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Lecture13.13

Lecture13.14


14. Probability
9
Lecture14.1

Lecture14.2

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Lecture14.4

Lecture14.5

Lecture14.6

Lecture14.7

Lecture14.8

Lecture14.9


15. Construction
7
Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5

Lecture15.6

Lecture15.7

NCERT Solutions – Arithmetic Progressions Exercise 5.1 – 5.4
Q.1 In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
(ii) The amount of air present in the cylinder when a vacuum pump removes 14 of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.
(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.
Sol. (i) According to the statement, the fare for journey of 1km, 2 km, 3, km, 4 km…. are respectively Rs 15, Rs (15 + 8), Rs (15 + 2 × 8), Rs (15 + 3 × 8) …
⇒ 15, 23, 31, 39 ….
Since, here the terms continually increases by the same number 8. So, the list forms an AP.
(ii) Let the amount of air present in the cylinder be x units.
Therefore, According to the statement, the the list giving the air present in the cylinder is given by
x,x−x4=3x4,3x4−14×3x4
=12x−3x16
=9x16,....
Here, 3x4−x=−x4
and 9x16−3x4=9x−12x16=−3x16
⇒ 3x4−x≠9x16−3x4
Therefore, These numbers do not form an AP.
(iii) According to question, the cost of digging a well for the first metre and its succeeding metres in rupees is given by 150, 200, 250, 300, ….
Since, here the terms continually increase by the same number 50. So, the list forms an AP.
(iv) According; to the question, the amount of money in the account in the first year and its succeeding years in rupees is given by
⇒ 10000,10000×[1+8100], 10000×[1+8100]2.....
i.e., 10000,10000×108100, 10000×108100×108100,....
i.e., 10000, 10800, 11664, ….
Since, 10800 – 10000 ≠ 11664 – 10800
Therefore The list does not form an AP.
Q.2 Write first four terms of the AP, when the first term a and common difference d are given as follows :
(i) a = 10, d = 10 (ii) a = – 2, d = 0
(iii) a = 4, d = – 3 (iv) a = – 1, d=12
(v) a = – 1.25, d = – 0.25
Sol. We know that if the first term is a and the common difference is d, then
a, a + d, a + 2d, a + 3d, …. represents an AP for different values of a and d.
(i) Putting a = 10, d = 10 in a, a + d, a + 2d, a + 3d, …., we get the required AP as 10, 10 + 10, 10 + 20, 10 + 30, …..
i.e., 10, 20 , 30 , 40 ….
(ii) Putting a = – 2, d = 0 in a, a + d , a + 2d, a + 3d, ….., we get the required AP as – 2, – 2 + 0 – 2 + 2 × 0, – 2 + 3 × 0, …..
i.e., – 2, – 2, – 2, – 2, ….
(iii) Putting a = 4, d = – 3 in a, a + d, a + 2d, a + 3d, … we get the required AP as 4, 4 – 3, 4 – 6, 4 – 9, …
i.e., 4, 1 , – 2, – 5, …
(iv) Putting a = – 1, d=12 in a + 0, a + 2d, a + 3d, …
−1,−1+12,−1+22,−1+32,.....
i.e., −1,−12,0,12....
(v) Putting a = – 1.25, d = – 0.25 in a, a + d, a + 2d, a + 3d, …. we get the required AP as
– 1.25, – 1.25 – 0.25, – 1.25 – 0.50, – 1.25 – 0.75, …
i.e., – 1.25, – 1.50, – 1.75, – 2.00, …
Q.3 For the following APs, write the first term and the common difference:
(i) 3, 1,–1, –3, ….
(ii) –5, –1, 3, 7, …
(iii) 13,53,93,133....
(iv) 0.6, 1.7, 2.8, 3.9…
Sol. (i) The given AP is 3, 1, – 1, – 3…..
Clearly, a = 3 and d = 1 – 3 = – 2.
(ii) The given AP is – 5, –1, 3, 7, ….
Clearly, a = – 5 and d = –1 – (–5) = – 1 + 5 = 4
(iii) The given AP is 13,53,93,133....
Clearly, a=13andd=53−13=43
(iv) The given AP is 0.6, 1.7, 2.8, 3.9,…..
Clearly , a = 0.6 and d = 1.7 – 0.6 = 1.1
Q.4 Which of the following are APs ? If they form an AP, find the common difference d. and write three more terms.
(i) 2, 4, 8, 16, …
(ii) 2,52,3,72,....
(iii) – 1.2, – 3.2, – 5.2, – 7.2, …
(iv) – 10, – 6, – 2, 2, ….
(v) 3,3+2–√,3+22–√,2,3+32–√,...
(vi) 0.2, 0.22, 0.222, 0.2222, …
(vii) 0, – 4, – 8, – 12, …..
(viii) −12,−12,−12,−12,....
(ix) 1, 3, 9, 27, ….
(x) a, 2a, 3a, 4a, …
(xi) a,a2,a3,a4....
(xii) 2–√,8–√,18−−√,32−−√,...
(xiii) 3–√,6–√,9–√,12−−√,...
(xiv) 12,32,52,72....
(xv) 12,52,72,73….
Sol. (i) Here , a2−a1=4−2=2
and, a3−a2=8−4=4
⇒ a2−a1≠a3−a2
Thus, 2, 4, 8, 16, ….. does not form an AP
(ii) Here, a2−a1=52−2=5−42=12 ,
a3−a2=3−52=6−52=12 ,
and, a4−a3=72−3=7−62=12
i.e., an+1−an is same every time, so the given list of numbers forms an AP.
So, d=12
The next three terms after the last given term are
72+12=7+12=82=4 ,
4+12=412=92 ,
and 92+12=9+12=102=5
(iii) Here, a2−a1 = – 3.2 – (–1.2) = – 3.2 + 1.2 = –2,
a3−a2 = – 5.2 – (–3.2) = – 5.2 + 3.2 = – 2
and a4−a3 = – 7.2 – (–5.2) = – 7.2 + 5.2 = – 2
i.e., an+1−an is same every time, so the given list of numbers forms an AP.
