Sequences, Series and Progressions
A sequence is a finite or infinite list of numbers
following a specific pattern. For example, 1, 2, 3, 4, 5,… is the sequence, an
infinite sequence of natural numbers.
A series is
the sum of the elements in the corresponding sequence. For example,
1+2+3+4+5….is the series of natural numbers. Each number in a sequence or a
series is called a term.
A progression is
a sequence in which the general term can be can be expressed using a
mathematical formula.
Arithmetic Progression
An
arithmetic progression (AP) is a progression in which the difference between
two consecutive terms is constant.
Example: 2, 5, 8, 11, 14…. is an arithmetic progression.
Common
Difference
The difference between two
consecutive terms in an AP, (which is constant) is the “common
difference“(d) of an A.P. In the progression: 2,
5, 8, 11, 14 …the common difference is 3.
As it is the difference between any two consecutive terms, for any A.P, if the
common difference is:
positive,
the AP is increasing.
zero, the AP
is constant.
negative,
the A.P is decreasing.
Finite and Infinite AP
A finite AP is an A.P in which
the number of terms is finite. For example the A.P: 2, 5, 8……32, 35,
38
An infinite A.P
is an A.P in which the number of terms is infinite. For example: 2,
5, 8, 11…..
A finite A.P will have the last
term, whereas an infinite A.P won’t.
General Term of AP
The nth term of an AP
The
nth term of an A.P is given by Tn= a+(n−1)d, where a is
the first term, d is a common difference and n is
the number of terms.
The general
form of an AP
The
general form of an A.P is: (a,
a+d,a+2d,a+3d……) where a is
the first term and d is a common
difference. Here, d=0, OR d>0, OR d<0
Sum of Terms
in an AP
The formula
for the sum to n terms of an AP
The
sum to n terms of an A.P is given by:
Sn= n/2(2a+(n−1)d)
Where a is
the first term, d is the common difference
and n is the number of terms.
The
sum of n terms of an A.P is also given by
Sn= n/2(a+l)
Where a is
the first term, l is the last term of the
A.P. and n is the number of terms.
Arithmetic Mean (A.M)
The
Arithmetic Mean is the simple average of a given set of numbers. The arithmetic
mean of a set of numbers is given by:
A.M= Sum of
terms/Number of terms
The
arithmetic mean is defined for any set of numbers. The numbers need not
necessarily be in an A.P.
Basic Adding
Patterns in an AP
The
sum of two terms that are equidistant from either end of an AP is constant.
For example: in an A.P: 2,5,8,11,14,17…
T1+T6=2+17=19
T2+T5=5+14=19 and so on….
Algebraically, this can be represented as
Tr+T(n−r)+1=constant
Sum of first
n natural numbers
The sum of
first n natural numbers is given
by:
Sn=n(n+1)/2
This
formula is derived by treating the sequence of natural numbers as an A.P where
the first term (a) = 1 and the common difference (d) = 1.
All the formulas related to Arithmetic Progression
class 10 are tabulated below:
|
First term |
a |
|
Common difference |
d |
|
General form of AP |
a, a + d, a + 2d, a + 3d,…. |
|
nth term |
an = a + (n – 1)d |
|
Sum of first n terms |
Sn = (n/2) [2a + (n – 1)d] |
|
Sum of all terms of AP |
S = (n/2)(a + l) |
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.
Solution:
We
can write the given condition as;
Taxi
fare for 1 km = 15
Taxi
fare for first 2 kms = 15+8 = 23
Taxi
fare for first 3 kms = 23+8 = 31
Taxi
fare for first 4 kms = 31+8 = 39
And
so on……
Thus,
15, 23, 31, 39 … forms an A.P. because every next 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.
Solution:
Let
the volume of air in a cylinder, initially, be V litres.
In
each stroke, the vacuum pump removes 1/4th of air remaining in
the cylinder at a time. Or we can say, after every stroke, 1-1/4 = 3/4th part
of air will remain.
Therefore,
volumes will be V,
3V/4 , (3V/4)2 , (3V/4)3…and
so on
Clearly,
we can see here, the adjacent terms of this series do not have the common
difference between them. Therefore, this series 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.
