NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.3
In this page we have NCERT Solutions for Class 10th Maths: Chapter 5 - Arithmetic Progressions for
EXERCISE 5.3 . Hope you like them and do not forget to like , social_share
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The sum of the first n terms of an AP is given by :
$S =\frac {n}{2} [2a +(n-1)d]$
If l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP is given by :
$S =\frac {n}{2} [a +l]$
If a ,b,c are in A.P.then
$b= \frac {a+c}{2}$
And b is called the Arithmetic Mean
NCERT Solutions Class 10 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) 2, 7, 12 ,…, to 10 terms
For this A.P.,
a= 2
d=a2-a1= 7 - 2 = 5
n= 10
We know that,
$S_n= \frac {n}{2} [2a+(n- 1)d]$
$S_{10}= \frac {10}{2} [2(2)+(10- 1) \times 5]$
= 245
(ii) -37, -33, -29 ,…, to 12 terms
For this A.P.,
a= -37
d=a2-a1= (-33) - (-37)
= - 33 + 37 = 4
n= 12
We know that,
$S_n= \frac {n}{2} [2a+(n- 1)d]$
$S_{12}= \frac {12}{2} [-74+(12 - 1) \times 4]$
= -180
(iii) 0.6, 1.7, 2.8 ,…, to 100 terms
For this A.P.,
a= 0.6
d=a2-a1= 1.7 - 0.6 = 1.1
n= 100
We know that,
$S_n= \frac {n}{2} [2a+(n- 1)d]$
$S_{100}=\frac {100}{2} [1.2 +(100-1) \times 1.1]$
= 5505
Question 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, 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 We will be using primarily these formula
$S_n= \frac {n}{2} [a+ a_n]$
$S_n= \frac {n}{2} [2a+ (n-1)d]$
$a_n=a+ (n- 1)d$
Question 4.
How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636? Answer
Let there be n terms of this A.P.
For this A.P.,a= 9
d=a2-a1= 17 - 9 = 8
$S_n= \frac {n}{2} [2a+ (n-1)d]$
$636 = \frac {n}{2} [18+ (8- 1) \times 8]$
$636 =n(4n+ 5)$
$4n^2+ 5n- 636 = 0$
(4n+ 53) (n- 12) = 0
Now n can neither be negative nor fractional.
so,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 that,
a= 5
$a_n = 45$
Sn= 400
Now
$S_n= \frac {n}{2} [a+ a_n]$
$400 = \frac {n}{2} (5+ 45)$
n= 16
$a_n=a+ (n- 1)d$
45 = 5 + (16 - 1)d
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,
a= 17
$a_n = 350$
d= 9
Now
$a_n=a+ (n- 1)d$
350 = 17 + (n- 1)9
n= 38
Now Sum is given by
$S_n= \frac {n}{2} [a+ a_n]$
$S_{38}= \frac {38}{2} (17+ 350)$
= 6973
Question 7.
Find the sum of first 22 terms of an AP in which d= 7 and 22ndterm is 149. Answer
d= 7
$a_{22}= 149$
S22= ?
$a_n=a+ (n- 1)d$
$149 =a+ 21 \times 7$
a= 2
$S_n= \frac {n}{2} [a+ a_n]$
$= \frac {22}{2} (2+149)=1661$
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
a3= 18
d=a3-a2= 18 - 14 = 4
a2=a+d
14 =a+ 4
a= 10
$S_n= \frac {n}{2} [2a+ (n-1)d]$
$S_{51}= \frac {51}{2} [20+ (51 - 1) \times 4]$
= 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
$S_7= 49$
$S_{17}= 289$
Let a and d be the first term and common difference
Now
$S_n= \frac {n}{2} [2a+ (n-1)d]$
Then
$S_7 = \frac {7}{2}[2a+ (n- 1)d]$
$49 = \frac {7}{2} [2a+16d]$
a+ 3d= 7 ...(i)
Similarly,
$S_{17}= \frac {17}{2} [2a+ (17 - 1)d]$
$289 = \frac {17}{2} (2a+ 16d)$
a+ 8d= 17 ...(ii)
Solving linear equation (1) and (2)
d= 2,a=1
Now
$S_n= \frac {n}{2} [2a+ (n-1)d]$
$= \frac {n}{2} [2+ (n- 1) \times2]$
$= n^2$
Question 10.
Show that a1,a2… ,an, … form an AP where anis 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
In these type of question, we can find the various term by substituting the values of n and check for Common difference
(i)$a_n= 3 + 4n$
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
We can see that difference of two terms is same
a2-a1= 11 - 7 = 4
a3-a2= 15 - 11 = 4
a4-a3= 19 - 15 = 4
i.e., $a_{n+1}- a_{n}$is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.
$S_n= \frac {n}{2} [2a+ (n-1)d]$
$S_{15}= \frac {15}{2} [14+ (15- 1) \times 4]$
= 525
(ii)$a_n= 9 - 5n$
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
We can see that difference of two terms is same
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 A.P. with common difference as -5 and first term as 4.