So, d = – 2
The next three terms after the last given term are
– 7.2 – 2 = – 9.2,
– 9.2 – 2 = – 11.2,
and, – 11.2 – 2 = – 13.2
(iv) Here, a2−a1 = – 6 – (–10) = – 6 +10 = 4,
a3−a2 = – 2 – (– 6) = – 2 + 6 = 4,
and a4−a3 = 2 – (– 2) = 2 + 2 = 4
i.e., an+1−an is same every time, so the given list of number forms an AP.
So, d = 4
The next three terms after the last given term are
2 + 4 = 6 ,
6 + 4 = 10 ,
and, 10 + 4 = 14
(v) Here, a2−a1=(3+2–√)−3=2–√
and a3−a2=(3+22–√)−(3+2–√)=2–√
i.e., an+1−an is same every time, so the given list of numbers forms an AP.
So, d=2–√
The next three terms after the last given term are
(3+32–√)+2–√=3+42–√
(3+42–√)+2–√=3+52–√
and (3+52–√)+2–√=3+62–√
(vi) Here, a2−a1 = 0.22 – 0.2 = 0.22 – 0.20 = 0.02
and, a3−a2 = 0.222 – 0.22
= 0.222 – 0.220 = 0.002
Therefore a2−a1≠a3−a2
Thus, the given list of numbers does not form an AP.
(vii) Here, a2−a1 = – 4 – 0 = – 4,
a3−a2 = – 8 – (– 4) = – 8 + 4 = – 4,
and a4−a3 = 12 – (– 8) = – 12 + 8 = – 4,
i.e., an+1−an is same every time, so the given list of numbers forms an AP.
So, d = – 4
The next three terms after the last given term are
– 12 + (– 4) = – 12 – 4 = – 16
– 16 + (– 4) = – 16 – 4 = – 20
and – 20 + (– 4) = – 20 – 4 = – 24.
(viii) Here, a2−a1=−12−(−12)
=−12+12=0
a3−a2=−12−(−12)
=−12+12=0
and a4−a3=−12−(−12)
=−12+12=0
i.e., an+1−an is same every time, so the given list of numbers forms an AP.
So, d = 0
The next three terms after the last given term are
−12,−12and−12
(ix) Here , a2−a1=3?1=2
and a3−a2=9−3=6
⇒ a2−a1≠a3−a2
Thus , the given list of number does not form an AP.
(x) Here, a2−a1 = 2a – a = a
a3−a2 = 3a – 2a = a
a4−a3 = 4a – 3a = a
i.e., an+1−an is same every time, so the given list of numbers forms an AP.
So, d = a
The next three terms after the last given term are
4a + a = 5a,
5a + a = 6a
and, 6a + a = 7a
(xi) Here, a2−a1=a2−a=a(a−1)
and a3−a2=a3−a2=a2(a−1)
⇒ a2−a1≠a3−a2
Thus, the given list of number does not form an AP.
(xii) Here, a2−a1=8–√−2–√=4×2−−−−√−2–√
=22–√−2–√=2–√
and, a3−a2=18−−√−8–√=9×2−−−−√−4×2−−−−√
=32–√−22–√=2–√
i.e., an+1−an is same every time, so the given list of numbers forms an AP.
So, d=2–√
The next three terms after the last given term are
32−−√+2–√=42–√+2–√=52–√
=25×2−−−−−√=50−−√
50−−√+2–√=52–√+2–√=62–√
=36×2−−−−−√=72−−√
and =72−−√+2–√=62–√+2–√=72–√
=49×2−−−−−√=98−−√
(xiii) Here, a2−a1=6–√−3–√=3×2−−−−√−3–√
=3–√(2–√−1)
and, a3−a2=9–√−6–√=3−3×2−−−−√
=3–√(3–√−2–√)
⇒ a2−a1≠a3−a2
Thus, the given list of numbers does not form an AP.
(xiv) Here, a2−a1=32−12=9−1=8
and a3−a2=52−32=25−9=16
⇒ a2−a1≠a3−a2
Thus , the given list of number does not form an AP.
(xv) Here, a2−a1=52−12=25−1=24,
a3−a2=72−52=49−25=24,
and a4−a3=73−72=73−49=24.
i.e., an+1−an is same every time, so the given list of numbers forms an AP.
So, d = 24
The next three terms after the last given term are
73 + 24 = 97,
97 + 24 = 121,
and 121 + 24 = 145.
Q.1 Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP :
Sol. Blanks may be filled as under :
(i) an=a+(n−1)d=7+(8−1)3
=7+7×3=7+21=28
(ii) an=a+(n−1)d
⇒ 0=−18+(10−1)d
⇒ 18=9d
⇒ d=189=2
(iii) an=a+(n−1)d
⇒ −5=a+(18−1)(−3)
⇒ −5=a+17(−3)
⇒ −5=a−51
⇒ a=−5+51=46
(iv) an=a+(n−1)d
⇒ 3.6=−18.9+(n−1)2.5
⇒ 3.6+18.9=(n−1)2.5
⇒ 22.5=(n−1)2.5
⇒ n−1=22.52.5
⇒ n−1=9
⇒ n=9+1=10
(v) an=a+(n−1)d=3.5+(105−1)×0
=3.5+0=3.5
Q.2 Choose the correct choice in the following and justify :
(i) 30th term of the 10, 7, 4, ….. is
(a) 97 (b) 77 (c) – 77 (d) – 87
(ii) 11th term of the −3,−12,2...,is
(a) 28 (b) 22 (c) – 38 (d) −4812
Sol. (i) Here, a = 10, d = 7 – 10 = – 3, n = 30
We know that an=a+(n−1)d
Therefore, a30 = 10 + (30 – 1) (–3) = 10 + 29 (– 3)
= 10 – 87 = – 77
So, (C) is the correct choice.
