These NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions 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) \(\frac { 1 }{ 15 }\), \(\frac { 1 }{ 12 }\), \(\frac { 1 }{ 10 }\), …….., to 11 terms.
Solution:
(i) We have
2, 7, 12, … to 10 terms
Here, a = 2, d = 7 – 2 = 5 and n = 10
Therefore, the sum of 10 terms of the AP 2, 7, 12, ….. is 245
(ii) We have,
– 37, – 33, – 29, … to 12 terms
Here, a = – 37, d = – 33 – (- 37) = 4 and n = 12
Therefore, the sum of 12 terms of the AP – 37, – 33, – 29,… is – 180.
(iii) We have,
0.6,1.7, 2.8,… to 100 terms
Here, a = 0.6, d = 1.7 – 0.6 = 1.1 and n = 100
Therefore, the sum of 100 terms of the AP 0.6, 1.7,2.8,. .. is 5505.
(iv) We have,
Therefore, the sum of 11 terms of the AP \(\frac { 1 }{ 15 }\), \(\frac { 1 }{ 12 }\), \(\frac { 1 }{ 10 }\), ….. is \(\frac { 33 }{ 20 }\).
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)
Solution:
(i) We have
7 + 10\(\frac { 1 }{ 2 }\) + 14 + … + 84
Therefore 7 + 10\(\frac { 1 }{ 2 }\) + 14 + … + 84 = 1046\(\frac { 1 }{ 2 }\)
(ii) We have
34 + 32 + 30 + … + 10
Therefore, 34 + 32 + 30 + … + 10 = 286
(iii) We have
-5 + (-8) + (-11) + ….. + (-230)
Therefore, -5 + (-8) + (-11) + ….. + (-230) is – 8390.
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.
Solution:
(i) We have
a = 5, d = 3, an = 50
But, We know that
Therefore, the value of n = 16 and Sn = 440.
(ii) We have a = 7, a13 = 35
But, We know that
a13 = a + 12d
or 35 = 7 + 12 d
or d = \(\frac { 28 }{ 12 }\) = \(\frac { 7 }{ 3 }\)
Therefore, the value of d = \(\frac { 7 }{ 3 }\) and S13 = 273.
(iii) We have a12 = 37, d = 3
But, We know that
a12 = a + 11d
or 37 = a + 11 x 3
or 37 = a ÷ 33
or a = 37 – 33 = 4
Again we know that
Therefore, the value of a = 4 and S12 = 246.
(iv) We have a3 = -15, S10 = 125
But, We know that
Multiplying equation (i) by 10 then subtracting from equation (ii), we get,
Putting the value of d = – 1 in equation, (i) we get,
a + 2(-1) = 15
a – 2 = 15 ⇒ a = 17
Now, a = a+ (n – 1)d
a10 = 17 + (10 – 1) (- 1)
a10 = 17 – 9 = 8
Therefore, d = – 1 and a10 = 8
(v) We have d = 5, S9 = 75
But, We know that
Therefore, the value of a = \(\frac { -35 }{ 3 }\) and a9 = \(\frac { 85 }{ 3 }\).
(vi) We have a = 2, d = 8, Sn = 90
But, We know that
But number of terms cannot be in negative
∴ n = \(\frac { -9 }{ 2 }\) is neglected.
So, n = 5
Again we know that
an = a + (n – 1)d
∴ a5 = 2 + (5 – 1) x 8
= 2 + 4 x 8
or a5 = 2 + 32 = 34
Therefore, the value of n = 5 and a5 = 34.
(vii) We have a = 8, an = 62, Sn = 210
But, We know that
an = a + (n – 1)d
62 = 8 + (n – 1) d
∴ (n – 1)d = 62 – 8 = 54
or a5 = 2 + 32 = 34 … (i)
Again
Therefore, the value of n = 6 and d = \(\frac { 54 }{ 5 }\).
(viii) We have an = 4, d = 2, Sn = – 14
But, We know that
an = a + (n – 1)d
or 4 = a+ (n – 1)2
or 4 = a + 2n – 2
∴ a + 2n = 6
∴ a = 6 – 2n … (i)
Again, we know that
Sn = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
or – 14 = \(\frac { n }{ 2 }\)[2 x a + (n – 1) x 2]
or – 14 x 2 = n (2a + 2n – 2)
or – 28 = n[2(6 – 2n) + 2n – 2]
[Putting the value of a from equation (i)]
or – 28 = n (12 – 4n + 2n – 2)
or – 28 = n (- 2n +10)
or – 28 = – 2n² + 10n
or 2n² – 10n – 28 = 0
or n² – 5n – 14 = 0
or n² – 7n + 2n – 14 = 0
or n (n – 7) + 2 (n – 7) = 0
or (n + 2) (n – 7) = 0
∴ n = 7 and n= -2
But, number of terms cannot be in negative
∴ n = – 2 is neglected.
So, n = 7
Putting the value of n in equation (i), we get,
a = 6 – 2n = 6 – 2 x 7
∴ a = 6 – 14 = – 8
Therefore, the value of a = – 8 and n = 7.
(ix) We have a = 3, n = 8, S = 192
But, We know that
Therefore, the value of d = 6.
(x) We have l = 28, S = 144, and n = 9
But, We know that
an = l
or a + (n – 1)d = 28
or a + (9 – 1)d = 28
or a + 8d = 28 … (i)
Again we know that
Multiply equation (i) by 2 then subtract equation (ii) from equation (i) we get,
Putting the value of d in equation (i)
a + 8 d = 28
or a + 8 x 3 = 28
∴ a = 28 – 24 = 4
Therefore, the value of a = 4.
Question 4.