Solution:
We
can write the given condition as;
Cost
of digging a well for first metre = Rs.150
Cost
of digging a well for first 2 metres = Rs.150+50 = Rs.200
Cost
of digging a well for first 3 metres = Rs.200+50 = Rs.250
Cost
of digging a well for first 4 metres =Rs.250+50 = Rs.300
And
so on..
Clearly,
150, 200, 250, 300 … forms an A.P. with a common difference of 50 between each
term.
(iv) The amount of money in the
account every year, when Rs 10000 is deposited at compound interest at 8% per
annum.
Solution:
We
know that if Rs. P is deposited at r% compound interest per annum for
n years, the amount of money will be:
P(1+r/100)n
Therefore,
after each year, the amount of money will be;
10000(1+8/100),
10000(1+8/100)2, 10000(1+8/100)3……
Clearly,
the terms of this series do not have the common 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 difference are given as follows:
(i) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = – 3
(iv) a = -1 d = 1/2
(v) a = – 1.25, d = – 0.25
Solutions:
(i) a = 10, d = 10
Let
us consider, the Arithmetic Progression series be a1, 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
And
so on…
Therefore,
the A.P. series will be 10, 20, 30, 40, 50 …
And
First four terms of this A.P. will be 10, 20, 30, and 40.
(ii) a = – 2, d = 0
Let
us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …
a1 = a = -2
a2 = a1+d = – 2+0 = – 2
a3 = a2+d = – 2+0 = – 2
a4 = a3+d = – 2+0 = – 2
Therefore,
the A.P. series will be – 2, – 2, – 2, – 2 …
And,
First four terms of this A.P. will be – 2, – 2, – 2 and – 2.
(iii) a = 4, d = – 3
Let
us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …
a1 = a = 4
a2 = a1+d = 4-3 = 1
a3 = a2+d = 1-3 = – 2
a4 = a3+d = -2-3 = – 5
Therefore,
the A.P. series will be 4, 1, – 2 – 5 …
And,
first four terms of this A.P. will be 4, 1, – 2 and – 5.
(iv) a = – 1, d = 1/2
Let
us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …
a2 = a1+d = -1+1/2 = -1/2
a3 = a2+d = -1/2+1/2 = 0
a4 = a3+d = 0+1/2 = 1/2
Thus,
the A.P. series will be-1, -1/2, 0, 1/2
And
First four terms of this A.P. will be -1, -1/2, 0 and 1/2.
(v) a = – 1.25, d = – 0.25
Let
us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …
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
Therefore,
the A.P series will be 1.25, – 1.50, – 1.75, – 2.00 ……..
And
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 …
Solutions
(i)
Given series,
3,
1, – 1, – 3 …
First
term, a = 3
Common
difference, d =
Second term – First term
⇒ 1 – 3 = -2
⇒ d = -2
(ii) Given series, – 5, – 1, 3, 7
…
First
term, a = -5
Common
difference, d =
Second term – First term
⇒ ( – 1)-( – 5) = – 1+5 = 4
(iii) Given series, 1/3, 5/3,
9/3, 13/3 ….
First
term, a = 1/3
Common
difference, d =
Second term – First term
⇒ 5/3 – 1/3 = 4/3
(iv) Given series, 0.6, 1.7, 2.8,
3.9 …
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 difference d and 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 …
Solution
(i)
Given to us,
2,
4, 8, 16 …
Here,
the common difference is;
a2 – a1 = 4 – 2 = 2
a3 – a2 = 8 – 4 = 4
a4 – a3 = 16 – 8 = 8
Since, an+1 – an or
the common difference is not the same every time.
Therefore,
the given series are not forming an A.P.
(ii) Given, 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
Since, an+1 – an or
the common difference is same every time.
Therefore, d =
1/2 and the given series are in A.P.
The
next three terms are;
a5 = 7/2+1/2 = 4
a6 = 4 +1/2 = 9/2
a7 = 9/2 +1/2 = 5
(iii) Given, -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
Since, an+1 – an or
common difference is same every time.