(ii) Here, a = – 3, d=−12−(−3)
=12+3=−1+62
=52,n=11
We know that an=a+(n−1)d
Therefore, a11=−3+(11−1)52
=−3+10×52=−3+25=22
So, (B) is the correct choice.
Q.3 In the following APs, find the missing terms in the boxes :
(i) 2, , 26 (ii) , 13, , 3
(iii) 5, , , 912 (iv) –4, , , , , 6
(v) , 38, , , , –22.
Sol. (i) Let 2, , 26, be a + (a + d) and (a + 2d)
Therefore, a = 2 and a + 2d = 26
⇒ 2 + 2d = 26
⇒ 2d = 26 – 2 = 24
⇒ d=242=12
Thus the missing term = a + d = 2 + 12 = 14
(ii) Let , 13, , 3 be a, a + d, a + 2d and a + 3d
Therefore, a + d = 13 … (1)
and, a + 3d = 3 … (2)
(2) – (1) gives, 2d = – 10
⇒ d = – 5
Putting d = – 5 in (1), we get
a – 5 = 13
⇒ a = 13 + 5 = 18
Therefore, The missing terms are a, i.e., 18
and , a + 2d, i.e., 18 + 2 (– 5) = 18 – 10 = 8
(iii) Let 5, , , 912, be a, a +d, a +2d and a + 3d .
Therefore, a = 5 … (1)
and , a+3d=912=192 … (2)
(2) – (1) gives, 3d=192−5=19−102=92
⇒ d=13×92=32
Therefore, The missing terms are a + d, i.e.,
5+32=5+112=612
and a + 2d, 5 + 2 × 32 = 5 + 3 = 8
(iv) Let – 4, , , , , 6 be a, a + d, a + 2 d, a + 3 d, a + 4 d and a + 5 d.
Therefore, a = – 4 … (1)
and, a + 5d = 6 … (2)
(2) – (1) gives , 5d = 10 d = 2
Therefore, The missing terms are
a + d = – 4 + 2 = – 2
a + 2d = – 4 + 2 × 2 = – 4 + 4 = 0,
a + 3d = – 4 + 3 × 2 = – 4 + 6 = 2,
and, a + 4d = – 4+ 4 × 2 = – 4 + 8 = 4
(v) Let , 38, , , , – 22 be a, a + d, a + 2 d, a + 3 d, a + 4 d and a + 5 d.
Therefore, a + d = 38 … (1)
and a + 5d = – 22 … (2)
(2) – (1) gives, 4d = – 60
⇒ d = – 15
Putting d = – 15 in (1), we get
a –15 = 38
⇒ 38 + 15 = 53
Therefore, The missing terms are a, i.e., 53.
a + 2d, i.e., 53 + 2 × (– 15 )= 53 – 30 = 23,
a + 3d, i.e., 53 + 3 × (– 15) = 53 – 45 = 8,
and a + 4 d, i.e., 53 + 4 × (– 15) = 53 – 60 = – 7
Q.4 Which terms of the AP : 3, 8, 13, 18, …. is 78 ?
Sol. Clearly the given list is an AP.
We have, a = 3, d = 8 – 3 = 5
Let 78 be the nth term of the given AP then,
an=78
⇒ a+(n−1)d=78
Therefore, 3+(n−1)5=78
⇒ (n−1)5=78−3
⇒ 5(n−1)=75
⇒ n−1=15
⇒ n=15+1
⇒ n=16
Thus, 78 is the 16th term of the given list.
Q.5 Find the number of terms in each of the following APs :
(i) 7, 13, 19 …., 205
(ii) 18, 1512, 13, …, – 47
Sol. (i) Clearly it forms an AP with first term a = 3 and common difference d = 13 – 7 = 6
(i) Let there be n terms in the given list. Then , nth term = 205.
⇒ a + (n –1) d = 205
⇒ 7 + (n –1) 6 = 205
⇒ 6 (n – 1) = 205 – 7
⇒ 6 (n – 1) = 198
⇒ n−1=1986=33
⇒ n = 33 + 1 = 34
Thus the given list contains 34 terms.
(ii) Let there be n terms in the given list.
18, 1512, 13, …. , – 47. Clearly, it forms an AP with first term a = 18 and common difference
d=1512−18
=312−18=31−362=−52
Then, nth term = – 47.
⇒ a+(n−1)d=−47
⇒ 18+(n−1)(−52)=−47
⇒ (−52)(n−1) =−47−18=−65
⇒ n−1=−65×−25
=−13×−2=26
⇒ n=26+1=27
There are 27 terms in the given AP.
Q.6 Check whether – 150 is a term of the AP : 11 , 8, 5, 2, ….
Sol. Here, a2−a1=8−11=−3
a3−a2=5−8=−3
a4−a3=2−5=−3
As an+1−an is same every time, so the given list of numbers is an AP.
Now , a = 11 , d = – 3
Let – 150 be the nth term of the given AP.
We know that an=a+(n−1)d
⇒ −150=11+(n−1)(−3)
⇒ −3(n−1)=−150−11=−161
⇒ n−1=1613
⇒ n=1613+1=1643
But n should be a positive integer. So, our assumption was wrong and so – 150 is not a term of the given AP.
Q.7 Find the 31st term of an AP whose 11 th term is 38 and the 16th term is 73.
Sol. Let a be the first term and d be the common difference.
Now, an=a+(n−1)d
Therefore, a11=a+10d=38 … (1)
and a16=a+15d=73 … (2)
Subtracting (1) from (2) we get
5d=35
⇒ d=355=7
and then from (1),
a + 10 × 7 = 38
⇒ a = 38 – 70 = – 32
Therefore, a31=a+30d=−32+30×7
=−32+210=178
Q.8 An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Sol. Let a be the first term and d the common difference.
Now an=a+(n−1)d
a3=a+2d=12 … (1)
a50=a+49d=106 … (2)
Subtracting (1) from (2) , we get –
47d=94
⇒ d=9447=2
and then from (1)
a + 2 × 2 = 12
⇒ a = 12 – 4 = 8
Therefore, a29 = a + 28 d = 8 + 28 × 2
= 8 + 56 = 64
Q.9 If the 3rd and 9th terms of an AP are 4 and 8 respectively, which term of this AP is zero?
