NCERT Solutions for Class 10th: Ch 1 Arithmetic Progressions Maths
Page No: 99
Exercise 5.1
1. In which of the following situations, does the list of numbers involved make as 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.
Answer
It can be observed that
Taxi fare for 1st km = 15
Taxi fare for first 2 km = 15 + 8 = 23
Taxi fare for first 3 km = 23 + 8 = 31
Taxi fare for first 4 km = 31 + 8 = 39
Clearly 15, 23, 31, 39 … forms an A.P. because every term is 8 more than the preceding term.
(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
Answer
Let the initial volume of air in a cylinder beVlitres. In each stroke, the vacuum pump removes 1/4 of air remaining in the cylinder at a time. In other words, after every stroke, only 1 - 1/4 = 3/4th part of air will remain.
Therefore, volumes will beV, 3V/4 , (3V/4)2, (3V/4)3...
Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.
(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.
Answer
Cost of digging for first metre = 150
Cost of digging for first 2 metres = 150 + 50 = 200
Cost of digging for first 3 metres = 200 + 50 = 250
Cost of digging for first 4 metres = 250 + 50 = 300
Clearly, 150, 200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term.
(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.
Answer
We know that if RsPis deposited atr% compound interest per annum for n years, our money will be
Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.
2. Write first four terms of the A.P. when the first term a and the common differenced are given as follows
(i)a= 10,d= 10
(ii)a= -2,d= 0
(iii)a= 4,d= - 3
(iv)a= -1d= 1/2
(v)a= - 1.25,d= - 0.25
Answer
(i)a= 10,d= 10
Let the series bea1,a2,a3,a4,a5…
a1=a= 10
a2=a1+d= 10 + 10 = 20
a3=a2+d= 20 + 10 = 30
a4=a3+d= 30 + 10 = 40
a5=a4+d= 40 + 10 = 50
Therefore, the series will be 10, 20, 30, 40, 50 …
First four terms of this A.P. will be 10, 20, 30, and 40.
(ii)a= - 2,d= 0
Let the series bea1, a2,a3,a4…
a1=a= -2
a2=a1+d= - 2 + 0 = - 2
a3=a2+ d = - 2 + 0 = - 2
a4=a3+d= - 2 + 0 = - 2
Therefore, the series will be - 2, - 2, - 2, - 2 …
First four terms of this A.P. will be - 2, - 2, - 2 and - 2.
(iii)a= 4,d= - 3
Let the series bea1,a2,a3,a4…
a1=a= 4
a2=a1+d= 4 - 3 = 1
a3=a2+d= 1 - 3 = - 2
a4=a3+d= - 2 - 3 = - 5
Therefore, the series will be 4, 1, - 2 - 5 …
First four terms of this A.P. will be 4, 1, - 2 and - 5.
(iv)a= - 1,d= 1/2
Let the series bea1,a2,a3,a4…a1=a= -1
a2=a1+d= -1 + 1/2 = -1/2
a3=a2+d= -1/2 + 1/2 = 0
a4=a3+d= 0 + 1/2 = 1/2
Clearly, the series will be-1, -1/2, 0, 1/2
First four terms of this A.P. will be -1, -1/2, 0 and 1/2.
(v)a= - 1.25,d= - 0.25
Let the series bea1,a2,a3,a4…
a1=a= - 1.25
a2=a1+d= - 1.25 - 0.25 = - 1.50
a3=a2+d= - 1.50 - 0.25 = - 1.75
a4=a3+d= - 1.75 - 0.25 = - 2.00
Clearly, the series will be 1.25, - 1.50, - 1.75, - 2.00 ……..
First four terms of this A.P. will be - 1.25, - 1.50, - 1.75 and - 2.00.
3. For the following A.P.s, write the first term and the common difference.
(i) 3, 1, - 1, - 3 …
(ii) -5, - 1, 3, 7 …
(iii) 1/3, 5/3, 9/3, 13/3 ....
(iv) 0.6, 1.7, 2.8, 3.9 …
Answer
(i) 3, 1, - 1, - 3 …
Here, first term,a= 3
Common difference,d= Second term - First term
= 1 - 3 = - 2
(ii) - 5, - 1, 3, 7 …
Here, first term,a= - 5
Common difference,d= Second term - First term
= ( - 1) - ( - 5) = - 1 + 5 = 4
(iii) 1/3, 5/3, 9/3, 13/3 ....
Here, first term,a= 1/3
Common difference,d= Second term - First term
= 5/3 - 1/3 = 4/3
(iv) 0.6, 1.7, 2.8, 3.9 …
Here, first term,a= 0.6
Common difference,d= Second term - First term
= 1.7 - 0.6
= 1.1
4. Which of the following are APs? If they form an A.P. find the common differencedand write three more terms.
(i) 2, 4, 8, 16 …
(ii) 2, 5/2, 3, 7/2 ....
(iii) -1.2, -3.2, -5.2, -7.2 …
(iv) -10, - 6, - 2, 2 …
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
(vii) 0, - 4, - 8, - 12 …
(viii) -1/2, -1/2, -1/2, -1/2 ....