Therefore, d = -2 and
the given series are in A.P.
Hence,
next three terms are;
a5 = – 7.2-2 = -9.2
a6 = – 9.2-2 = – 11.2
a7 = – 11.2-2 = – 13.2
(iv) Given, -10, – 6, – 2, 2 …
Here, the terms and their difference are;
a2 – a1 = (-6)-(-10) = 4
a3 – a2 = (-2)-(-6) = 4
a4 – a3 = (2 -(-2) = 4
Since, an+1 – an or
the common difference is same every time.
Therefore, d = 4 and
the given numbers are in A.P.
Hence,
next three terms are;
a5 = 2+4 = 6
a6 = 6+4 = 10
a7 = 10+4 = 14
(v) Given, 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
Since, an+1 – an or
the common difference is same every time.
Therefore, d = √2 and
the given series forms a A.P.
Hence,
next three 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
Since, an+1 – an or
the common difference is not same every time.
Therefore,
and the given series doesn’t forms a A.P.
(vii) 0, -4, -8, -12 …
Here,
a2 – a1 = (-4)-0 = -4
a3 – a2 = (-8)-(-4) =
-4
a4 – a3 = (-12)-(-8) =
-4
Since, an+1 – an or
the common difference is same every time.
Therefore, d = -4 and
the given series forms a A.P.
Hence,
next three 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
Since, an+1 – an or
the common difference is same every time.
Therefore, d = 0 and
the given series forms a A.P.
Hence,
next three 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
Since, an+1 – an or
the common difference is not same every time.
Therefore,
and the given series doesn’t form a A.P.
(x) a, 2a, 3a, 4a …
Here,
a2 – a1 = 2a–a = a
a3 – a2 = 3a-2a = a
a4 – a3 = 4a-3a = a
Since, an+1 – an or
the common difference is same every time.
Therefore, d = a and the
given series forms a A.P.
Hence,
next three 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(a-1)
a3 – a2 = a3 – a2 = a2(a-1)
a4 – a3 = a4 – a3 = a3(a-1)
Since, an+1 – an or
the common difference is not same every time.
Therefore,
the given series doesn’t forms a 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
Since, an+1 – an or
the common difference is same every time.
Therefore, d = √2 and
the given series forms a A.P.
Hence,
next three 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)
Since, an+1 – an or
the common difference is not same every time.
Therefore,
the given series doesn’t form a 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
Since, an+1 – an or
the common difference is not same every time.
Therefore,
the given series doesn’t form a 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
Since, an+1 – an or
the common difference is same every time.
Therefore, d =
24 and the given series forms a A.P.
Hence,
next three terms are;
a5 = 73+24 = 97
a6 = 97+24 = 121
a7 = 121+24 = 145
Exercise 5.2
Page: 105
1. Fill in the blanks in the
following table, given that a is the first term, d the common difference and an the nth term of the A.P.
Solutions:
(i) Given,
First
term, a = 7
Common
difference, d =
3
Number
of terms, n =
8,
We
have to find the nth term, an =
?
As
we know, for an A.P.,
an = a+(n−1)d
Putting
the values,
=>
7+(8 −1) 3
=>
7+(7) 3
=>
7+21 = 28
Hence, an = 28
(ii) Given,
First
term, a = -18
Common
difference, d =
?
Number
of terms, n =
10
Nth
term, an = 0
As
we know, for an A.P.,
an = a+(n−1)d
Putting
the values,
0
= − 18 +(10−1)d
18
= 9d
d =
18/9 = 2
Hence,
common difference, d = 2
(iii) Given,
First
term, a = ?
Common
difference, d =
-3
Number
of terms, n =
18
Nth
term, an = -5
As
we know, for an A.P.,
an = a+(n−1)d
Putting
the values,
−5
= a+(18−1) (−3)
−5
= a+(17) (−3)
−5
= a−51
a =
51−5 = 46
Hence, a = 46
(iv) Given,
First
term, a = -18.9
Common
difference, d =
2.5
Number
of terms, n =
?
