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Learn Exercise 5.3 with Free Lessons & Tips

Question 1:

A spiral is made up of successive semicircles, with centres alternatively 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 the figure. What is the total length of such a spiral made of thirteen consecutive semicircles? (Take $\pi =\frac{22}{7}$ ).

[Hint: Length of the consecutive semicircles is $l_{1}, l_{2}, l_{3, }, l_{4}, .............$ with centres at A, B respectively]

Solution 1:

Length of semicircle=circumference/2 = 2Πr/2=Πr

There are 13 semicircles with radius

0.5,1,1.5,2,2.5,3,3.5,4,4.5,5,5.5,6,6.5

I.e. in AP whose 1St term=0.5 &

common difference=0.5

Now, total length of spiral=Π(0.5+1+1.5+2+2.5+3+3.5+4+4.5+5+5.5+6+6.5)

As we know sum of an AP is=

no. of terms(1St term+last term)/2= 13(0.5+6.5)/2=13×7/2=91/2

Now put value

Total length of spiral =(22/7)×(91/2)=11×13=143cm i.e. ans

Comments

Aman Kumar Gupta

Question 2:

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). In how many rows are the 200 logs placed and how many logs are in the top row?

Solution 2:

the number of logs stacked row wise form an arthimetic progression:

20,19,18,...

a=20

d=t2-t1=19-20= -1

Sn=(n/2)*(2a+(n-1)d)

200=(n/2)*(2*20+(n-1)(-1))

200=(n/2)(40-n+1)

400=n(41-n)

400=41n-n*n

n*n-41n+400=0

n*n-16n-25n+400=0

n(n-16)-25(n-16)=0

(n-16)(n-25)=0

n=16 or n=25

tn=a+(n-1)d

t₁₆=20+(16-1)(-1)

t₁₆=5

Checking for n=25

t₂₅=20+(25-1)(-1) = -5

Seq is

20,19,18,....5,4,3,2,1,0,1,2,3,4

This is not applicable

Answer

16 rows

5 logs on top.

Comments

Venkatesan Iyer

Question 3:

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 × 5 + 2 × (5 + 3)]

Solution 3:

Distance ran by him one way to complete the race=5+(5+3)+(5+3+3)+(5+3+3+3)+(5+3+...9tim) =5+8+11+14+.....+32,which is in arithmetic progression;

first term;a=5,common difference;d=8-5=3&nth term is a_n=32. S_{n=n/2[2a+(n-1)d]}

Sum to n- terms = 10/2[10+(10-1)3] = 5[10+27]=5x37=185;but he completes his run to and fro,so multiply 185 by 2=370m.

Comments

Sherine Das

Question 4:

Find the sums given below :

$(i) 7+10\tfrac{1}{2}+14...............+84$

$(ii) 34 + 32 + 30 + . . . + 10$

$(iii) –5 + (–8) + (–11) + . . . + (–230)$

Solution 4:

(i)a = 7

l = 84

$d=a_{2}-a_{1}=10\frac{1}{2}-7=\frac{7}{2}$

$l = a + (n -1)d$

$84=7+(n-1)\frac{7}{2}$

22 = n − 1

n = 23

$S_{n}=\frac{n}{2}[a+l]$

On substituting the values, we get

$S_{n}=\frac{23}{2}[7+84]=1046\frac{1}{2}$

(ii) a = 34

d = a₂ − a₁ = 32 − 34 = −2

l = 10

Let 10 be the n^th term of this A.P.

$l = a + (n -1) d$

10 = 34 + (n − 1) (−2)

−24 = (n − 1) (−2)

12 = n − 1

n = 13

$S_{n}=\frac{n}{2}[a+l]$

$S_{n}=\frac{13}{2}[34+10]=286$

(iii)

a = −5

l = −230

d = a₂ − a₁ = (−8) − (−5)

= − 8 + 5 = −3

Let −230 be the n^th term of this A.P.