(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…
Answer
(i) 2, 4, 8, 16 …
Here,
a2-a1= 4 - 2 = 2
a3-a2= 8 - 4 = 4
a4-a3= 16 - 8 = 8
⇒an+1-anis not the same every time.
Therefore, the given numbers are forming an A.P.
(ii) 2, 5/2, 3, 7/2 ....
Here,
a2-a1= 5/2 - 2 = 1/2
a3-a2= 3 - 5/2 = 1/2
a4-a3= 7/2 - 3 = 1/2
⇒an+1-anis same every time.
Therefore,d= 1/2 and the given numbers are in A.P.
Three more terms are
a5= 7/2 + 1/2 = 4
a6= 4 + 1/2 = 9/2
a7= 9/2 + 1/2 = 5
(iii) -1.2, - 3.2, -5.2, -7.2 …
Here,
a2-a1= ( -3.2) - ( -1.2) = -2
a3-a2= ( -5.2) - ( -3.2) = -2
a4-a3= ( -7.2) - ( -5.2) = -2
⇒an+1-anis same every time.
Therefore,d= -2 and the given numbers are in A.P.
Three more terms are
a5= - 7.2 - 2 = - 9.2
a6= - 9.2 - 2 = - 11.2
a7= - 11.2 - 2 = - 13.2
(iv) -10, - 6, - 2, 2 …
Here,
a2-a1= (-6) - (-10) = 4
a3-a2= (-2) - (-6) = 4
a4-a3= (2) - (-2) = 4
⇒an+1-anis same every time.
Therefore,d= 4 and the given numbers are in A.P.
Three more terms are
a5= 2 + 4 = 6
a6= 6 + 4 = 10
a7= 10 + 4 = 14
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
Here,
a2-a1= 3 + √2- 3 = √2
a3-a2= (3 + 2√2) - (3 + √2) = √2
a4-a3= (3 + 3√2) - (3 + 2√2) = √2
⇒an+1-anis same every time.
Therefore,d= √2and the given numbers are in A.P.
Three more terms are
a5= (3 + √2)+ √2= 3 + 4√2
a6= (3 + 4√2) + √2= 3 + 5√2
a7= (3 + 5√2) + √2= 3 + 6√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
Here,
a2-a1= 0.22 - 0.2 = 0.02
a3-a2= 0.222 - 0.22 = 0.002
a4-a3= 0.2222 - 0.222 = 0.0002
⇒an+1-anis not the same every time.
Therefore, the given numbers are forming an A.P.
(vii) 0, -4, -8, -12 …
Here,
a2-a1= (-4) - 0 = -4
a3-a2= (-8) - (-4) = -4
a4-a3= (-12) - (-8) = -4
⇒an+1-anis same every time.
Therefore,d= -4 and the given numbers are in A.P.
Three more terms are
a5= -12 - 4 = -16
a6= -16 - 4 = -20
a7= -20 - 4 = -24
(viii) -1/2, -1/2, -1/2, -1/2 ....
Here,
a2-a1= (-1/2) - (-1/2) = 0
a3-a2= (-1/2) - (-1/2) = 0
a4-a3= (-1/2) - (-1/2) = 0
⇒an+1-anis same every time.
Therefore,d= 0 and the given numbers are in A.P.
Three more terms are
a5= (-1/2) - 0 = -1/2
a6= (-1/2) - 0 = -1/2
a7= (-1/2) - 0 = -1/2
(ix) 1, 3, 9, 27 …
Here,
a2-a1= 3 - 1 = 2
a3-a2= 9 - 3 = 6
a4-a3= 27 - 9 = 18
⇒an+1-anis not the same every time.
Therefore, the given numbers are forming an A.P.
(x)a, 2a, 3a, 4a…
Here,
a2-a1= 2a-a=a
a3-a2= 3a- 2a=a
a4-a3= 4a- 3a=a
⇒an+1-anis same every time.
Therefore,d=aand the given numbers are in A.P.
Three more terms are
a5= 4a+a= 5a
a6= 5a+a= 6a
a7= 6a+a= 7a
(xi)a,a2,a3,a4…
Here,
a2-a1=a2-a= (a- 1)
a3-a2=a3-a2=a2(a- 1)
a4-a3=a4-a3=a3(a- 1)
⇒an+1-anis not the same every time.
Therefore, the given numbers are forming an A.P.
(xii) √2, √8, √18, √32...
Here,
a2-a1= √8- √2= 2√2- √2= √2
a3-a2= √18- √8= 3√2- 2√2= √2
a4-a3= 4√2- 3√2= √2
⇒an+1-anis same every time.
Therefore,d= √2and the given numbers are in A.P.
Three more terms are
a5= √32+ √2= 4√2+ √2= 5√2= √50
a6= 5√2+√2= 6√2= √72
a7= 6√2+ √2= 7√2= √98
(xiii) √3, √6, √9, √12...