Nth
term, an = 3.6
As
we know, for an A.P.,
an = a +(n −1)d
Putting
the values,
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) Given,
First
term, a = 3.5
Common
difference, d =
0
Number
of terms, n =
105
Nth
term, an = ?
As
we know, for an A.P.,
an = a+(n −1)d
Putting
the values,
an = 3.5+(105−1) 0
an = 3.5+104×0
an = 3.5
Hence, an = 3.5
2. Choose the correct choice in
the following and justify:
(i) 30th term
of the A.P: 10,7, 4, …, is
(A) 97 (B) 77 (C) −77 (D) −87
(ii)
11th term of the A.P. -3, -1/2,
,2 …. is
(A) 28 (B) 22 (C) – 38 (D)
-48(1/2)
Solutions:
(i)
Given here,
A.P.
= 10, 7, 4, …
Therefore,
we can find,
First
term, a = 10
Common
difference, d = a2 − a1 = 7−10 = −3
As
we know, for an A.P.,
an = a +(n−1)d
Putting
the values;
a30 = 10+(30−1)(−3)
a30 = 10+(29)(−3)
a30 = 10−87 = −77
Hence,
the correct answer is option C.
(ii)
Given here,
A.P.
= -3, -1/2, ,2 …
Therefore,
we can find,
First
term a = – 3
Common
difference, d = a2 − a1 = (-1/2) -(-3)
⇒(-1/2) + 3 = 5/2
As
we know, for an A.P.,
an = a+(n−1)d
Putting
the values;
a11 = -3+(11-1)(5/2)
a11 = -3+(10)(5/2)
a11 = -3+25
a11 = 22
Hence, the answer is option B.
Solutions:
(i) For the given A.P., 2,2 , 26
The first and third term are;
a = 2
a3 =
26
As we know, for an A.P.,
an = a+(n −1)d
Therefore, putting the values here,
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 the given A.P., , 13, ,3
a2 =
13 and
a4 =
3
As we know, for an A.P.,
an = a+(n−1) d
Therefore, putting the values here,
a2 = a +(2-1)d
13 = a+d ………………. (i)
a4 = a+(4-1)d
3 = a+3d ………….. (ii)
On subtracting
equation (i) from (ii), we get,
– 10 = 2d
d = – 5
From equation (i), putting the value of
d,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 the given A.P.,
a = 5 and
a4 =
19/2
As we know, for an A.P.,
an = a+(n−1)d
Therefore, putting the values here,
a4 = a+(4-1)d
19/2 = 5+3d
(19/2) – 5 = 3d
3d = 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 the given A.P.,
a = −4 and
a6 =
6
As we know, for an A.P.,
an = a +(n−1) d
Therefore, putting the values here,
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 the given A.P.,
a2 =
38
a6 =
−22
As we know, for an A.P.,
an = a+(n −1)d
Therefore, putting the values here,
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 = 4d
d = −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?
Solutions:
Given the A.P. series as3, 8, 13, 18, …
First term, a = 3
Common difference, d = a2 − a1 =
8 − 3 = 5
Let the nth term
of given A.P. be 78. Now as we know,
an = a+(n−1)d
Therefore,
78 = 3+(n −1)5
75 = (n−1)5
(n−1) = 15
n = 16
Hence, 16th term 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, 15(1/2),13…-47
Solutions:
(i) Given, 7, 13, 19, …, 205 is the A.P
Therefore
First term, a =
7
Common difference, d = a2 − a1 =
13 − 7 = 6
Let there are n terms in
this A.P.
an =
205
As we know, for an A.P.,
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)
18, 15(1/2),13…-47
First term, a = 18
Common difference, d = a2-a1 =
15(1/2-18)
d = (31-36)/2 = -5/2
Let there are n terms in this A.P.
an = -47
As we know, for an A.P.,
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, …
Solution:
For the given series, A.P. 11, 8, 5, 2..
First term, a =
11
Common difference, d = a2−a1 =
8−11 = −3
Let −150 be the nth term
of this A.P.