Sol. Let a be the first term and d the common difference.
a3 = a + 2d = 4 … (1)
a9 = a + 8d = – 8 … (2)
Subtracting (1) from (2) , we get
6d = – 12 ⇒ d=−126=−2
and then from (1)
a + 2 × (– 2) = 4
⇒ a = 4 + 4 = 8
Let an=0
⇒ a + (n – 1) d = 0
⇒ 8 + (n – 1) (– 2) = 0
⇒ (n – 1) (– 2) = – 8
⇒ n−1=−8−2=4
⇒ n = 4 + 1 = 5
Thus , 5th term is zero.
Q.10 The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Sol. Let a be the first term and d the common difference.
It is given a17−a10=7
⇒ (a + 16d) – (a + 9d) = 7 [Since,an=a+(n−1)d]
⇒ 7d = 7
⇒ d = 1
Thus, the common difference is 1
Q.11 Which term of the AP : 3, 15, 27, 39, ….. will be 132 more than its 54th term?
Sol. Here, a = 3, d = 15 – 3 = 12. Then.
a54 = a + 53d = 3 + 53 × 12 = 3 + 636 = 639
Let an be 132 more than its 54th term
i.e., an=a54 + 132 = 639 + 132 = 771[Since,an=a+(n−1)d]
⇒ a + (n – 1)d = 771
⇒ 3 + (n – 1)12 = 771
⇒ 12(n – 1) = 771 – 3
⇒ 12(n– 1) = 768
⇒ n−1=76812=64
⇒ n = 64 + 1 = 65
Thus, 65th term is 132 more than its 54th term.
Q.12 Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th terms?
Sol. Let the two APs be a1,a2,a3,....an,...andb1,b2,b3...,bn,...
Also, let d be the common difference of two APs. Then,
an=a1+(n−1)d
and bn=b1+(n−1)d
⇒ an−bn=[a1+(n−1)d]
−[b1+(n−1)d]
⇒ an−bn=a1−b1foralln∉N
⇒ a100−b100=a1−b1=100 [Given]
Now, a1000−b1000=a1−b1
⇒ a1000−b1000=100 [Since a1−b1=100]
Hence, the difference between 1000th terms is the same as the difference between 100th terms, i.e., 100.
Q.13 How many threedigit numbers are divisible by 7 ?
Sol. 994 is the last 3digit number divisible by 7. Thus, we have to determine the number of terms in the list 105, 112, 119, …. 994.
Clearly, it form an AP with first term a = 105 and common difference d = 112 – 105 = 7
Let there be n terms in the AP then, nth term = 994.
As an=a+(n−1)d
⇒ 105 + (n – 1)7 = 994
⇒ 7(n – 1) = 994 – 105
⇒ 7 (n – 1) = 889
⇒ n−1=8897=127
⇒ n = 127 + 1 = 128
Hence, there are 128 numbers of three digit which are divisible by 7.
Q.14 How many multiples of 4 lie between 10 are 250 ?
Sol. We observe that 12 is the first integer between 10 and 250, which is a multiple of 4 (i.e., divisible by 4). Also, when we divide 250 by 4, the remainder is 2.
Therefore, 250 – 2 = 248 is the largest integer divisible by 4 (i.e., multiple of 4) and lying between 10 and 250. Thus, we have to find the number of terms in an AP with first term = 12, last term = 248, and common difference = 4 (as the numbers are divisible by 4).
Let there be n terms in the AP then.
an=248
⇒ 12 + (n – 1)4 = 248
⇒ 4(n – 1) = 248 – 12 ⇒ 4(n – 1) = 236
⇒ n−1=2364=59
⇒ n = 59 + 1 = 60
Hence , there are 60 multiples of 4 between 10 and 250.
Q.15 For what value of n, are the nth terms of the APs : 63, 65, 67, …. and 3, 10, 17,… are equal?
Sol. If nth terms of the APs 63, 65, 67, …. and 3, 10, 17, ….. are equal. Then,
63 + (n – 1) 2 = 3 + (n – 1) 7
[Since In 1st AP, a = 63, d = 65 – 63 = 2 and in 2nd AP , a = 3, d = 10 – 3 = 7]
⇒ 7(n – 1) – 2 (n – 1) = 63 – 3
⇒ (n – 1) (7 – 2) = 60
⇒ 5(n – 1) = 60
⇒ n−1=605=12
⇒ n = 12 + 1 = 13
Hence, the 13th terms of the two given APs are equal.
Q.16 Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Sol. Let a be the first term and d the common difference.
Hence, a3=16anda7−a5=12
⇒ a + 2d = 16 … (1)
and, (a + 6d) – (a + 4d) = 12 ⇒ 2d = 12
⇒ d = 6 … (2)
Using (2) in (1), we get
a + 2 × 6 = 16
⇒ a = 16 – 12 = 4
Thus, the required AP is 4, 4 + 6, 4 + 2 × 6, 4 + 3 × 6, … i.e., 4, 10, 16, 22, ….
Q.17 Find the 20th term from the last term of the AP : 3, 8, 13, …. 253.
Sol. We have , l = Last term = 253
and d = Common ifference = 8 – 3 = 5
Therefore, 20th term from the end = l = (20 – 1)d
= l – 19 d
= 253 – 19 × 5
= 253 – 95 = 158
Q.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.
Sol. Let a be the first term and d the common difference.
Here, a4+a8=24
⇒ (a + 3d) + (a + 7d) = 24
⇒ 2a + 10d = 24
a6+a10=44
⇒ (a + 5d) + (a + 9d) = 44
⇒ 2a + 14d = 44 ⇒ a + 7d = 22 … (2)
Substracting (1) from (2), we get
2d = 10 ⇒ d = 5
and then from (1)
a + 25 = 12 ⇒ a = – 13
The first three terms are a, (a + d) and (a + 2d)
Putting values of a and d, we get – 13, (– 13 + 5) and (– 13 + 2 × 5)
i.e., – 13, – 8 and – 3
Q.19 Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?