Here,
a2-a1= √6- √3= √3 × 2-√3= √3(√2- 1)
a3-a2= √9- √6= 3 - √6= √3(√3- √2)
a4-a3= √12- √9= 2√3- √3× 3 = √3(2 - √3)
⇒an+1-anis not the same every time.
Therefore, the given numbers are forming an A.P.
(xiv) 12,
32, 52, 72…
Or, 1, 9, 25, 49 …..
Here,
a2−a1= 9 − 1 = 8
a3−a2= 25 − 9 = 16
a4−a3= 49 − 25 = 24
⇒an+1-anis not the same every time.
Therefore, the given numbers are forming an A.P.
(xv) 12, 52,
72, 73 …
Or
1, 25, 49, 73 …
Here,
a2−a1= 25 − 1 = 24
a3−a2= 49 − 25 = 24
a4−a3= 73 − 49 = 24
i.e.,ak+1−akis same every time.
⇒an+1-anis same every time.
Therefore,d= 24 and the given numbers are in A.P.
Three
more terms are
a5= 73+ 24 = 97
a6= 97 + 24 = 121
a7= 121 + 24 = 145
Page No: 105
Exercise 5.2
1. Fill in the blanks in
the following table, given thatais the first term,dthe common difference andanthenthterm of the A.P.
a
d
n
an
(i)
7
3
8
…...
(ii)
− 18
…..
10
0
(iii)
…..
− 3
18
− 5
(iv)
− 18.9
2.5
…..
3.6
(v)
3.5
0
105
…..
Answer
(i)a= 7,d= 3,n= 8,an= ?
We
know that,
For
an A.P.an=a+ (n− 1)d
= 7
+ (8 − 1) 3
= 7
+ (7) 3
= 7
+ 21 = 28
Hence,an= 28
(ii) Given that
a= −18,n= 10,an= 0,d= ?
We
know that,
an=a+ (n− 1)d
0 =
− 18 + (10 − 1)d
18
= 9d
d= 18/9 = 2
Hence,
common difference,d= 2
(iii) Given that
d= −3,n= 18,an= −5
We
know that,
an=a+ (n− 1)d
−5
=a+ (18 − 1) (−3)
−5
=a+ (17) (−3)
−5
=a− 51
a= 51 − 5 = 46
Hence,a= 46
(iv)a=
−18.9,d= 2.5,an= 3.6,n= ?
We
know that,
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) = 22.5/2.5
n- 1 = 9
n= 10
Hence,n= 10
(v)a=
3.5,d= 0,n= 105,an= ?
We
know that,
an=a+ (n− 1)d
an= 3.5 + (105 − 1) 0
an= 3.5 + 104 × 0
an= 3.5
Hence,an= 3.5
Page No: 106
Choose the correct
choice in the following and justify
(i) 30thterm of the A.P: 10, 7, 4, …, is
(A)97 (B)77 (C)−77 (D.)−87
(ii) 11thterm of the A.P. -3, -1/2, ,2 .... is
(A) 28 (B) 22 (C) - 38 (D)
Answer
(i) Given that
A.P.
10, 7, 4, …
First
term,a= 10
Common
difference,d=a2−a1=
7 − 10 =
−3
We
know that,an=a+ (n−
1)d
a30= 10 + (30 − 1) (−3)
a30= 10 + (29) (−3)
a30= 10 − 87 = −77
Hence,
the correct answer is option C.
(ii) Given that A.P. is -3, -1/2, ,2 ...
First terma= - 3
Common difference,d=a2−a1= (-1/2) - (-3)
= (-1/2) + 3 = 5/2
We know that,an=a+ (n− 1)d
a11= 3 + (11 -1)(5/2)
a11= 3 + (10)(5/2)
a11= -3 + 25
a11= 22
Hence, the answer is option B.
3. In the following APs find the missing term in the boxes.
Answer
(i) For this A.P.,
a= 2
a3= 26
We know that,an=a+ (n− 1)d
a3= 2 + (3 - 1)d
26 = 2 + 2d
24 = 2d
d= 12
a2= 2 + (2 - 1) 12
= 14
Therefore, 14 is the missing term.
(ii) For this A.P.,
a2= 13 and
a4= 3
We know that,an=a+ (n− 1)d
a2=a+ (2 - 1)d
13 =a+d...(i)
a4=a+ (4 - 1)d
3 =a+ 3d...(ii)
On subtracting(i)from(ii), we get
- 10 = 2d
d= - 5
From equation(i), we get
13 =a+ (-5)
a= 18
a3= 18 + (3 - 1) (-5)
= 18 + 2 (-5) = 18 - 10 = 8
Therefore, the missing terms are 18 and 8 respectively.
(iii) For this A.P.,
a= 5 and
a4= 19/2
We know that,an=a+ (n− 1)d
a4=a+ (4 - 1)d
19/2 =5+ 3d
19/2 - 5 = 3d3d = 9/2
d = 3/2
a2=a+ (2 - 1)d
a2=5+ 3/2
a2= 13/2
a3=a+ (3 - 1)d
a3=5+ 2×3/2
a3=8
Therefore, the missing terms are 13/2 and 8 respectively.
