Arithmetic Progressions
Sn= Sum of n term =2n[2a+(n−1)d]
Sequence, series & progression1.0Sequence
In our daily life, we come across the arrangement of numbers or objects in an order such as arrangement of students in a row as per their roll numbers, arrangement of books in the library, etc. An arrangement of numbers depends on the given rule
Thus, a sequence is an ordered arrangement of numbers according to a given rule.
Terms of a sequence : The individual numbers that form a sequence are the terms of a sequence.
For example : 2, 4, 6, 8, 10,.... forming a sequence are called the first, second, third, fourth, fifth,.... terms of the sequence.
The terms of a sequence in successive order is denoted by ' Tn ' or ' an '. The nth term ' Tn ' is called the general term of the sequence.
Series
The sum of terms of a sequence is called the series of the corresponding sequence. T1+T2+T3+…. is an infinite series, where as T1+T2+T3+…+Tn−1+Tn is a finite series of n terms. Usually, the series of finite number of n terms is denoted by Sn.
Sn=T1+T2+T3+…+Tn−2+Tn−1+Tn
Numerical Ability 1
Write the first five terms of the sequence, whose nth term is an={1+(−1)n}n.
Solution:
an={1+(−1)n}n
Substituting n=1,2,3,4 and 5 , we get
a1={1+(−1)1}1=0;a2={1+(−1)2}2=4;
a3={1+(−1)3}3=0;a4={1+(−1)4}4=8;
a5={1+(−1)5}5=0
Thus, the required terms are : 0,4,0,8 and 0 .
2.0Progression
It is not always possible to write each and every sequence of some rule.
For example : Sequence of prime numbers 2,3,5,7,11,… cannot be expressed explicitly by stating a rule and we do not have any expression for writing the general term of this sequence.
The sequence that follows a certain pattern is called a progression. Thus, the sequence 2,3 , 5,7,11,… is not a progression. In a progression, we can always write the nth term.
Consider the following collection of numbers :
(i) 1,3,5,7…
(ii) 21,31,41,51,…….
From the above collection of numbers, we observe that
(i) Each term is greater than the previous by 2.
(ii) In each term the numerator is 1 and the denominator is obtained by adding 1 to the preceding denominator.
Thus, we observe that the collection of numbers given in (i) and (ii) follow a certain pattern and so both are progressions.3.0Arithmetic progressions
- We know about some sequences which are arithmetic progressions as.
(i) Sequence of even numbers:
2,4,6,8,10,
(ii) Sequence of odd numbers: 1,3,5,7,9, ...
(iii) Sequence of multiples of 5: 5,10,15,20,25,
(iv) Sequence of number which are divisible by 7 and give remainders 1 in each case:
8,15,22,29,36,
An arithmetic progression is list of numbers (terms) in which the first term is given and each term, other than the first term is obtained by adding a fixed number 'd' to the preceding term.
The fixed number ' d ' is known as the common difference of the arithmetic progression. Its value can be positive, negative or zero.
The first term is denoted by 'a' or 'a1' and the last term by ' ℓ '.
e.g., Consider a sequence 6,10,14,18,22, .....
Here, a1=6,a2=10,a3=14,a4=18,a5=22
a2−a1=10−6=4
a3−a2=14−10=4
a4−a3=18−14=4
A sequence of non-zero numbers a1,a2,a3,..,an is said to be a geometric sequence or GP, if a1a2=a2a3=⋯ i.e. if anan+1=a constant for all n. e.g., 3 , 9,27,81,… SPOT LIGHT
Therefore, the sequence is an arithmetic progression in which the first term a=6 and the common difference d=4.
Symbolic form : Let us denote the first term of an AP by a1, second term by a2,…..nth term by an and the common difference by d. Then the AP becomes a1,a2,a3,…,an.
So, a2−a1=a3−a2=…=an−an−1=d.
General form : In general form, an arithmetic progression with first term 'a' and common difference ' d ' can be represented as follows: a,a+d,a+2d,a+3d,a+4d,….