$l = a + (n -1) d$

−230 = − 5 + (n − 1) (−3)

−225 = (n − 1) (−3)

(n − 1) = 75

n = 76

$S_{n}=\frac{n}{2}[a+l]$

$S_{n}=\frac{76}{2}[(-5)+(-230)]=-8930$

Comments

Tasneem

Question 5:

If the sum of the first n terms of an AP is $4n-n^{2}$ , what is the first term (that is $S_{1}$ )? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

Solution 5:

It is given that

$S_{n} = 4n - n^{2}$

First term, $a = S_{1} = 4(1) - (1)2 = 4 - 1=3$

Sum of first two terms, $S_{2}$ $= 4(2) - (2)^{2} = 8 - 4=4$

Second term, $a_{2} = S_{_{2}} - S_{1} = 4 - 3=1$

$d = a_{2}- a=1 - 3=-2$

$a_{n}= a + (n -1)d$

= 3 + (n - 1) (-2)

= 3 - 2n + 2

= 5 - 2n

Therefore, $a_{3} = 5 - 2(3) = 5 - 6=-1$

$a_{10} = 5 - 2(10) = 5 - 20=-15$

Hence, the sum of first two terms is 4.

The second term is 1. 3^rd, 10^th, and n^th terms are −1, −15, and 5 − 2n respectively.

Comments

Tasneem

Question 6:

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 6:

Formula: $S_{n}=\frac{n}{2}[2a+(n-1)d]$

(i)2, 7, 12 ,…, to 10 terms

a = 2

d = a₂ − a₁ = 7 − 2 = 5

n = 10

$S_{10}=\frac{10}{2}[2\times 2+(10-1)5]$

$S_{n}=245$

(ii)−37, −33, −29 ,…, to 12 terms

a = −37

d = a₂ − a₁ = (−33) − (−37)

= 4

n = 12

$S_{12}=\frac{12}{2}[2\times (-37)+(12-1)4]$

$S_{n}=-180$

(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

$S_{100}=\frac{100}{2}[2\times (0.6)+(100-1)1.1]$

$S_{n}= 5505$

(iv). …….. , to 11 terms

For this A.P.,

n = 11

$d= \frac{1}{12}-\frac{1}{15}$ $=\frac{1}{60}$

$S_{11}=\frac{11}{2}[2\left ( \frac{1}{15} \right )+(11-1)\frac{1}{60}]$

$S_{n}=\frac{33}{20}$

Comments

Tasneem

Question 7:

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 7:

(i) Given, $a = 5, d=3, a_{n} = 50$

$a_{n}= a + (n -1)d$

∴ $50 = 5 + (n -1)3$

$45 = (n -1)3$

$15 = n - 1$

n = 16

$S_{n}=\frac{n}{2}[a+a_{n}]$

$S_{16}=\frac{16}{2}[5+50]$ $=440$

(ii)Given, $a = 7, a_{13}= 35$

$a_{n}= a + (n -1)d$

$\therefore a_{13} = a + (13 -1) d$

35 = 7 + 12 d

$35 - 7 = 12d$

28 = 12d

$d=\frac{7}{3}$

$S_{13}=\frac{12}{2}[7+35] =273$

(iii) Given, $a_{12}$ = 37, d = 3

a₁₂ = a + (12 − 1)3

37 = a + 33

a = 4

$S_{13}=\frac{12}{2}[4+37] =246$

(iv) Given, $a_{3}= 15, S_{10} = 125$

$a_{3} = a + (3 -1)d$

$15 = a + 2d....... (i)$

$S_{n}=\frac{n}{2}[2a+(n-1)d]$

$S_{10}=\frac{10}{2}[2a+(10-1)d]$

$25=2a+9d........(ii)$

On multiplying equation (1) by 2, we obtain

30 = 2a + 4d (iii)

On subtracting equation (iii) from (ii), we obtain

$-5 = 5d$

d = −1

From equation (i),

$15 = a + 2(-1)$

$15 = a - 2$

a = 17

$a_{10} = a + (10 -1)d$

$a_{10}= 17 + (9) (-1)$

$17 - 9 = 8$

(v)Given, $d = 5, S_{9} = 75$

$S_{n}=\frac{n}{2}[2a+(n-1)d]$

$S_{9}=\frac{9}{2}[2a+(9-1)5]$

$75=\frac{9}{2}[2a+40]$

$25 = 3(a + 20)$

25 = 3a + 60

$3a = 25 - 60$

$a=-\frac{35}{5}$

$a_{n}= a + (n -1)d$

$a_{9} = a + (9 - 1) (5)$

$=\frac{-35}{3}+8(5)$

$=\frac{85}{3}$

(vi) Given, $a = 2, d=8, S_{n} = 90$

$S_{n}=\frac{n}{2}[2a+(n-1)d]$

$90= \frac{n}{2}[4+(n-1)8]$

$90 = n [2 + (n - 1)4]$

$90 = n [2 + 4n - 4]$

$90 = n (4n - 2) = 4n^{2} -2n$

$4n^{2} - 2n - 90=0$

$4n^{2} - 20n + 18n - 90=0$

$4n (n - 5) + 18 (n - 5) = 0$

$(n -5) (4n + 18) = 0$

Either n - 5 = 0 or 4n + 18 = 0

n = 5 or $n= -\frac{18}{4}=-\frac{9}{2}$

However, n can neither be negative nor fractional.