(iv) For
this A.P.,
a= −4 and
a6= 6
We
know that,
an=a+ (n− 1)d
a6= a + (6 − 1) d
6 =
− 4 + 5d
10
= 5d
d= 2
a2=a+d= − 4 + 2 = −2
a3=a+ 2d= − 4 + 2 (2) = 0
a4=a+ 3d= − 4 + 3 (2) = 2
a5=a+ 4d= − 4 + 4 (2) = 4
Therefore,
the missing terms are −2, 0, 2, and 4 respectively.
(v)
For
this A.P.,
a2= 38
a6= −22
We
know that
an=a+ (n− 1)d
a2=a+ (2 − 1)d
38
=a+d...(i)
a6=a+ (6 − 1)d
−22
=a+ 5d...(ii)
On
subtracting equation(i)from(ii), we get
−
22 − 38 = 4d
−60
= 4dd= −15
a=a2−d= 38 − (−15) = 53
a3=a+ 2d= 53 + 2 (−15) = 23
a4=a+ 3d= 53 + 3 (−15) = 8
a5=a+ 4d= 53 + 4 (−15) = −7
Therefore,
the missing terms are 53, 23, 8, and −7 respectively.
4. Which term of the A.P. 3, 8, 13, 18, … is 78?
Answer
3, 8, 13, 18, …
For this A.P.,
a= 3
d=a2−a1= 8 − 3 = 5
Letnthterm of this A.P. be 78.
an=a+ (n− 1)d
78 = 3 + (n−
1) 5
75 = (n−
1) 5
(n− 1) =
15
n= 16
Hence, 16thterm of this A.P. is 78.
5. Find the number of terms in each of the following A.P.
(i) 7, 13, 19, …, 205
(ii) 18,, 13,...., -47
Answer
(i) For
this A.P.,
a= 7
d=a2−a1= 13 − 7 =
6
Let
there arenterms in this A.P.
an= 205
We
know that
an=a+ (n− 1)d
Therefore,
205 = 7 + (n− 1) 6
198
= (n− 1) 6
33
= (n− 1)n= 34
Therefore,
this given series has 34 terms in it.
(ii) For this A.P.,
a= 18
Let there are n terms in this A.P.
an= 205
an=a+ (n− 1)d
-47 = 18 + (n- 1) (-5/2)
-47 - 18 = (n- 1) (-5/2)
-65 = (n- 1)(-5/2)
(n- 1) = -130/-5
(n- 1) = 26
n= 27
Therefore, this given A.P. has 27 terms in it.
6. Check whether -150 is a term of the A.P. 11, 8, 5, 2, …
Answer
For this A.P.,
a= 11
d=a2−a1= 8 − 11 = −3
Let −150 be thenthterm of this A.P.
We know that,
an=a+ (n− 1)d
-150 = 11 + (n- 1)(-3)
-150 = 11 - 3n+ 3
-164 = -3n
n= 164/3
Clearly,nis not an integer.
Therefore, - 150 is not a term of this A.P.
7. Find the 31stterm of an A.P. whose 11thterm is 38 and the 16thterm is 73.
Answer
Given that,
a11=
38
a16=
73
We know that,
an=a+ (n− 1)d
a11=a+ (11 − 1)d
38 =a+
10d...(i)
Similarly,
a16=a+ (16 − 1)d
73 =a+ 15d...(ii)
On subtracting(i)from(ii), we get
35 = 5d
d= 7
From equation(i),
38 =a+ 10 ×
(7)
38 − 70 =a
a= −32
a31=a+ (31 − 1)d
=
− 32 + 30 (7)
=
− 32 + 210
=
178
Hence, 31stterm is 178.
8. An A.P. consists of 50
terms of which 3rdterm is 12 and the last term is 106.
Find the 29thterm.
Answer
Given that,
a3= 12
a50=
106
We know that,
an=a+ (n− 1)d
a3=a+ (3 − 1)d
12 =a+
2d...(i)
Similarly,a50=a+ (50 − 1)d
106 =a+
49d...(ii)
On subtracting(i)from(ii), we get
94 = 47d
d= 2
From equation(i), we get
12 =a+ 2 (2)
a= 12 − 4
= 8
a29=a+ (29 − 1)d
a29=
8 + (28)2
a29=
8 + 56 = 64
Therefore, 29thterm is 64.
9. If the 3rdand the 9thterms of an A.P. are 4 and − 8
respectively. Which term of this A.P. is zero.
Answer
Given that,
a3= 4
a9= −8
We know that,
an=a+ (n− 1)d
a3=a+ (3 − 1)d
4 =a+ 2d...(i)
a9=a+ (9 − 1)d
−8 =a+
8d...(ii)
On subtracting equation(i)from(ii), we get,
−12 = 6d
d= −2
From equation(i), we
get,
4 =a+ 2 (−2)
4 =a− 4
a= 8
Letnthterm of this A.P. be zero.
an=a+ (n− 1)d
0 = 8 + (n−
1) (−2)
0 = 8 − 2n+ 2
2n= 10
n= 5
Hence, 5thterm of this A.P. is 0.