Sol. The annual salary drawn by Subha Rao in the years 1995, 1996, 1997, etc. is Rs 5,000, Rs 5,200, Rs 5,400…. Rs 7,000.
The list of these nos. is 5000, 5200, 5400, …. 7000.
It form an AP [since a2−a1=a3−a2=200]
Let an=7000
⇒ 7000 = a + (n – 1) d
⇒ 7000 = 5000 + (n – 1) (200)
⇒ 200 (n – 1) = 7000 – 5000
⇒ n−1=2000200=10
⇒ n = 10 + 1 = 11
Thus, in the 11th year (i.e., in 2005) of his service, Subba Rao drew an annual salary of Rs. 7,000.
Q.20 Ramkali saves Rs 5 in the first week of a year and then increased her weekly savings by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.
Sol. Ramkali’s savings in the subsequent weeks are respectively Rs 5, Rs 5 + Rs 1.75, Rs 5 + 2 × Rs 1.75, Rs 5 + 3 × Rs 1.75…
And in the nth week her saving will be Rs 5 + (n – 1) × Rs 1.75
⇒ 5 + (n – 1) × 1.75 = 20.75
⇒ (n – 1) × 1.75 = 20.75 – 5 = 15.75
Therefore n−1=15.751.75=9
⇒ n = 9 + 1 = 10
Q.1 Find the sum of the following APs :
(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) 115,112,110,.... to 11 terms.
Sol. (i) Let a be the first term and d be the common difference of the given AP then, we have
a = 2 and d = 7 – 2 = 5
We have to find the sum of 10 terms of the given AP.
Putting a = 2, d = 5, n = 10 in Sn=n2[2a+(n−1)d]
we get
S10=102[2×2+(10−1)5]
=5(4+9×5)
= 5(4 + 45) = 5 × 49 = 245
(ii) Let a be the first term and d be the common difference of the given AP. Then we have
a = – 37, d = – 33 – (– 37) = – 33 + 37 = 4
We have to find the sum of 12 terms of the given AP.
Putting a = – 37, d = 4, n = 12 in
Sn=n2[2a+(n−1)d],weget
S12=122[2×−37+12(12−1)4]
= 6 (– 74 + 11 × 4)
= 6 (– 74 + 44) = 6 × (– 30) = – 180
(iii) Let a be the first term and d be the common difference of the given AP. Then, we have
a = 0.6, d = 1.7 – 0.6 = 1.1
We have to find the sum of 100 terms of the given AP
Putting a = 0.6 d = 1.1, n = 100 in
Sn=n2[2a+(n−1)d],weget
S100=1002[2×0.6+(100−1)1.1]
= 50(1.2 + 99 × 1.1)
= 50 (1.2 + 108.9)
= 50 × 110.1 = 5505
(iv) Let a be the first term and d be the common difference of the given AP. Then we have
a=115,d=112−115=5−460=160
We have to find the sum of 11 terms of the given AP.
Putting a=115,d=160,n=11in
Sn=n2[2a+(n−1)d],weget
S11=112[2×115+(11−1)160]
=112(215+10×160)
=112(215+16)
=112×4+530
=112×930=3320
Q.2 Find the sums given below :
(i) 7+1012+14+....+84
(ii) 34 + 32 + 30 + …… + 10
(iii) – 5 + (– 8) + (– 11) + …. + (– 230)
Sol. (i) Here, the last term is given. We will first have to find the number of terms.
a=7,d=1012−7=312=72,ℓ=an=84
Therefore, 84 = a + (n – 1) d
⇒ 84=7+(n−1)72
⇒ 72(n−1)=84−7
⇒ 72(n−1)=77
⇒ n−1=77×27
⇒ n – 1 = 22
⇒ n = 23
We know that
Sn=n2(a+ℓ)
⇒ S23=232(7+84)=232×91
=20932=104612
(ii) Here, the last term is given. We will first have to find the number of terms.
a = 34, d = 32 – 34 = – 2, l = an=10
Therefore 10 = a + (n – 1)d
⇒ 10 = 34 + (n – 1) (– 2)
⇒ (– 2) (n – 1) = 10 – 34
⇒ (– 2) (n –1) = – 24
⇒ n – 1 = 12
⇒ n = 12 + 1 = 13
Using Sn=n2(a+ℓ), we have
S13=132(34+10)=132×44 = 13 × 22 = 286
(iii) Here the last term is given. We will first have to find the number of terms.
a = – 5, d = – 8 – (–5) = – 8 + 5 = – 3, l = an=−230
Therefore – 230 = a + (n – 1) d
⇒ – 230 = – 5 + (n – 1) (– 3)
⇒ (– 3) (n – 1) = – 230 + 5
⇒ (– 3) (n – 1) = – 225
⇒ n−1=−225−3
⇒ n – 1 = 75
⇒ n = 75 + 1 = 76
Using Sn=n2(a+ℓ), we have
S76=762(−5−230) = 38 × – 235 = – 8930
Q.3 In an AP :
(i) Given a = 5, d = 3, an=50,findnandSn.
(ii) Given a = 7, a13=35,finddandS13
(iii)Given a12=37,d=3,findaandS12
(iv)Given a3=15,S10=125,finddanda10
(v) Given d=5,S9=75,findaanda9
(vi) Given a = 2, d = 8 , Sn=90,findnandan
(vii) Given a = 8, an=62,Sn=210, find n and d.
(viii) Given an=4,d=2,Sn=−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.
