NCERT Solutions class 10 maths chapter 5
Arithmetic Progression
NCERT Solutions class 10 maths chapter 5 Exercise 5.3
Question 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) 1/15, 1/12, 1/10, …… , to 11 terms
Answer
(i) Given, 2, 7, 12 ,…, to 10 terms
For this A.P.,
a = 2
d = a2 − a1 = 7−2 = 5
n = 10
sum of nth term in AP series is,
Sn = n/2 [2a +(n-1)d]
S10 = 10/2 [2(2)+(10 -1)×5]
= 5[4+(9)×(5)]
= 5 × 49 = 245
(ii) Given, −37, −33, −29 ,…, to 12 terms
first term, a = −37
And common difference, d = a2− a1
d= (−33)−(−37)
= − 33 + 37 = 4
n = 12
sum of nth term in AP series is,
Sn = n/2 [2a+(n-1)d]
S12 = 12/2 [2(-37)+(12-1)×4]
= 6[-74+11×4]
= 6[-74+44]
= 6(-30) = -180
(iii) Given, 0.6, 1.7, 2.8 ,…, to 100 terms
first term, a = 0.6
d = a2 − a1 = 1.7 − 0.6 = 1.1
n = 100
sum of nth term in AP series is,
Sn = n/2[2a +(n-1)d]
S12 = 50/2 [1.2+(99)×1.1]
= 50[1.2+108.9]
= 50[110.1]
= 5505
(iv) Given, 1/15, 1/12, 1/10, …… , to 11 terms
For this A.P.,
a = 1/5
d = a2 –a1 = (1/12)-(1/5) = 1/60
n = 11
sum of nth term in AP series is,
Sn = n/2 [2a + (n – 1) d] = 11/2
= 33/20
Question 2
Find the sums given below (i) 7 + 10(ii) 34 + 32 + 30 + ……….. + 10
(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)
Answer
(i) 7 + 10a = 7
an = 84
d =d =
an = a(n-1)d
84 = 7+(n – 1) x22 = n−1
n = 23
sum of n term is;
Sn = n/2 (a + l) , l = 84
Sn = 23/2 (7+84)
Sn = (23×91/2) = 2093/2
= 1046.5(ii) Given, 34 + 32 + 30 + ……….. + 10
first term, a = 34
d = a2−a1 = 32−34 = −2
nth term, an= 10
10 be the nth term of this A.P.
an= a +(n−1)d
10 = 34+(n−1)(−2)
−24 = (n −1)(−2)
12 = n −1
n = 13
sum of n terms is;
Sn = n/2 (a +l) , l = 10
= 13/2 (34 + 10)
= (13×44/2) = 13 × 22
= 286
(iii)Given, (−5) + (−8) + (−11) + ………… + (−230)
First term, a = −5
nth term, an= −230
Common difference, d = a2−a1 = (−8)−(−5)
⇒d = − 8+5 = −3
−230 be the nth term of this A.P
an= a+(n−1)d
−230 = − 5+(n−1)(−3)
−225 = (n−1)(−3)
(n−1) = 75
n = 76
And, Sum of n term,
Sn = n/2 (a + l)
= 76/2 [(-5) + (-230)]
= 38(-235)
= -8930
Question 3
In an AP(i) Given a = 5, d = 3, an = 50, find n and Sn.
(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, Sn = 90, find n and an.
(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.