10. If 17thterm of an A.P. exceeds its 10thterm
by 7. Find the common difference.
Answer
We know that,
For an A.P.,an=a+ (n− 1)d
a17=a+ (17 − 1)d
a17=a+ 16d
Similarly,a10=a+ 9d
It is given that
a17−a10= 7
(a+ 16d)
− (a+ 9d) = 7
7d= 7
d= 1
Therefore, the common
difference is 1.
11. Which term of the A.P.
3, 15, 27, 39, … will be 132 more than its 54thterm?
Answer
Given A.P. is 3, 15,
27, 39, …
a= 3
d=a2−a1= 15 − 3 = 12
a54=a+ (54 − 1)d
=
3 + (53) (12)
=
3 + 636 = 639
132 + 639 = 771
We have to find the
term of this A.P. which is 771.
Letnthterm be 771.
an=a+ (n− 1)d
771 = 3 + (n−
1) 12
768 = (n−
1) 12
(n− 1) =
64
n= 65
Therefore, 65thterm was 132 more than 54thterm.
Or
Letnthterm be 132 more than 54thterm.
n= 54 + 132/2
= 54 + 11 = 65thterm
12. Two APs have the same
common difference. The difference between their 100thterm
is 100, what is the difference between their 1000thterms?
Answer
Let the first term of
these A.P.s bea1anda2respectively and the common difference of these A.P.s bed.
For first A.P.,
a100=a1+ (100 − 1)d
=a1+ 99d
a1000=a1+ (1000 − 1)d
a1000=a1+ 999d
For second A.P.,
a100=a2+ (100 − 1)d
=a2+ 99d
a1000=a2+ (1000 − 1)d
=a2+ 999d
Given that, difference
between100thterm
of these A.P.s = 100
Therefore, (a1+ 99d) − (a2+ 99d) = 100
a1−a2= 100 ...(i)
Difference between
1000thterms of these A.P.s
(a1+
999d) − (a2+ 999d) =a1−a2
From equation(i),
This difference,a1−a2= 100
Hence, the difference
between 1000thterms of these A.P. will be 100.
13. How many three digit
numbers are divisible by 7?
Answer
First three-digit
number that is divisible by 7 = 105
Next number = 105 + 7 =
112
Therefore, 105, 112,
119, …
All are three digit
numbers which are divisible by 7 and thus, all these are terms of an
A.P. having first term as 105 and common difference as 7.
The maximum possible
three-digit number is 999. When we divide it by 7, the remainder will
be 5. Clearly, 999 − 5 = 994 is the maximum possible
three-digit number that is divisible by 7.
The series is as
follows.
105, 112, 119, …,
994
Let 994 be thenth
term of this A.P.
a= 105
d= 7
an= 994
n= ?
an=a+ (n− 1)d
994 = 105 + (n−
1) 7
889 = (n−
1) 7
(n− 1) =
127
n= 128
Therefore, 128
three-digit numbers are divisible by 7.
Or
Three digit numbers which are divisible by 7 are 105, 112, 119, .... 994 .
These numbers form an AP witha= 105 andd= 7.
Let number of three-digit numbers divisible by 7 ben,an= 994
⇒a+ (n- 1)d= 994
⇒ 105 + (n- 1) × 7 = 994
⇒7(n- 1) = 889
⇒n- 1 = 127
⇒n= 128
14. How many multiples of 4
lie between 10 and 250?
Answer
First multiple of 4
that is greater than 10 is 12. Next will be 16.
Therefore, 12, 16, 20,
24, …
All these are divisible
by 4 and thus, all these are terms of an A.P. with first term as 12
and common difference as 4.
When we divide 250 by
4, the remainder will be 2. Therefore, 250 − 2 = 248 is
divisible by 4.
The series is as
follows.
12, 16, 20, 24, …,
248
Let 248 be thenthterm of this A.P.
a= 12
d= 4
an= 248
an=a+ (n- 1)d
248 = 12 + (n- 1) × 4
236/4 =n- 1
59 =n- 1
n= 60
Therefore, there are 60 multiples of 4 between 10 and 250.
Or
Multiples of 4 lies between 10 and 250 are 12, 16, 20, ...., 248.
These numbers form an AP witha= 12 andd= 4.
Let number of three-digit numbers divisible by 4 ben,an= 248
⇒a+ (n- 1)d= 248
⇒ 12 + (n- 1) × 4 = 248
⇒4(n- 1) = 248
⇒n- 1 = 59
⇒n= 60
15. For what value ofn, are thenthterms of
two APs 63, 65, 67, and 3, 10, 17, … equal?