As we know, for an A.P.,
an = a+(n−1)d
-150 = 11+(n -1)(-3)
-150 = 11-3n +3
-164 = -3n
n = 164/3
Clearly, n is not an integer
but a fraction.
Therefore, – 150 is not a term of this A.P.
7.
Find the 31st term of an A.P. whose 11th term
is 38 and the 16th term is 73.
Solution:
Given that,
11th term, a11 =
38
and 16th term, a16 =
73
We know that,
an = a+(n−1)d
a11 = a+(11−1)d
38 = a+10d ………………………………. (i)
In the same way,
a16 = a +(16−1)d
73 = a+15d ………………………………………… (ii)
On subtracting
equation (i) from (ii), we get
35 = 5d
d = 7
From equation (i), we can write,
38 = a+10×(7)
38 − 70 = a
a = −32
a31 = a +(31−1) d
= − 32 + 30 (7)
= − 32 + 210
= 178
Hence, 31st term is 178.
8.
An A.P. consists of 50 terms of which 3rd term is 12 and the
last term is 106. Find the 29th term.
Solution: Given that,
3rd term, a3 =
12
50th term, a50 =
106
We know that,
an = a+(n−1)d
a3 = a+(3−1)d
12 = a+2d ……………………………. (i)
In the same way,
a50 = a+(50−1)d
106 = a+49d …………………………. (ii)
On subtracting
equation (i) from (ii), we get
94 = 47d
d = 2 = common difference
From equation (i), we can write now,
12 = a+2(2)
a = 12−4 = 8
a29 = a+(29−1) d
a29 =
8+(28)2
a29 =
8+56 = 64
Therefore, 29th term is 64.
9.
If the 3rd and the 9th terms of an A.P. are 4
and − 8 respectively. Which term of this A.P. is zero.
Solution:
Given that,
3rd term, a3 =
4
and 9th term, a9 =
−8
We know that,
an = a+(n−1)d
Therefore,
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 will get here,
−12 = 6d
d = −2
From equation (i), we can write,
4 = a+2(−2)
4 = a−4
a = 8
Let nth term 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, 5th term of this A.P.
is 0.
10.
If 17th term of an A.P. exceeds its 10th term
by 7. Find the common difference.
Solution:
We know that, for an A.P series;
an = a+(n−1)d
a17 = a+(17−1)d
a17 = a +16d
In the same way,
a10 = a+9d
As it is given in the question,
a17 − a10 =
7
Therefore,
(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 54th term?
Solution:
Given A.P. is 3, 15, 27, 39, …
first term, a = 3
common difference, d = a2 − a1 =
15 − 3 = 12
We know that,
an = a+(n−1)d
Therefore,
a54 = a+(54−1)d
⇒3+(53)(12)
⇒3+636
= 639
a54 =
639+132=771
We have to find the term of this A.P. which
is 132 more than a54, i.e.771.
Let nth term be
771.
an = a+(n−1)d
771 = 3+(n −1)12
768 = (n−1)12
(n −1) = 64
n = 65
Therefore, 65th term was 132
more than 54th term.
Or another method is;
Let nth term be 132 more than 54th term.
n = 54 + 132/2
= 54 + 11 = 65th term
12.
Two APs have the same common difference. The difference between their 100th term
is 100, what is the difference between their 1000th terms?
Solution:
Let, the first term of two APs be a1 and a2 respectively
And the common difference of these APs
be d.
For the first A.P.,we know,
an = a+(n−1)d
Therefore,
a100 = a1+(100−1)d
= a1 + 99d
a1000 = a1+(1000−1)d
a1000 = a1+999d
For second A.P., we know,
an = a+(n−1)d
Therefore,
a100 = a2+(100−1)d
= a2+99d
a1000 = a2+(1000−1)d
= a2+999d
Given that, difference between 100th term
of the two APs = 100
Therefore, (a1+99d)
− (a2+99d) = 100
a1−a2 =
100……………………………………………………………….. (i)
Difference between 1000th terms
of the two APs
(a1+999d) − (a2+999d)
= a1−a2
From equation (i),
This difference, a1−a2 =
100
Hence, the difference between 1000th terms
of the two A.P. will be 100.