Sol. (i) We have a = 5, d = 3 and an=50
⇒ a + (n – 1)d = 50
⇒ 5 + (n – 1) 3 = 50
⇒ 3(n – 1) = 50 – 5
⇒ n−1=453=15
⇒ n = 15 + 1 = 16
Putting n = 16, a = 5 and ℓ=an=50inSn=n2(a+ℓ)
We get
S16=162(5+50)=8×55=440
Hence, n = 16 and S16=440
(ii) We have a = 7 and a13=35
Let d be the common difference of the given AP. Then,
⇒ a13=35
⇒ a + 12 d = 35
⇒ 7 + 12 d = 35 [Since a = 7]
⇒ 12d = 35 – 7 = 28
⇒ d=2812=73
Putting n = 13, a = 7 and ℓ=a13=35in
Sn=n2(a+ℓ),weget
S13=132(7+35)
=132×42
= 13 × 21 = 273
Hence, d=73andS13=273
(iii) We have a12=37,d=3
Let a be the first term of the given AP. Then,
a12=37
⇒ a + 11d = 37
⇒ a + 11(3) = 37
⇒ a + 11(3) = 37 [Since d = 3]
⇒ a = 37 – 33 = 4
Putting n = 12, a = 4 and ℓ=a12=37in
Sn=n2(a+ℓ),weget
S12=122(4+37)=6×41=246
Hence, , a = 4 and S12=246
(iv) We have, a3=15,S10=125
Let a be the first term and d the common difference of the given AP. Then,
a3=15andS10=125
⇒ a + 2d = 15 … (1)
and 102 [2a + (10 – 1)d] = 125
⇒ 5(2a + 9d) = 125
⇒ 2a + 9d = 25 … (2)
2 × (1) – (2) gives,
2(a + 2d) – (2a + 9d) = 2 × 15 – 25
⇒ 4d – 9d = 30 – 25
⇒ – 5d = 5
⇒ d=−55=−1
Now, a10=a+9d=(a+2d)+7d
= 15 + 7 (– 1) [Using (1)]
= 15 – 7 = 8
Hence, d = – 1 and a10=8
(v) We have d = 5 , S9=75
Let a be the first term of the given AP. Then,
S9=75
⇒ 92[2a+(9−1)5]=75
⇒ 92(2a+40)=75
⇒ 9a + 180 = 75
⇒ 9a = 75 – 180
⇒ 9a = – 105
⇒ a=−1059=−353
Now a9=a+8d=−353+8×5
=−35+1203=853
Hence, a=−353anda9=853
(vi) We have, a = 2, d = 8 , Sn=90
Sn=90
⇒ n2[2×2+(n−1)8]=90
⇒ n2(4+8n−8)=90
⇒ n2(8n−4)=90
⇒ n(4n−2)=90
⇒ 4n2−2n−90=0
Therefore, n=−(−2)±(−2)2−4×4×(−90)√2×4
=2±4+1440√8
=2±1444√8
=2±388
=408,−368=5,−92
But n cannot be negative
Therefore, n = 5
Now an=a+(n−1)d
⇒ a5=2+(5−1)8=2+32=34
Hence, n = 5 and an=34
(vii) We have , a = 8, an=62,Sn=210
Let d be the common difference of the given AP.
Now, Sn=210
⇒ n2(a+ℓ)=210
⇒ n2(8+62)=210 [Since a=8,an=62]
⇒ n2×70=210
⇒ n=210×270=3×2=6
and an=62 ⇒ a6=62
⇒ a + 5d = 62
⇒ 8 + 5d = 62 [since a = 8]
⇒ 5d = 62 – 8 = 54
⇒ d=545
Hence, d=545 and n = 6
(viii) We have an=4,d=2,Sn=−14
Let a be the first term of the given AP. Then.
an=4
⇒ a + (n – 1)2 = 4 [since d = 2]
⇒ a = 4 – 2 (n – 1) … (1)
and Sn=−14
⇒ n2(a+ℓ)=−14 [since ℓ=an]
⇒ n (a + 4) = – 28
⇒ n[4 – 2 (n – 1) + 4] = – 28
⇒ n (4 – 2n + 2 + 4) = – 28
⇒ n(– 2n + 10) = – 28
⇒ n (– n + 5) = – 14
⇒ −n2+5n=−14
⇒ n2−5n−14=0
⇒ (n – 7) (n + 2) = 0
⇒ n = 7 or – 2
But n cannot be negative
n = 7
Putting n = 7 in (1), we get
a = 4 – 2 (7 – 1) = 4 – 2 × 6
= 4 – 12 = – 8
Hence, n = 7 and a = – 8
(ix) We have, a = 3, n = 8, S = 192
Let d be the common difference of the given AP.
Sn=n2[2a+(n−1)d]
⇒ 192=82[2×3+(8−1)d]
⇒ 192 = 4(6 + 7d)
⇒ 48 = 6 + 7d
⇒ 7d = 48 – 6
⇒ 7d = 42
⇒ d=427=6
Hence, d = 6
(x) We have l = 28, S = 144, n = 9
Let a be the first term of the given AP.
S = 144
⇒ n2(a+ℓ)=144
⇒ 92(a+28)=144
⇒ a+28=144×29
⇒ a + 28 = 32
⇒ a = 32 – 28 = 4
Hence, a = 4
Q.4 How many terms of the AP : 9 , 17, 25, … must be taken to give a sum of 636 ?
Sol. Let the first term be a = 9 and common difference d = 17 – 9 = 8. Let the sum of n terms be 636. Then,
Sn=636
⇒ n2[2a+(n−1)d]=636
⇒ n2[2×9+(n−1)8]=636
⇒ n2(18+8n−8)=636
⇒ n2(8n+10)=636
⇒ n(4n + 5) = 636
⇒ 4n2+5n−636=0
Therefore, n=−5±25−4×4−636√2×4
=−5±25+10176√8
=−5±10201√8
=−5±1018=968,−1068
12,−534
But n cannot be negative
Therefore, n = 12
Thus, the sum of 12 terms is 636.
Q.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.
Sol. Let a be the first term and d the common difference of the AP such that.
a = 5, l = 45 and S = 400
Therefore, S = 400
⇒ n2(a+ℓ)=400
⇒ n(5+45)=400×2
⇒ n(50) = 400 × 2
⇒ n=400×250=8×2=16
and l = 45 ⇒ a + (n – 1) d = 45
⇒ 5 + (16 – 1)d = 45
⇒ 15d = 45 – 5 = 40
⇒ a=4015=83
Hence, the number of term is 16 and the common difference is 83.