Answer
(i) a = 5, d = 3, an = 50
As we know that nth term in an AP,
an = a +(n −1)d,
we get,
⇒ 50 = 5+(n -1)×3
⇒ 3(n -1) = 45
⇒ n -1 = 15
⇒ n = 16
Now, sum of n terms,
Sn = n/2 (a +an)
Sn = 16/2 (5 + 50) = 440
(ii) a = 7, a13 = 35
As we know,that nth term in an AP,
an = a+(n−1)d,
we get,
⇒ 35 = 7+(13-1)d
⇒ 12d = 28
⇒ d = 28/12 = 2.33
Now, Sn = n/2 (a+an)
S13 = 13/2 (7+35) = 273
(iii)Given that, a12 = 37, d = 3
nth term in an AP,
an = a+(n −1)d,
⇒ a12 = a+(12−1)3
⇒ 37 = a+33
⇒ a = 4
Now, sum of nth term,
Sn = n/2 (a+an)
Sn = 12/2 (4+37)
= 246
(iv) Given that, a3 = 15, S10 = 125
nth term in an AP,
an = a +(n−1)d,
a3 = a+(3−1)d
15 = a+2d ………………………….. (i)
Sum of the nth term,
Sn = n/2 [2a+(n-1)d]
S10 = 10/2 [2a+(10-1)d]
125 = 5(2a+9d)
25 = 2a+9d ……………………….. (ii)
On multiplying equation (i) by (ii), we get;
30 = 2a+4d ………………………………. (iii)
By subtracting equation (iii) from (ii), we get,
−5 = 5d
d = −1
From equation (i),
15 = a+2(−1)
15 = a−2
a = 17 = First term
a10 = a+(10−1)d
a10 = 17+(9)(−1)
a10 = 17−9 = 8
(v) Given that, d = 5, S9 = 75
As, sum of n terms in AP is,
Sn = n/2 [2a +(n -1)d]
Therefore, the sum of first nine terms are;
S9 = 9/2 [2a +(9-1)5]
25 = 3(a+20)
25 = 3a+60
3a = 25−60
a = -35/3
nth term AP
an = a+(n−1)d
a9 = a+(9−1)(5)
= -35/3+8(5)
= -35/3+40
= (35+120/3) = 85/3
(vi) Given that, a = 2, d = 8, Sn = 90
As, sum of n terms in an AP is,
Sn = n/2 [2a +(n -1)d]
90 = n/2 [2a +(n -1)d]
⇒ 180 = n(4+8n -8) = n(8n-4) = 8n2-4n
⇒ 8n2-4n –180 = 0
⇒ 2n2–n-45 = 0
⇒ 2n2-10n+9n-45 = 0
⇒ 2n(n -5)+9(n -5) = 0
⇒ (n-5)(2n+9) = 0
So, n = 5
∴ a5 = 8+5×4 = 34
(vii) Given that, a = 8, an = 62, Sn = 210
As, sum of n terms in an AP is,
Sn = n/2 (a + an)
210 = n/2 (8 +62)
⇒ 35n = 210
⇒ n = 210/35 = 6
Now, 62 = 8+5d
⇒ 5d = 62-8 = 54
⇒ d = 54/5 = 10.8
(viii) Given that, nth term, an = 4, d = 2, Sn = −14.
As we know,that the nth term in an AP,
an = a+(n −1)d,
4 = a+(n −1)2
4 = a+2n−2
a+2n = 6
a = 6 − 2n …………………………………………. (i)
As we know, the sum of n terms is;
Sn = n/2 (a+an)
-14 = n/2 (a+4)
−28 = n (a+4)
−28 = n (6 −2n +4) {From equation (i)}
−28 = n (− 2n +10)
−28 = − 2n2+10n
2n2 −10n − 28 = 0
n2 −5n −14 = 0
n2 −7n+2n −14 = 0
n (n−7)+2(n −7) = 0
(n −7)(n +2) = 0
Either n − 7 = 0 or n + 2 = 0
n = 7 or n = −2
However, n can neither be negative
Therefore, n = 7
From equation (i), we get
a = 6−2n
a = 6−2(7)
= 6−14
= −8
(ix) Given that, first term, a = 3,
n = 8
Sn = 192
As we know,
Sn = n/2 [2a+(n -1)d]
192 = 8/2 [2×3+(8 -1)d]
192 = 4[6 +7d]
48 = 6+7d
42 = 7d
d = 6
(x) Given that, l = 28,S = 144 and there are total of 9 terms.
Sum of n terms
Sn = n/2 (a + l)
144 = 9/2(a+28)
(16)×(2) = a+28
32 = a+28
a = 4
Question 4
How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?Answer
For A.P.,
First term, a = 9
Common difference, d = a2−a1 = 17−9 = 8
As, the sum of n terms, is;
Sn = n/2 [2a+(n -1)d]
636 = n/2 [2×a+(8-1)×8]
636 = n/2 [18+(n-1)×8]
636 = n [9 +4n −4]
636 = n (4n +5)
4n2 +5n −636 = 0
4n2 +53n −48n −636 = 0
n (4n + 53)−12 (4n + 53) = 0
(4n +53)(n −12) = 0
Either 4n+53 = 0 or n−12 = 0
n = (-53/4) or n = 12
n cannot be negative therefore, n = 12 only.
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 that,
first term, a = 5
last term, l = 45
Sum of the AP, Sn = 400
As we know
Sn = n/2 (a+l)
400 = n/2(5+45)
400 = n/2(50)
Number of terms, n =16
l = a+(n −1)d
45 = 5 +(16 −1)d
40 = 15d
Common difference, d = 40/15 = 8/3
Question 6
The first and the last term 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 that,
First term, a = 17
Last term, l = 350
Common difference, d = 9
l = a+(n −1)d
350 = 17+(n −1)9
333 = (n−1)9
(n−1) = 37
n = 38
Sn = n/2 (a+l)
S38 = 13/2 (17+350)
= 19×367
= 6973
Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.