13. How many three digit numbers are
divisible by 7?
Solution:
First three-digit number that is divisible by
7 are;
First number = 105
Second number = 105+7 = 112
Third number = 112+7 =119
Therefore, 105, 112, 119, …
All are three digit numbers are divisible by
7 and thus, all these are terms of an A.P. having first term as 105 and common
difference as 7.
As we know, the largest possible three-digit
number is 999.
When we divide 999 by 7, the remainder will
be 5.
Therefore, 999-5 = 994 is the maximum
possible three-digit number that is divisible by 7.
Now the series is as follows.
105, 112, 119, …, 994
Let 994 be the nth term of this A.P.
first term, a = 105
common difference, d = 7
an = 994
n = ?
As we know,
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.
14. How many multiples of 4 lie between 10
and 250?
Solution:
The first multiple of 4 that is greater than
10 is 12.
Next multiple will be 16.
Therefore, the series formed as;
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, now;
12, 16, 20, 24, …, 248
Let 248 be the nth term
of this A.P.
first term, a =
12
common difference, d =
4
an = 248
As we know,
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.
15.
For what value of n, are the nth terms
of two APs 63, 65, 67, and 3, 10, 17, … equal?
Solution:
Given two APs as; 63, 65, 67,… and 3, 10,
17,….
Taking first AP,
63, 65, 67, …
First term, a = 63
Common difference, d = a2−a1 =
65−63 = 2
We know, nth term of this
A.P. = an = a+(n−1)d
an=
63+(n−1)2 = 63+2n−2
an =
61+2n ………………………………………. (i)
Taking second AP,
3, 10, 17, …
First term, a = 3
Common difference, d = a2 − a1 =
10 − 3 = 7
We know that,
nth term
of this A.P. = 3+(n−1)7
an =
3+7n−7
an =
7n−4 ……………………………………………………….. (ii)
Given, nth term
of these A.P.s are equal to each other.
Equating both these equations, we get,
61+2n = 7n−4
61+4 = 5n
5n = 65
n = 13
Therefore, 13th terms of both
these A.P.s are equal to each other.
16.
Determine the A.P. whose third term is 16 and the 7th term
exceeds the 5th term by 12.
Solutions:
Given,
Third term, a3 =
16
As we know,
a +(3−1)d = 16
a+2d = 16
………………………………………. (i)
It is given that, 7th term
exceeds the 5th term by 12.
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 be4, 10, 16, 22, …
17.
Find the 20th term from the last term of the A.P. 3, 8, 13, …,
253.
Solution:
Given A.P. is3, 8, 13, …, 253
Common difference, d= 5.
Therefore, we can write the given AP in
reverse order as;
253, 248, 243, …, 13, 8, 5
Now for the new AP,
first term, a = 253
and common difference, d = 248 − 253 =
−5
n = 20
Therefore, using nth term formula, we get,
a20 = a+(20−1)d
a20 =
253+(19)(−5)
a20 =
253−95
a = 158
Therefore, 20th term from the
last term of the AP 3, 8, 13, …, 253.is 158.
18.
The sum of 4th and 8th terms of an A.P. is 24
and the sum of the 6th and 10th terms is 44.
Find the first three terms of the A.P.
Solution:
We know that, the nth term of the AP is;
an = a+(n−1)d
a4 = a+(4−1)d
a4 = a+3d
In the same way, we can write,
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 = 10
d = 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?
Solution:
It can be seen from the given question, that
the incomes of Subba Rao increases every year by Rs.200 and hence, forms an AP.
Therefore, after 1995, the salaries of each
year are;
5000, 5200, 5400, …
Here, first term, a =
5000
and common difference, d =
200
Let after nth year,
his salary be Rs 7000.
Therefore, by the nth term
formula of AP,
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 the nth week, her weekly
savings become Rs 20.75, find n.
Solution:
Given that, Ramkali saved Rs.5 in first week
and then started saving each week by Rs.1.75.
Hence,
First term, a = 5
and common difference, d = 1.75
Also given,
an =
20.75
Find, n = ?