Answer
63, 65, 67, …
a= 63
d=a2−a1= 65 −
63 = 2
nthterm of this A.P. =an=a+ (n− 1)d
an= 63 + (n− 1) 2 = 63
+ 2n− 2
an= 61 + 2n...(i)
3, 10, 17, …
a= 3
d=a2−a1= 10 − 3 = 7
nthterm of this A.P. = 3 + (n− 1) 7
an= 3 + 7n− 7
an= 7n− 4 ...(ii)
It is given that,nthterm of these A.P.s are equal to each other.
Equating both these
equations, we obtain
61 + 2n= 7n− 4
61 + 4 = 5n
5n= 65
n= 13
Therefore, 13thterms of both these A.P.s are equal to each other.
16. Determine the A.P. whose third term is 16 and the 7thterm exceeds the 5thterm by 12.
Answer
a3= 16
a+ (3 − 1)d= 16
a+ 2d= 16 ...(i)
a7−a5= 12
[a+ (7 − 1)d] − [a+ (5 − 1)d]= 12
(a+ 6d) − (a+ 4d) = 12
2d= 12
d= 6
From equation(i), we get,
a+ 2 (6) = 16
a+ 12 = 16
a= 4
Therefore, A.P. will be
4, 10, 16, 22, …
Page No: 107
17. Find the 20thterm from the last term of the A.P. 3, 8, 13, …, 253.
Answer
Given A.P. is
3, 8, 13, …, 253
Common difference for
this A.P. is 5.
Therefore, this A.P.
can be written in reverse order as
253, 248, 243, …,
13, 8, 5
For this A.P.,
a= 253
d= 248 −
253 = −5
n= 20
a20=a+ (20 − 1)d
a20=
253 + (19) (−5)
a20=
253 − 95
a= 158
Therefore, 20thterm from the last term is 158.
18. The sum of 4thand 8thterms of an A.P. is
24 and the sum of the 6thand 10thterms is 44.
Find the first three terms of the A.P.
Answer
We know that,
an=a+ (n− 1)d
a4=a+ (4 − 1)d
a4=a+ 3dSimilarly,
a8=a+ 7d
a6=a+ 5d
a10=a+ 9d
Given that,a4+a8= 24
a+ 3d+a+ 7d= 24
2a+ 10d= 24
a+ 5d= 12 ...(i)
a6+a10= 44
a+ 5d+a+ 9d= 44
2a+ 14d= 44
a+ 7d= 22 ...(ii)
On subtracting equation(i)from(ii), we get,
2d= 22 − 12
2d= 10d= 5
From equation(i), we get
a+ 5d= 12
a+ 5 (5) = 12
a+ 25 = 12
a= −13
a2=a+d= − 13 + 5 = −8
a3=a2+d= − 8
+ 5 = −3
Therefore, the first three terms of this A.P. are −13, −8,
and −3.
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?
Answer
It can be observed that
the incomes that Subba Rao obtained in various years are in A.P. as
every year, his salary is increased by Rs 200.
Therefore, the salaries
of each year after 1995 are
5000, 5200, 5400, …
Here,a= 5000
d= 200
Let afternthyear, his salary be Rs 7000.
Therefore,an=a+ (n− 1)d
7000 = 5000 + (n− 1) 200
200(n− 1)
= 2000
(n− 1) =
10
n= 11
Therefore, in 11th
year, his salary will be Rs 7000.
20. Ramkali saved Rs 5 in
the first week of a year and then increased her weekly saving by Rs
1.75. If in thenthweek, her week, her weekly
savings become Rs 20.75, findn.
Answer
Given that,
a= 5
d= 1.75
an=
20.75
n= ?
an=a+ (n− 1)d
20.75 = 5 + (n- 1) × 1.75
15.75 = (n- 1) × 1.75
(n- 1) = 15.75/1.75 = 1575/175
= 63/7 = 9
n- 1 = 9
n= 10
Hence,nis 10.
Page No: 112
Exercise 5.3
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,
Sn=n/2[2a + (n- 1)d]
S10= 10/2[2(2) + (10 - 1) × 5]
= 5[4 + (9) × (5)]
= 5 × 49 = 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,
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) 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,
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) 1/15, 1/12, 1/10, ...... , to 11 terms
For this A.P.,
2. Find the sums given
below
(i) 7 ++ 14 + .................. +84
(ii)+ 14 + ………… + 84(ii) 34 + 32 + 30 +
……….. + 10
(iii) − 5 + (−
8) + (− 11) + ………… + (− 230)
Answer
(i) For this A.P.,
a= 7
l= 84
d=a2−a1=- 7 = 21/2 - 7 = 7/2
Let
84 be thenthterm of this A.P.
l=a(n- 1)d
84 = 7 + (n- 1) × 7/2
77 = (n- 1) × 7/2
22
=n− 1n= 23
We
know that,
Sn=n/2 (a+l)
Sn= 23/2 (7 + 84)
= (23×91/2) = 2093/2
=
(ii) 34 + 32 +
30 + ……….. + 10
For
this A.P.,
a= 34
d=a2−a1= 32 − 34 =
−2
l= 10
Let
10 be thenthterm of this A.P.