Q.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?
Sol. Let a be the first term and d be the common difference. Let l be its last term. Then a = 17, ℓ=an=350, d = 9 .
ℓ=an=350
⇒ a + (n – 1) d = 350
⇒ 17 + (n – 1)9 = 350
⇒ 9(n – 1) = 350 – 17 = 333
⇒ n−1=3339=37
⇒ n = 37 +1 = 38
Putting a = 17, l = 350, n = 38
in Sn=n2(a+ℓ),weget
S38=382(17+350)
= 19 × 367 = 6973
Hence, there are 38 terms in the AP having their sum as 6973.
Q.7 Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Sol. Let a be the first term and d the common difference of the given AP then,
d = 7 and a22=149
⇒ a + (22 – 1) d = 149
⇒ a + 21 × 7 = 149
⇒ a = 149 – 147 = 2
Putting n = 22, a = 2 and d = 7 in
Sn=n2[2a+(n−1)d], we get
S22=222[2×2+(22−1)7]
= 11(4 + 21 × 7)
= 11(4 + 147)
= 11 × 151 = 1661
Hence, the sum of first 22 terms is 1661.
Q.8 Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Sol. Let a be the first term and d the common difference of the given AP. Then,
a2=14anda3=18
⇒ a + d = 14 and a + 2d = 18
Solving these equations , we get
d = 4 and a = 10
Putting a = 10, d = 4 and n = 51 in
Sn=n2[2a+(n−1)d], we get
S51=512[2×10+(51−1)×4]
=512[20+50×4]
=512(20+200)=512×220
= 51 × 110 = 5610
Q.9 If the sum of 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of n terms.
Sol. Let a be the first term and d the common difference of the given AP. Then.
S7=49andS17=289
⇒ 72[2a+(7−1)d]=49
⇒ 72(2a+6d)=49
⇒ a + 3d = 7 … (1)
and 172[2a+(17−1)d]=289
⇒ 172(2a+16d)=289
⇒ a + 8d = 17 … (2)
Solving these two equations, we get
⇒ 5d = 10 , d = 2 and a = 1
Therefore, Sn=n2[2a+(n−1)d]
=n2[2×1+(n−1)2]
=n2(2+2n−2)=n2×2n=n2
Q.10 Show that a1,a2....an.... form an AP where an is defined as below :
(i) an=3+4n (ii) an=9−5n
Also find the sum of the first 15 term in each case.
Sol. (i) We have, an=3+4n
Substituting n = 1, 2, 3, 4, … , n , we get
The sequence 7, 11, 15, 19, …. (3 + 4n) which is an AP with common difference 4.
Putting a = 7, d = 4 and n = 15 in
Sn=n2[2a+(n−1)d], we get
S15=152[2×7+(15−1)4]
=152(14+14×4)=152(14+56)
=152×70=15×35=525
(ii) We have, an=9−5n
Substituting n = 1, 2, 3, 4, …. n, we get
The sequence 4, – 1, – 6, – 11, …. (9 – 5n), which is an AP with common difference – 5.
Putting a = 4, d = – 5 and n = 15 in
Sn=n2[2a+(n−1)d], we get
S15=152[2×4+(15−1)(−5)]
=152(8+14×−5)
=152(8−70)=152×−62
= 15 × – 31 = – 465
Q.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.
Sol. According to the question,
Sn=4n−n2
S1=4×1−12
= 4 – 1 = 3
⇒ First term = 3
Now, sum of first two terms = S2=4×2−22
=8−4=4
Therefore Second term =S2−S1=4−3=1
=S3=4×3−32
= 12 – 9 = 3
Therefore Third term = S3−S2 = 3 – 4 = – 1
S9=4×9−92
= 36 – 81 = – 45
and, S10=4×10−102
= 40 – 100 = – 60
Therefore Tenth term = S10−S9
= – 60 – (– 45)
= – 60 + 45 = – 15
Also, Sn=4n−n2
and Sn−1=4(n−1)−(n−1)2
=4n−4−n2+2n−1
=−n2+6n−5
Therefore, nth term = Sn−Sn−1
=4n−n2−(−n2+6n−5)
=4n−n2+n2−6n+5=5−2n
Q.12 Find the sum of the first 40 positive integers divisible by 6.
Sol. The first positive integers divisible by 6 are 6, 12, 18, …. Clearly, it is an AP with first term a = 6 and common difference d = 6.
We want to find S10
Therefore, S40=402[2×6+(40−1)6]
= 20 (12 + 39 × 6)
= 20(12 + 234) = 20 × 246 = 4920
Q.13 Find the sum of the first 15 multiples of 8.
Sol. The first 15 multiples of 8 are 8 × 1, 8 × 2, 8 × 3, … 8 × 15 i.e., 8, 16, 24 …. 120, which is an AP.
Therefore Sum of 1st 15 multiples of
8=152(8+120)
[Sn=n2(a+ℓ)]
=152×128 = 15 × 64 = 960
Q.14 Find the sum of the odd numbers between 0 and 50.
Sol. The odd numbers between 0 and 50 are 1, 3, 5, 49. They form an AP and there are 25 terms.
Therefore, Their sum =252(1+49)
=252×50 = 25 × 25 = 625
Q.15 A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 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?
Sol. Here a = 200 , d = 50 and n = 30
Therefore, S=302[2×200+(30−1)50]
[SinceSn=n2(2a+(n−1)d]
= 15(400 + 29 × 50)
= 15(400 + 1450)
= 15 × 1850
= 27750
Hence, a delay of 30 days costs the contractor Rs 27750.
Q.16 A sum of Rs 700 is to be used to give seven each prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.