$S_n= \frac {n}{2} [2a+ (n-1)d]$
$S_{15}= \frac {15}{2} [2(4)+ (15- 1) (-5)]$
= -465
Question 11.
If the sum of the first n terms of an AP is $4n- n^2$, what is the first term (that isS1)? What is the sum of first two terms? What is the second term? Similarly find the 3rd, the 10thand thenthterms. Answer
Given
$S_n= 4n- n^2$
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
d=a2-a= 1 - 3 = -2
Now nth term is given by
$a_n=a+ (n- 1)d$
= 3 + (n- 1) (-2)
= 5 - 2n
Therefore,
a3= 5 - 2(3) = 5 - 6 = -1
a10= 5 - 2(10) = 5 - 20 = -15
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 …
it is an A.P whose first term is 6 and common difference is 6.
a= 6,d= 6 ,S40=?
$S_n= \frac {n}{2} [2a+ (n-1)d]$
$S_{40}= \frac {40}{2} [2(6)+ (40 - 1) 6]$
= 4920
Question 13.
Find the sum of first 15 multiples of 8. Answer
The multiples of 8 are
8, 16, 24, 32…
It is in an A.P., having first term as 8 and common difference as 8.
a= 8,d= 8 ,S15= ?
$S_n= \frac {n}{2} [2a+ (n-1)d]$
$S_{15}= \frac {15}{2} [16+ (15 - 1)8]$
= 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
It is an A.P.
a= 1
d= 2
$a_n = 49$ ,n=?
$S_n=?$
From nth term formula
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
It can be seen that these penalties are in an A.P. having first term as 200 and common difference as 50.
a= 200
d= 50
Penalty that has to be paid if he has delayed the work by 30 days =S30
$S_n= \frac {n}{2} [2a+ (n-1)d]$
$S_{30}= \frac {30}{2} [2(200) + (30 - 1) 50]$
= 27750
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 first prize is a,then seven prizes will be
a, a-20,a- 40,a-60,a-80,a-100,a-120
Now
$a+ a-20+a- 40+a-60+a-80+a-100+a-120=700$
$7a=1120$
$a= 160$
So prizes will be
160,140,120,100,80,60,40
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
It can be see that the number of trees planted by the students in various classes are in AP.
1, 2, 3, 4, 5............12
First term,a= 1
Common difference,d= 2 - 1 = 1
$S_n= \frac {n}{2} [2a+ (n-1)d]$
$S_{12}= \frac {12}{2} [2(1)+ (12 - 1)(1)]$
= 78
Number of trees planted by 3 sections of the classes = 3 × 78 = 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, 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? (Takeπ = 22/7) Answer
We know that perimeter of semi-circle is given by$\pi r$
Now
Lenght of first semi circle
$P_1= \pi (0.5) = \frac {\pi}{2}$cm
Lenght of second semi circle
$P_2= \pi (1) = \pi $cm
Lenght of Third semi circle
$P_3= \pi (1.5) = \frac {3 \pi}{2 }$cm
.........
So the series is
$\frac {\pi }{2}$, $\pi$, $\frac {3\pi}{2}$ , $2 \pi$, ....
$a= \frac {\pi}{2}$cm
$d=\pi- \frac {\pi}{2}= \frac {\pi}{2}$
$S_n= \frac {n}{2} [2a+ (n-1)d]$
Therefore, Sum of the length of 13 consecutive circles
$S_{13}= \frac {13}{2} [2(\frac {\pi}{2})+ (13- 1)\frac {\pi}{2}]$
= 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
It can be seen that the numbers of logs in rows are in an A.P.
20, 19, 18…
For this A.P.,
a= 20 ,d=a2-a1= 19 - 20 = -1
Let total of 200 logs be placed in n rows.
Sn= 200
$S_n= \frac {n}{2} [2a+ (n-1)d]$
$200= \frac {12}{2} [40+ (n- 1)(-1)]$
$400 =n(40 -n+ 1)$
$n^2- 41n+ 400 = 0$ Factoring the quadratic
(n- 16) (n- 25) = 0
Either (n- 16) = 0 or n- 25 = 0
n= 16 or n= 25
For n=16
an=a+ (n- 1)d
a16= 20 + (16 - 1) (-1)
a16= 5
Similarly,
a25= 20 + (25 - 1) (-1)
= -4
Clearly, 25th row is not possible as it is negative
So answer is 16 rows
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
Distance for First potato = $2 \times 5 =10$
Distance for second potato = $2 \times (5 + 3)=16$
Distance for third potato = $2 \times (5 + 2 \times 3)=22$
Distance for Fourth potato = $2 \times (5 + 3\times 3)=28$
So the series will be
10, 16, 22, 28, 34,......
a= 10
d= 16 - 10 = 6
S10=?
$S_n= \frac {n}{2} [2a+ (n-1)d]$
$S_{10}= \frac {10}{2}[20+ (10- 1)6]$
= 370 m
Summary
NCERT Solutions for Class 10th Maths: Chapter 5 - Arithmetic Progressions Ex 5.3 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail.You can download this as pdf
This chapter 5 has total 4 Exercise 5.1 ,5.2,5.3 and 5.4. This is the third exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below