Therefore, n = 5

$a_{n}= a + (n -1)d$

$a_{5} = 2 + (5 - 1)8$

= 2 + (4) (8)

= 2 + 32 = 34

(vii) Given, $a = 8, a_{n} = 62, S_{n} = 210$

$S_{n}=\frac{n}{2}[a+a_{n}]$

$210=\frac{n}{2}[8+62]$

n = 6

$a_{n}= a + (n -1)d$

$62 = 8 + (6 - 1)d$

62 - 8 = 5d

54 = 5d

$d=\frac{54}{5}$

(viii) Given, $a_{n} = 4, d=2, S_{n}= -14$

$a_{n}= a + (n -1)d$

4 = a + (n - 1)2

4 = a + 2n - 2

a + 2n = 6

a = 6 - 2n.............. (i)

$S_{n}=\frac{n}{2}[a+a_{n}]$

$-14=\frac{n}{2}[a+4]$

$-28 = n (a + 4)$

$-28 = n (6 - 2n + 4)$ {From equation (i)}

$-28 = n (- 2n + 10)$

$-28 = - 2n^{2} + 10n$

$2n^{2} -10n - 28=0$

$n^{2}- 5n -14 = 0$

$n^{2} - 7n + 2n - 14=0$

$n (n - 7) + 2(n - 7) = 0$

$(n - 7) (n + 2) = 0$

Either n=- 7 = 0 or n + 2 = 0

n = 7 or n = -2

However, n can neither be negative nor fractional.

Therefore, n = 7

From equation (i), we obtain

$a = 6 - 2n$

$a = 6 - 2(7)$

= 6 - 14

= -8

(ix)Given, $a = 3, n = 8, S = 192$

$S_{n}=\frac{n}{2}[2a+(n-1)d]$

$192=\frac{8}{2}[2(3)+(8-1)d]$

192 = 4 [6 + 7d]

48 = 6 + 7d

42 = 7d

d = 6

(x)Given, l = 28, S = 144 and there are total of 9 terms.

$S_{n}=\frac{n}{2}[a+l]$

$144=\frac{9}{2}[a+28]$

$(16) \times (2) = a + 28$

32 = a + 28

a = 4

Comments

Tasneem

Question 8:

How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

Solution 8:

Let there be n terms of this A.P

For this A.P, a = 9

$d = a_{2} - a_{1}$ = 17 - 9 = 8

$S_{n}=\frac{n}{2}[2a+(n-1)d]$

$636=\frac{n}{2}[2a+(n-1)8]$

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

$n=-\frac{53}{4}$ or n = 12

n cannot be $-\frac{53}{4}$ . As the number of terms can neither be negative nor fractional.

Therefore, n = 12

Comments

Tasneem

Question 9:

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 9:

Given a=5 l=45 Sn=400

Sn= n/2(a+l)

400 =(n/2) (5+45)

400=50n/2=25n

n=400/25

=16 = number of terms

To find common difference 'd'

a₁₆=45 ⇒5+15d=45

15d=40

d= 8/3

Comments

Bhavani Ramesh

Question 10:

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 10:

Let a be the first term, d be the common difference & l be the last team.Now, a = 17, d = 9, l = 350.

∴ 17 + ( n - 1 )9 = 350

⇒ ( n - 1 )9 = 333 ∴ n - 1 = 333/9

⇒ n = 37 + 1 = 38.

Now, S38 = 38/2 ( 17 + 350 ) = 19 X 367 = 6973.

∴ there are 38 terms in the AP & their sum is 6973.