As we know, by the nth term
formula,
an = a+(n−1)d
Therefore,
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, n is 10.
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
Solutions:
(i) Given, 2, 7, 12 ,…, to 10 terms
For this A.P.,
first term, a = 2
And common difference, d = a2 − a1 =
7−2 = 5
n = 10
We know that, the formula for 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
For this A.P.,
first term, a = −37
And common difference, d = a2− a1
d= (−33)−(−37)
= − 33 + 37 = 4
n = 12
We know that, the formula for 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
For this A.P.,
first term, a = 0.6
Common difference, d = a2 − a1 =
1.7 − 0.6 = 1.1
n = 100
We know that, the formula for 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.,
First term, a = 1/5
Common difference, d = a2 –a1 =
(1/12)-(1/5) = 1/60
And number of terms n = 11
We know that, the formula for sum of nth term
in AP series is,
Sn = n/2 [2a + (n –
1) d]
sn=(11/2){[2(1/5)]+[((11-1)1)/60]}
= 11/2(2/15
+ 10/60)
= 11/2
(9/30)
= 33/20
2.
Find the sums given below:
(i) 7+10(1/2)+14+…..+84
(ii) 34 + 32 +
30 + ……….. + 10
(iii) − 5 + (− 8) + (− 11) + …………
+ (− 230)
Solutions:
(i)
For this
given ap., (i) 7+10(1/2)+14+…..+84,
First term,
a = 7
nth term,
an = 84
common difference , d= a2-a1=10(1/2)-7=(21/2)-7=7/2
Let 84 be
the nth term
of this A.P., then as per the nth term formula,
an = a(n-1)d
84 =
7+(n – 1)×7/2
77 =
(n-1)×7/2
22
= n−1
n = 23
We know
that, sum of n term is;
Sn = n/2
(a + l) , l = 84
Sn =
23/2 (7+84)
Sn = (23×91/2) = 2093/2
Sn =1046(1/2)
(ii) Given, 34 + 32 + 30 + ……….. + 10
For this A.P.,
first term, a = 34
common difference, d = a2−a1 =
32−34 = −2
nth term, an= 10
Let 10 be the nth term
of this A.P., therefore,
an= a +(n−1)d
10 = 34+(n−1)(−2)
−24 = (n −1)(−2)
12 = n −1
n = 13
We know that, 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)
For this A.P.,
First term, a = −5
nth term, an= −230
Common difference, d = a2−a1 =
(−8)−(−5)
⇒d =
− 8+5 = −3
Let −230 be the nth term
of this A.P., and by the nth term formula we know,
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
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.
Solutions:
(i) Given that, a =
5, d = 3, an = 50
As we know, from the formula of the nth
term in an AP,
an = a +(n −1)d,
Therefore, putting the given values, 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) Given that, a =
7, a13 = 35
As we know, from the formula of the nth
term in an AP,
an = a+(n−1)d,
Therefore, putting the given values, 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
As we know, from the formula of the nth term
in an AP,
an = a+(n −1)d,
Therefore, putting the given values, we get,
⇒ 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
As we know, from the formula of the nth
term in an AP,
an = a +(n−1)d,
Therefore, putting the given values, we get,
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 will 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
As we know, the nth term can
be written as;
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 (as n only be a
positive integer)
∴ 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, common difference, d = 2, sum of n terms, Sn =
−14.
As we know, from the formula of the nth term
in an AP,
an = a+(n −1)d,
Therefore, putting the given values, we get,
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 nor fractional.
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,
Number of terms, n = 8
And sum of n terms, S =
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 formula,
Sn = n/2
(a + l)
144 = 9/2(a+28)
(16)×(2) = a+28
32 = a+28
a = 4
4.
How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
Solutions:
Let there be n terms of the
AP. 9, 17, 25 …
For this 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 or
fraction, 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.
Solution:
Given that,
first term, a =
5
last term, l =
45
Sum of the AP, Sn =
400
As we know, the sum of AP formula is;
Sn = n/2
(a+l)
400 = n/2(5+45)
400 = n/2(50)
Number of terms, n =16
As we know, the last term of AP series can be
written as;
l = a+(n −1)d
45 = 5 +(16 −1)d
40 = 15d
Common difference, 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?