l=a+ (n− 1)d
10
= 34 + (n− 1) (−2)
−24
= (n− 1) (−2)
12
=n− 1
n= 13
Sn=n/2 (a+l)
= 13/2 (34 + 10)
= (13×44/2) = 13 × 22
= 286
(iii) (−5)
+ (−8) + (−11) + ………… +
(−230)
For
this A.P.,
a= −5
l= −230
d=a2−a1= (−8) −
(−5)
= −
8 + 5 = −3
Let
−230 be thenthterm of this A.P.
l=a+ (n− 1)d
−230
= − 5 + (n− 1) (−3)
−225
= (n− 1) (−3)
(n− 1) = 75
n= 76
And,
Sn=n/2 (a+l)
= 76/2[(-5) + (-230)]
= 38(-235)
= -8930
3. In an AP
(i) Givena= 5,d= 3,an= 50, findnandSn.
(ii) Givena=
7,a13= 35, finddandS13.
(iii) Givena12= 37,d= 3, findaandS12.
(iv) Givena3= 15,S10= 125, finddanda10.
(v) Givend= 5,S9= 75, findaanda9.
(vi) Givena=
2,d= 8,Sn= 90, findnandan.
(vii) Givena=
8,an= 62,Sn=
210, findnandd.
(viii) Givenan= 4,d= 2,Sn= − 14, findnanda.
(ix) Givena=
3,n= 8,S= 192, findd.
(x) Givenl= 28,S= 144 and there are total 9 terms. Finda.
Answer
(i) Given that,a= 5,d= 3,an= 50
Asan=a+ (n− 1)d,
⇒ 50 = 5 + (n- 1) × 3
⇒ 3(n- 1) = 45
⇒n- 1 = 15
⇒n= 16
Now,Sn=n/2 (a+an)
Sn= 16/2 (5 + 50) = 440
(ii) Given that,a= 7,a13= 35
Asan=a+ (n− 1)d,
⇒ 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
Asan=a+ (n− 1)d,
⇒a12=a+ (12 − 1)3
⇒ 37
=a+ 33
⇒a= 4
Sn=n/2 (a+an)
Sn=12/2 (4 + 37)
= 246
(iv) Given that,a3= 15,S10= 125
Asan=a+ (n− 1)d,
a3=a+ (3 − 1)d
15
=a+ 2d...(i)
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)
On
subtracting equation(iii)from(ii), we get
−5
= 5d
d= −1
From
equation(i),
15
=a+ 2(−1)
15
=a− 2a= 17
a10=a+ (10 − 1)d
a10= 17 + (9) (−1)
a10= 17 − 9 = 8
(v) Given that,d= 5,S9= 75
AsSn=n/2[2a+ (n- 1)d]
S9= 9/2[2a+ (9 - 1)5]
25
= 3(a+ 20)
25
= 3a+ 60
3a= 25 − 60a= -35/3
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
AsSn=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
⇒ (2n- 9)(2n+ 9) = 0
So,n= 5 (as it is positive integer)
∴a5= 8 + 5 × 4 = 34
(vii) Given that,a= 8,an= 62,Sn=
210
AsSn=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,an= 4,d= 2,Sn= −14
an=a+ (n− 1)d
4 =a+ (n− 1)2
4 =a+ 2n− 2
a+ 2n= 6
a= 6 − 2n...(i)
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
Eithern− 7 = 0 orn+ 2 = 0
n= 7 orn= −2
However,ncan neither be negative nor fractional.
Therefore,n= 7
From
equation(i), we get
a= 6 − 2n
a= 6 − 2(7)
=
6 − 14
=
−8
(ix) Given that,a= 3,n= 8,S= 192
AsSn=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.
Sn=n/2 (a+l)
144 = 9/2 (a+ 28)
(16)
× (2) =a+ 28
32
=a+ 28
a= 4
Page No: 113
4. How many terms of the AP. 9, 17, 25 … must be taken to give a
sum of 636?
Answer
Let there benterms of this A.P.
For this A.P.,a= 9
d=a2−a1= 17 − 9 = 8
AsSn=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 orn− 12 = 0
n= (-53/4) orn= 12
ncannot be (-53/4). As the number of terms can neither be negative nor fractional,
therefore,n= 12 only.
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
l= 45
Sn= 400
Sn=n/2 (a+l)
400 =n/2 (5 + 45)
400 =n/2 (50)
n= 16
l = a +(n− 1)d
45 = 5 + (16 − 1)d
40 = 15d
d= 40/15 = 8/3
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
l= 350
d= 9
Let there benterms in the A.P.
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.
7. Find the sum of first
22 terms of an AP in whichd= 7 and 22ndterm is
149.