Comments

Manoj Kumar

Question 11:

Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Solution 11:

Given d=7 and a₂₂=149, n=22

$a_{n} = a + (n - 1)d$

$a_{22} = a + (22 - 1)d$

149 = a + 21 × 7

149 = a + 147

a = 2

We have to find S₂₂= $\frac{22}{2}(2+149)$

= $11\times 151$

=1661

Comments

Bhavani Ramesh

Question 12:

Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Solution 12:

Given a₂=a+d=14→(1)

$a_{3}$ =a+2d=18→(2)

(2)-(1)⇒d=4

Substitute d=4in (1) we get a=10

We have to find S₅₁,

S₅₁=51/2(2*10+50*4)

=51/2*(20+200)

=51*220/2

=51*110

=5610

∴ the sum of first 51 terms of the AP is 5610

Comments

Bhavani Ramesh

Question 13:

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 13:

Let the A.P be a,a+d,a+2d,...

Sn=(n/2)(2a+(n-1)d)

S7=(7/2)(2a+(7-1)d)=49

2a+6d=2*49/7

2a+6d =14 eq1

S₁₇=(17/2)(2a+(17-1)d)=289

2a+16d=289*2/17

2a+16d=34 eq2

2a+6d= 14 eq1

Subtract

10d=20

d=2

Substitute this in eq1

2a+6*2=14

2a=14-12=2

a=1

d=2

Sn= (n/2)(2*1+(n-1)2)

Answer

Sn=(n/2)(2n)

Comments

Venkatesan Iyer

Question 14:

Show that $a_{1},a_{2},....................,a_{n},.............$ from an A.P where $a_{n}$ 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

Solution 14:

i) $a_{n} = 3 + 4n$

$a_{1} = 3 + 4(1) = 7$

$a_{2} = 3 + 4(2) = 3 + 8=11$

$a_{3} = 3 + 4(3) = 3 + 12=15$

$a_{2} - a_{1} = 11 - 7=4$

$a_{3} - a_{2} = 15 - 11=4$

$\therefore a_{2}-a_{1}= a_{3}-a_{2}$

This forms an AP with common difference as 4 and first term as 7.

$S_{n}=\frac{n}{2}[2a+(n-1)d]$

$S_{15}=\frac{15}{2}[2(7)+(15-1)4]$

= 525

(ii) $a_{n} = 9 - 5n$

$a_{1} = 9 - 5 × 1=9 - 5=4$

$a_{2} = 9 - 5 × 2=9 - 10=-1$

$a_{3} = 9 - 5 × 3=9 - 15=- 6$

$a_{2} - a_{1} = - 1 - 4=-5$

$a_{3} - a_{2} = - 6 - (-1) = -5$

$\therefore a_{2}-a_{1}= a_{3}-a_{2}$

This forms 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

Comments

Tasneem

Question 15:

Find the sum of the first 40 positive integers divisible by 6.

Solution 15:

The first positive integers divisible by 6 are 6, 12, 18,..,which forms an AP with first term a = 6 and common difference d = 6.

∴ S40 = 40/2 [ 2 X 6 + ( 40 - 1 ) 6 ]

= 20 ( 12 + 39 X 6 ) = 20 X 246 = 4920.

Comments

Manoj Kumar

Question 16:

Find the sum of the first 15 multiples of 8.

Solution 16:

The multiples of 8 are 8, 16, 24, 32…

This in an A.P

Therefore, a = 8

d = 8

$S_{15}=\frac{15}{2}[2\times 8+(15-1)8] =960$

Comments

Tasneem

Question 17:

Find the sum of the odd numbers between 0 and 50. 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?

Solution 17:

The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49

This forms an A.P.

a = 1

d = 2

l = 49

$l = a + (n - 1) d$

$49 = 1 + (n -1)2$

$48 = 2(n - 1)$

n - 1 = 24

n = 25

$S_{n}=\frac{n}{2}[a+l]$

$S_{25}=\frac{25}{2}[1+49]$

= 625

Comments

Tasneem

Question 18:

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.

Solution 18:

Let the cost of 1^st prize be P.

Cost of 2^nd prize = P - 20

And cost of 3^rd prize = P - 40

These prizes are in an A.P

d = −20

Given, $S_{7}=700$

$\frac{7}{2}\left [ 2a+(7-1)d \right ]=700$

a + 3(-20) = 100

a -60 = 100

a = 160

Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.

Comments

Tasneem

Question 19:

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?

Solution 19:

The number of trees planted by the students is in an 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

Therefore, 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.

Comments

Tasneem

All Exercises - Chapter 5 - Arithmetic Progression

Exercise 5.1

Exercise 5.2

Exercise 5.3

Exercise 5.4

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