Question 7
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.Answer
Given,
Common difference, d = 7
22nd term, a22 = 149
Sum of first 22 term, S22 = ?
By nth term,
an = a+(n−1)d
a22 = a+(22−1)d
149 = a+21×7
149 = a+147
a = 2 = First term
Sum of n terms,
Sn = n/2(a+an)
S22 = 22/2 (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 that,
Second term, a2 = 14
Third term, a3 = 18
Common difference, d = a3−a2 = 18−14 = 4
a2 = a+d
14 = a+4
a = 10
Sum of n terms
Sn = n/2 [2a + (n – 1)d]
S51 = 51/2 [2×10 (51-1) 4]
= 51/2 [2+(20)×4]
= 51 × 220/2
= 51 × 110
= 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 that,
S7 = 49
S17 = 289
Sum of nth term;
Sn = n/2 [2a + (n – 1)d]
Therefore,
S7= 7/2 [2a +(n -1)d]
S7 = 7/2 [2a + (7 -1)d]
49 = 7/2 [2a +16d]
7 = (a+3d)
a + 3d = 7 …………………………………. (i)
S17 = 17/2 [2a+(17-1)d]
289 = 17/2 (2a +16d)
17 = (a+8d)
a +8d = 17 ………………………………. (ii)
Subtracting equation (i) from equation (ii),
5d = 10
d = 2
From equation (i), we can write it as;
a+3(2) = 7
a+ 6 = 7
a = 1
Hence,
Sn = n/2[2a+(n-1)d]
= n/2[2(1)+(n – 1)×2]
= n/2(2+2n-2)
= n/2(2n)
= n2
Question 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 terms in each case.
Answer
(i) an = 3+4n
a1 = 3+4(1) = 7
a2 = 3+4(2) = 3+8 = 11
a3 = 3+4(3) = 3+12 = 15
a4 = 3+4(4) = 3+16 = 19
the common difference between the terms
a2 − a1 = 11−7 = 4
a3 − a2 = 15−11 = 4
a4 − a3 = 19−15 = 4
Now, the sum of nth term is;
Sn = n/2[2a+(n -1)d]
S15 = 15/2[2(7)+(15-1)×4]
= 15/2[(14)+56]
= 15/2(70)
= 15×35
= 525
(ii) an = 9−5n
a1 = 9−5×1 = 9−5 = 4
a2 = 9−5×2 = 9−10 = −1
a3 = 9−5×3 = 9−15 = −6
a4 = 9−5×4 = 9−20 = −11
the common difference between the terms
a2 − a1 = −1−4 = −5
a3 − a2 = −6−(−1) = −5
a4 − a3 = −11−(−6) = −5
Now, the sum of nth term is;
Sn = n/2 [2a +(n-1)d]
S15 = 15/2[2(4) +(15 -1)(-5)]
= 15/2[8 +14(-5)]
= 15/2(8-70)
= 15/2(-62)
= 15(-31)
= -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, the10th and the nth terms.Answer
Given
Sn = 4n−n2
First term, a = S1 = 4(1) − (1)2 = 4−1 = 3
Sum of first two terms = S2= 4(2)−(2)2 = 8−4 = 4
Second term, a2 = S2 − S1 = 4−3 = 1
Common difference, d = a2−a = 1−3 = −2
Nth term, an = a+(n−1)d
= 3+(n −1)(−2)
= 3−2n +2
= 5−2n
Therefore, a3 = 5−2(3) = 5-6 = −1
a10 = 5−2(10) = 5−20 = −15
Hence, the sum of first two terms is 4. The second term is 1.
The 3rd, the 10th, and the nth terms are −1, −15, and 5 − 2n respectively.
Question 12
Find the sum of first 40 positive integers divisible by 6.Answer
The positive integers that are divisible by 6 are 6, 12, 18, 24 ….
a = 6
d = 6
S40 = ?
By the formula of sum of n terms, we know,
Sn = n/2 [2a +(n – 1)d]
Therefore, n = 40, we get,
S40 = 40/2 [2(6)+(40-1)6]
= 20[12+(39)(6)]
= 20(12+234)
= 20×246
= 4920
Question 13
Find the sum of first 15 multiples of 8.Answer
The multiples of 8 are 8, 16, 24, 32…
Therefore, a = 8
d = 8
S15 = ?
sum of nth term, we know,
Sn = n/2 [2a+(n-1)d]
S15 = 15/2 [2(8) + (15-1)8]
= 15/2[6 +(14)(8)]
= 15/2[16 +112]
= 15(128)/2
= 15 × 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, 9 … 49.