Solution:
Given that,
First term, a =
17
Last term, l =
350
Common difference, d =
9
Let there be n terms in the
A.P., thus the formula for last term can be written as;
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 =
38/2 (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 which d = 7 and 22nd term
is 149.
Solution:
Given,
Common difference, d = 7
22nd term, a22 =
149
Sum of first 22 term, S22 = ?
By the formula of 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
8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and
18 respectively.
Solution:
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 = First term
Sum of n terms;
Sn = n/2 [2a +
(n – 1)d]
S51 =
51/2 [2×10 (51-1) 4]
= 51/2 [20+(50)×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 first n terms.
Solution:
Given that,
S7 =
49
S17 = 289
We know, Sum of n terms;
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 +6d]
7 = (a+3d)
a + 3d = 7
…………………………………. (i)
In the same way,
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
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.
Solutions:
(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
We can see here, the common difference
between the terms are;
a2 − a1 =
11−7 = 4
a3 − a2 =
15−11 = 4
a4 − a3 =
19−15 = 4
Hence, ak +
1 − ak is the same value every time.
Therefore, this is an AP with common difference as 4 and first term as 7.
Now, we know, 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
We can see here, the common difference
between the terms are;
a2 − a1 =
−1−4 = −5
a3 − a2 =
−6−(−1) = −5
a4 − a3 =
−11−(−6) = −5
Hence, ak +
1 − ak is same every time. Therefore,
this is an A.P. with common difference as −5 and first term as 4.
Now, we know, 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
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.
Solution:
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
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.
12.
Find the sum of first 40 positive integers divisible by 6.
Solution:
The positive integers that are divisible by 6
are 6, 12, 18, 24 ….
We can see here, that this series forms an
A.P. whose first term is 6 and common difference is 6.
a = 6
d = 6
S40 = ?
By the formula of sum of n terms, we know,
Sn = n/2 [2a +(n –
1)d]
Therefore, putting n = 40, we get,
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.
Solution:
The multiples of 8 are 8, 16, 24, 32…
The series is in the form of AP, having first
term as 8 and common difference as 8.
Therefore, a = 8
d = 8
S15 =
?
By the formula of sum of nth term, we know,
Sn = n/2 [2a+(n-1)d]
S15 =
15/2 [2(8) + (15-1)8]
= 15/2[16 +(14)(8)]
= 15/2[16 +112]
= 15(128)/2
= 15 × 64
= 960
14.
Find the sum of the odd numbers between 0 and 50.
Solution:
The odd numbers between 0 and 50 are 1, 3, 5,
7, 9 … 49.
Therefore, we can see that these odd numbers
are in the form of A.P.
Hence,
First term, a = 1
Common difference, d = 2
Last term, l = 49
By the formula of 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
By the formula of sum of nth term, we know,
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 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:
We can see, that the given penalties are in
the 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
By the formula of sum of nth term, we know,
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.
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.
Solution:
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., having common difference as −20 and first term as 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
{[2a+6*(-20)]}/2=100
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?
Solution:
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)
Solution:
We know,
Perimeter of
a semi-circle = πr
Therefore,
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.
Hence we got
a series here, as,
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
By the sum
of n term formula, 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
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?
Solution:
We can see
that the numbers of logs in rows are in the form of an A.P.20, 19, 18…
For the
given A.P.,
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 as the numbers cannot be negative.
Therefore,
200 logs can be placed in 16 rows and the number of logs in the 16th row 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)]
Solution:
The
distances of potatoes from the bucket are 5, 8, 11, 14…, which is in the form
of AP.
Given, the
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 w.r.t distances of potatoes, could be written as;
10, 16, 22,
28, 34,……….
Hence, the first term, a = 10 and d = 16−10 = 6
S10 =?
By the
formula of sum of n terms, we know,
S10 = 10/2 [2(10)+(10 -1)(6)]
= 5[20+54]
= 5(74)
= 370