Answer
d= 7
a22=
149
S22=
?
an=a+ (n− 1)d
a22=a+ (22 − 1)d
149 =a+ 21 ×
7
149 =a+ 147
a= 2
Sn=n/2 (a+an)
= 22/2 (2 + 149)
= 11 × 151
= 1661
8. Find the sum of first
51 terms of an AP whose second and third terms are 14 and 18
respectively.
Answer
Given that,
a2=
14
a3=
18
d=a3−a2= 18 − 14 = 4
a2=a+d
14 =a+ 4
a= 10
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
9. If the sum of first 7
terms of an AP is 49 and that of 17 terms is 289, find the sum of
firstnterms.
Answer
Given that,
S7=
49
S17= 289
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)
Similarly,
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),
a+ 3(2) = 7
a +6 = 7
a =1
Sn=n/2[2a+ (n- 1)d]
=n/2[2(1) + (n- 1) × 2]
=n/2 (2 + 2n- 2)
=n/2 (2n)
=n2
10. Show thata1,a2… ,an, … form an AP whereanis 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
It
can be observed that
a2−a1= 11 − 7 = 4
a3−a2= 15 − 11 = 4
a4−a3= 19 − 15 = 4
i.e.,ak+ 1−akis same every time. Therefore, this is an AP with common difference
as 4 and first term as 7.
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
It
can be observed that
a2−a1= − 1 − 4 = −5
a3−a2= − 6 − (−1) = −5
a4−a3= − 11 − (−6) = −5
i.e.,ak+ 1−akis same every time. Therefore, this is an A.P. with common difference
as −5 and first term as 4.
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
11. If the sum of the firstnterms of an AP is 4n−n2,
what is the first term (that isS1)? What is the
sum of first two terms? What is the second term? Similarly find the
3rd, the10thand thenthterms.
Answer
Given that,
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
d=a2−a= 1 − 3 = −2
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. 3rd, 10th,
andnthterms are −1, −15, and 5 −
2nrespectively.
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 can be observed that
these are making an A.P. whose first term is 6 and common difference
is 6.
a= 6
d= 6
S40=?
Sn=n/2[2a+ (n- 1)d]
S40= 40/2[2(6) + (40 - 1) 6]
= 20[12 + (39) (6)]
= 20(12 + 234)
= 20 × 246
= 4920
13. Find the sum of first
15 multiples of 8.
Answer
The multiples of 8 are
8, 16, 24, 32…
These are in an A.P.,
having first term as 8 and common difference as 8.
Therefore,a= 8
d= 8
S15= ?
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
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
Therefore, it can be
observed that these odd numbers are in 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
Sn=n/2 (a+l)
S25= 25/2 (1 + 49)
= 25(50)/2
=(25)(25)
= 625
15. A contract on construction job specifies a penalty for delay of completion beyond a certain dateas 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 observed 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
= 30/2[2(200) + (30 - 1) 50]
= 15[400 + 1450]
= 15 (1850)
= 27750
Therefore, the
contractor has to pay Rs 27750 as penalty.
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 1stprize beP.
Cost of 2ndprize =P− 20
And cost of 3rdprize =P− 40
It can be observed that
the cost of these prizes are in an A.P. having common difference as
−20 and first term asP.
a=P
d= −20
Given that,S7= 700
7/2[2a+ (7 - 1)d]= 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.
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 observed that
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
Sn=n/2[2a+ (n- 1)d]
S12= 12/2[2(1) + (12 - 1)(1)]
= 6 (2 + 11)
= 6 (13)
= 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.
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)
Answer
perimeter of semi-circle = πr
P1= π(0.5) = π/2 cm
P2= π(1) = π cm
P3= π(1.5) = 3π/2 cm
P1,P2,P3are the lengths of the semi-circles
π/2, π, 3π/2, 2π, ....
P1= π/2 cm
P2= π cm
d=P2-P1 =π - π/2 = π/2
First term =P1 =a=π/2 cm
Sn=n/2[2a+ (n- 1)d]
Therefor, Sum of the length of 13 consecutive circles
S13=13/2[2(π/2) + (13 - 1)π/2]
=13/2[π + 6π]
=13/2(7π)
=13/2 × 7 ×22/7
= 143 cm
Page No: 114
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 observed 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 a total of 200 logs
be placed innrows.
Sn= 200
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 orn− 25 = 0
n= 16 orn= 25
an=a+ (n− 1)d
a16=
20 + (16 − 1) (−1)
a16=
20 − 15
a16=
5
Similarly,
a25=
20 + (25 − 1) (−1)
a25=
20 − 24
= −4
Clearly, the number of
logs in 16throw is 5. However, the number of logs in 25throw is negative, which is not possible.
Therefore, 200 logs can
be placed in 16 rows and the number of logs in the 16throw is 5.
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
The distances of
potatoes from the bucket are 5, 8, 11, 14…
Distance run by the
competitor for collecting these potatoes are two times of the
distance at which the potatoes have been kept. Therefore, distances
to be run are
10, 16, 22, 28, 34,……….
a= 10
d= 16 −
10 = 6
S10=?
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.
Go To Chapters
No comments:
Post a Comment