How many terms of AP: 9, 17, 25, ….. must be taken to give a sum of 636?
Solution:
We have, 9,17,25 Here,
a = 9 and d = 17 – 9 = 8
Let n terms of this AP must be taken to give a sum of 636.
We know that
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.
Solution:
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?
Solution:
Question 7.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:
Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
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.
Solution:
We have
S7 and S17 = 289
But, we know that
Again, we know that
Subtracting equation (ii) from equation (i), we get,
3d – 8d = 7 – 17
or – 5d = – 10
Putting the value of d in equation (i), we get,
a + 3d = 7
or a + 3 x 2 = 7
∴ a = 7 – 6 = 1
Therefore, sum of n terms of this AP is
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.
Solution:
(i) We have
a = 3 + 4n
Put n = 1 in this equation
∴ a1 = 3 + 4 x 1 = 7
Now, put n = 2 in this equation
a2 = 3 + 4 x 2 = 11
Again, put n = 3 in this equation
a3 = 3 + 4 x 3 = 15
Therefore, the required AP is 7, 11, 15, ….
Here, a = 7, d = 11 – 7 = 4
We know that
Therefore, sum of 15 terms of the sequence an = 3 + 4n is 525.
(ii) We have,
∴ an = 9 – 5n
a1 = 9 – 5 x 1 = 4
a2 = 9 – 5 x 2 = – 1
and a3= 9 – 5 x 3 = – 6
∴ AP is 4,-1,-6,…
∴ Common difference (d) = – 1 – 4 = – 5
Therefore, sum of 15 terms of the sequence an = 9 – 5n is – 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.
Solution:
Sn = 4n – n²
Put n = 1, 2, 3, …
S1 = 4(1) – 1
S1 = 3,
S2=4,
S3=3,
The second term is
a2 = S2 – S1
a2 = 4 – 3 ⇒ a2 = 1;
Third term,
a2= S3 – S2
a3 = 3 – 4 ⇒ a2 = – 1
Third term
a2 = S3 – S2
a3 = 3 – 4 ⇒ a3 = – 1;
⇒ – 60 + 45 = – 15
and nth term is
an = (4n – n2) – [(4n – 4) – (n – 1)²]
= 4n – n2 – 4n + 4 + n2 + 1 – 2n
= 5 – 2n
nth term is (5 – 2n).
Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
The positive integers divisible by 6 are
6,12,18, …….
Here, a = 6 and d = 12 – 6 = 6
∴ Sn = \(\frac { 40 }{ 2 }\)[2 x 6 + (40 – 1) x 6]
= 20 (12 + 39 x 6) = 20 (12 + 234)
∴ S40 = 20 x 246 = 4920
Question 13.
Find the sum of the first 15 multiples of 8.
Solution:
The first 15 multiples of 8 are 8,16, 24…
Here, a = 8,d = 16-8 = 8
Sn = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
⇒ S15 = \(\frac {15 }{ 2 }\)[2 x 8 + (15 – 1)8]
⇒ S15 = \(\frac { 15 }{ 2 }\)[16 + 112]
⇒ S15 = \(\frac { 15 }{ 2 }\) x 128
⇒ S15 = 15 x 64 = 960
∴ The sum of first 15 multiples of 8 is 960.
Question 14.
Find the sum of the odd numbers between 0 and 50.
Solution:
Let odd numbers between 0 and 50 be 1, 3, 5, 7,…….., 49.
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?
Solution:
We have given that penalty for delay of completion beyond a certain date are 200, 250, 300, ….
It is the form of an AP, where a = 200 d = 50
Sn = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
S30= \(\frac { 30 }{ 2 }\) [2 x 200 + (30 – 1)50]
= 15 [400 + 1450]
= (15 x 1850) = 27750
The contractor has to pay ₹ 27750 as penalty.
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.
Solution:
Let 1st prize be of ₹ a
2nd prize be ₹ (a – 20) and
3rd prize be ₹ (a – 20 – 20) = ₹ (a – 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, example 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?
Solution:
Let the trees be planted 1, 2, 3, 4, 5 , …… 12
Here, a = 1, d = 1, n = 12
Total number of trees planted by each section
S12 = \(\frac { 12 }{ 2 }\) [2a + (n – 1) d] = 6 [2 x 1 + (12 – 1) x 1]
= 6 [2 + 11] = 6 x 13 = 78
Total number of trees planted by 3 sections = 78 x 3 = 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 figure. What is the total length of such a spiral made up of thirteen consecutive semicircles?
(Take π = \(\frac { 22 }{ 7 }\))
[Hint:Length of successive semicircles is l1, l2, l3, l4, … with centres at A, B, respectively.]
Solution:
Length of successive semi circles is l1, l2, l3, l4, …
Series is 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm… we have to found ln(total length)
Here, l1 = 0.5 π
d = 1.0 – 0.5 = 0.5π
n = 13 (consective semi circles)
Sum of ln terms
∴ Total length of spiral of 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 Figure). In how many rows are the 200 logs placed and how many logs are in the top row?
Solution:
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.) 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 x 5 + 2 x (5 + 3)]
Solution:
Distance covered by the girl pick up the first potato = 10m
The distance covered by the girl to pick up the second potato = 8 x 2 = 16m
and the distance covered by the girl to pick up the third potato = 11 x 2 = 22m
So, we get the following series ;
10, 16, 22, 28, … upto 10th term.
We know that, [where a = 10, d = 6]
Total distance We know that
∴ Sn = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
= \(\frac { 10 }{ 2 }\)[2 x 10 + (10 – 1)6]
= 5 [20 + 54]
= 5 [74] = 370 m
Hence, total distance covered by the competitor is 370.