Hence,
a = 1
Common difference, d = 2
Last term, l = 49
for the last term, we know,
l = a+(n−1) d
49 = 1+(n−1)2
48 = 2(n − 1)
n − 1 = 24
n = 25 = Number of terms
sum of nth term, we know,
Sn = n/2(a +l)
S25 = 25/2 (1+49)
= 25(50)/2
=(25)(25)
= 625
Question 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.Answer
Since the penalties for each succeeding day by Rs 50 more than the preceding day Therefore penalities form of A.P. having first term as 200 and common difference as 50.
Therefore, a = 200 and d = 50
Penalty that has to be paid if contractor has delayed the work by 30 days = S30
sum of nth term,
Sn = n/2[2a+(n -1)d]
Therefore,
S30= 30/2[2(200)+(30 – 1)50]
= 15[400+1450]
= 15(1850)
= 27750
Therefore, the contractor has to pay Rs 27750 as penalty.
Question 16
A sum of Rs 700 is to be used to give seven cash 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.Answer
Let the cost of 1st prize be Rs. P.
Cost of 2nd prize = Rs. P − 20
And cost of 3rd prize = Rs. P − 40
We can see that the cost of these prizes are in the form of A.P..
Thus, a = P and d = −20
Given that, S7 = 700
By the formula of sum of nth term, we know,
Sn = n/2 [2a + (n – 1)d]
7/2 [2a + (7 – 1)d] = 700
a + 3(−20) = 100
a −60 = 100
a = 160
hence, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and 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
According to question the number of tree that each section planted
1, 2, 3, 4, 5………………..12
First term, a = 1
Common difference, d = 2−1 = 1
Sn = n/2 [2a +(n-1)d]
S12 = 12/2 [2(1)+(12-1)(1)]
= 6(2+11)
= 6(13)
= 78
Hence , number of trees planted by 1 section of the classes = 78
Number of trees planted by 3 sections of the classes = 3×78 = 234
Therefore, 234 trees will be planted by the students.
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, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)
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Answer
We know,
Perimeter of a semi-circle = πr
So,
P1 = π(0.5) = π/2 cm
P2 = π(1) = π cm
P3 = π(1.5) = 3π/2 cm
Where, P1, P2, P3 are the lengths of the semi-circles.
π/2, π, 3π/2, 2π, ….
P1 = π/2 cm
P2 = π cm
Common difference, d = P2 – P1 = π – π/2 = π/2
First term = P1= a = π/2 cm
the sum of n term we know,
Sn = n/2 [2a + (n – 1)d]
Therefor, Sum of the length of 13 consecutive circles is;
S13 = 13/2 [2(π/2) + (13 – 1)π/2]
= 13/2 [π + 6π]
=13/2 (7π)
= 13/2 × 7 × 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. In how many rows are the 200 logs placed and how many logs are in the top row?
Answer
the numbers of logs in rows are in the form of an A.P.20, 19, 18…
First term, a = 20 and common difference, d = a2−a1 = 19−20 = −1
Let a total of 200 logs be placed in n rows.
Thus, Sn = 200
By the sum of nth term formula,
Sn = n/2 [2a +(n -1)d]
S12 = 12/2 [2(20)+(n -1)(-1)]
400 = n (40−n+1)
400 = n (41-n)
400 = 41n−n2
n2−41n + 400 = 0
n2−16n−25n+400 = 0
n(n −16)−25(n −16) = 0
(n −16)(n −25) = 0
Either (n −16) = 0 or n−25 = 0
n = 16 or n = 25
By the nth term formula,
an = a+(n−1)d
a16 = 20+(16−1)(−1)
a16 = 20−15
a16 = 5
Similarly, the 25th term could be written as;
a25 = 20+(25−1)(−1)
a25 = 20−24
= −4
It can be seen, the number of logs in 16th row is 5 .
Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5.
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Question 20
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.
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
According to question the distance run by competitor are 2 x 5, 2 x(5 + 3), 2 x(5 + 3 + 3), 2 x(5 + 3 + 3 + 3)
10 , 16 ,22 ,28.....
Hence, the first term, a = 10 and d = 16−10 = 6
S10 =?
sum of n terms, we know,
S10 = 12/2 [2(20)+(n -1)(-1)]
= 5[20+54]
= 5(74)
= 370
Therefore, the competitor will run a total distance of 370 m.