Sol. Let the respective prizes be a + 60, a + 40, a + 20, a, a – 20, a – 40, a – 60
Therefore, The sum of the prizes is
a + 60 + a + 40 + a + 20 + a + a – 20 + a – 40 + a – 60 = 700
⇒ 7a = 700
⇒ a=7007=100
Therefore, The seven prizes are 100 + 60, 100 + 40, 100 + 20, 100, 100 – 20, 100 – 40, 100 – 60
or 160, 140, 120, 100, 80, 60, 40 (in Rs)
Q.17 In a school, students thought of planting trees in an 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?
Sol. Since there are three sections of each class, so the number of trees planted by class I, class II, class III,… class XII are 1 × 3, 2 × 3, 3 × 3, …. 12 × 3 respectively.
i.e., 3, 6, 9, … 36. Clearly, it form an AP.
The sum of the number of the trees planted by these classes.
=122(3+36)=6×39=234
Q.18 A spiral is made up of successive semicircles , with centres alternately at A and B, starting with cenre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, …. as shown in fig.
What is the total length of such a spiral made up of thirteen consecutive semicircles ?
(Takeπ=227)
Sol. Length of a semicircum ference = πr where r is the radius of the circle.
Therefore, Length of spiral made up of thirteen consecutive semicircles.
=(π×0.5+π×1.0+π+1.5+π×2.0+....+π×6.5)cm
=π×0.5(1+2+3+....+13)cm
=π×0.5×132(2×1+[13−1)×1]cm
=227×510×132×14cm=143cm
Q.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 figure). In how many rows are the 200 logs placed and how many logs are in the top row?
Sol. Clearly logs stacked in each row form a sequence 20 + 19 + 18 + 17 + …. It is an AP with a = 20, d = 19 – 20 = – 1.
Let Sn=200. Then
n2[2×20+(n−1)(−1)]=200
⇒ n(40 – n + 1) = 400
⇒ n2−41n+400=0
⇒ (n−16)(n−25)=0
⇒ n = 16 or 25
Here the common difference is negative.
The terms go on diminishing and 21st term becomes zero. All terms after 21st term are negative. These
negative terms when added to positive terms from 17th term to 20th term, cancel out each other and the sum remains the same.
Thus n = 25 is not valid for this problem. So we take n = 16.
Thus, 200 logs are placed in 16 rows.
Number of logs in the 16th row
=a16
= a + 15d
= 20 + 15(–1)
= 20 – 15 = 5
Q.20 In a potato race, a bucket is placed at the starting point, which is 5 cm 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 figure).
A competitor starts from the bucket, picks up the earest 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?
Sol. To pick up the first potato second potato, third potato, fourth potato, ….
The distance (in metres) run by the competitor are
2 × 5 ; 2 × (5 + 3), 2 × (5 + 3 + 3), 2 × (5 + 3 + 3 + 3), ….
i.e., 10, 16, 22, 28, ….
which is in AP with a = 10, d = 16 – 10 = 6
Therefore, The sum of first ten terms,
S10=102[(2×10+(10−1)×6)]
= 5(20 + 54) = 5 × 74 = 370
Therefore, The total distance the competitor has to run is 370 m.
Q.1 Which term of the AP : 121, 117, 113….., is its first negative terms?
Sol. 121, 117, 113, ….
a = 121, d = 117 – 121 = – 4
an=a+(n−1)d
= 121 + (n – 1) × – 4
= 121 – 4n + 4 = 125 – 4n
For the first negative term
an<0⇒125−4n<0
⇒ 125 < 4n
⇒ 1254<n
⇒ 3114<n
n is an integer and n>3114
⇒ The first negative term is 32nd term.
Q.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.
Sol. Let the AP be a – 4 d, a – 3 d, a – 2 d, a – d, a, a + d, a + 2 d, a + 3 d, ….
Then, a3=a−2d,a7=a−2d
⇒ a3+a7=a−2d+a−2d=6
⇒ 2a = 6
⇒ a = 3 … (1)
Also (a – 2d) (a + 2d) = 8
⇒ a2−4d2=8
⇒ 4d2=a2−8
⇒ 4d2=(3)2−8=9−8=1
⇒ d2=14
⇒ d2=±12
Taking d=12
S16=162[2×(a−4d)+(16−1)×d]
=8[2×(3−4×12)+15×12]
=8[2+152]=8×192=76
Taking d=−12
S16=162[2×(a−4d)+(16−1)×d]
=8[2×(3−4×12)+15×−12]
=8[2×5−152]
=8[20−152]
=8×52=20
Therefore, S16=20,76
Q.3 A ladder has rungs 25 cm apart (see figure). 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 212 m apart, what is the length of the wood required for the rungs?
Sol. Number of rungs, n=212m25cm
=250cm25cm=10
So, there are 10 rungs
The length of the wood required for rungs = Sum of 10 rungs
=102[25+45]
=5×70=350cm
Q.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.
Sol. Here a = 1, and d = 1
Therefore, Sn−1=x−12[2×1+(x−1−1)×1]e
=x−12(2+x−2)=(x−1)(x)2
=x2−x2
Sn=x2[2×1+(x−1)×1]=x2(x+1)
x2+x2
and, S49=492[2×1+(49−1)×1]
=492[2+48]=492×50
=49×25
According to the question,
Sn−1=S49−Sx
i.e., x2−x2=49×25−x2+x2
⇒ x2−x2+x2+x2=49×25
⇒ x2−x+x2+x2=49×25
⇒ x2=49×25
⇒ x=±7×5
Since x is a counting number , so taking positive square root, x = 7 × 5 = 35.
Q.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 14m and a tread of 12m (see figure). Calculate the total volume of concrete required to build the terrace.
Sol. Volume of concrete required to build the first step, second step, third step…. (inm3) are
14×12×50,(2×14)×12×50,(3×14)×12×50,....
i.e. 508,2×508,3×508,....
Therefore Total volume of concrete required.
=508+2×508+3×503+...
=508[1+2+3+...]
=508×152[2×1+[15−1)]×1] [Since, n = 15]
=508×152×16=750m3
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