Finite AP : An AP in which there are only a finite number of terms is called a finite AP. It may be noted that each such AP has a last term.
e.g.,
(a) The heights (in cm) of some students of a school standing in a queue in the morning assembly are 147,148,149,…,157.
(b) The minimum temperatures (in degree Celsius) recorded for a week in the month of January in a city arranged in ascending order are −3.1,−3.0,−2.9,−2.8,−2.7,−2.6,−2.5
Infinite AP : An AP in which the number of terms is not finite is called infinite AP. That means infinite AP does not have a last term.
The sum of cubes of first n natural numbers i.e., 13+23+33+…+n3 is usually written as Σn3.
∑n3=(2n(n+1))2=(∑n)2
e.g.,
(a) 1,2,3,4,…
(b) 100,70,40,10…
Least information required : To know about an AP, the minimum information we need to know is the first term a and the common difference d.
For instance if the first term a is 6 and the common difference d is 3 , then AP is 6,9,12,15,….
Similarly, when
a=−7,d=−2, the AP is −7,−9,−11,−13,…
a=1.0, d=0.1, the AP is 1.0,1.1,1.2,1.3,…
So, if we know what a and d are, we can list the AP.
- If a constant is added or subtracted from each term of an AP then the resulting sequence is also an AP with the same common difference.
In which of the following situations, the list of numbers obtained will be in the form of an arithmetic progression?
(i) Number of students left in the school auditorium from the total strength of 1000 students when they leave the auditorium in batches of 25 .
(ii) The amount of air present in the cylinder when a vacuum pump removes each time 1/3 of the air left in the cylinder.
(iii) Cash price of a particular brand of washing machine in the market is ₹12000. Sawitri buys one washing machine on monthly instalments of ₹1000 plus an interest at the rate of 2% per month on the balance amount. She makes the first instalment after one month. No amount is to be paid at the time of purchase.
Explanation:
(i) t1=1000,t2=1000−25=975,t3=975−25=950,t4=950−25=925 and so on.
Thus, the list of numbers is as below :
1000, 975, 950, 925, ......
Here, t2−t1=t3−t2=t4−t3=…...=−25.
Therefore, the above list of numbers forms an AP
(ii) Let us suppose that air present in the cylinder in the beginning is x units. Every time the vacuum pump removes the 31 of the air present in the cylinder.
Here, t1=x units,
t2=(x−31x) units =32x units,
t3={32x−31(32x)} units ={32x−92x} units
=96x−2x units =94x units,
t4={94x−31(94x)} units ={94x−274x} units
=278x units and so on.
In a finite AP the sum of the terms equidistant from the beginning and end is always same and is equal to the sum of first and last term.
Thus the list of numbers is as below : x,32x,94x,278x,………
Here, t2−t1=32x−x=−31x
t3−t2=94x−32x=94x−6x=−92x
t4−t3=278x−94x=278x−12x=−274x and so on.
i.e., t2−t1=t3−t2=t4−t3.
Obviously, the common differences of two consecutive terms are not same throughout. Hence, the list of numbers does not form an AP.
(iii) t1=₹1000+₹(12000×1002)=₹1000+₹240=₹1240,
t2=₹1000+₹(11000×1002)=₹1000+₹220=₹1220,
t3=₹1000+₹(10000×1002)=₹1000+₹200=₹1200,
t4=₹1000+₹(9000×1002)=₹1000+₹180=₹1180 and so on.
Thus, the list of the amounts of instalments is (in rupees) 1240, 1220, 1200, 1180, ....
t2−t1=1220−1240=−20
t3−t2=1200−1220=−20
t4−t3=1180−1200=−20 and so on.
Hence, the list form an AP.
Numerical Ability 2
Write first four terms of an AP, when the first term a and the common difference d are as follows : a=4,d=5.
Solution:
First term, a=4
Second term =4+d=4+5=9
Third term =9+d=9+5=14
Fourth term =14+d=14+5=19
Hence, first four terms of the given AP are 4, 9, 14, 19.
General term of an arithmetic progression
The formula for writing general term or the nth term of
an arithmetic progression is
an=a+(n−1)d
Where, a is the first term of arithmetic progression and d is the common difference of arithmetic progression.r^{\text {th }} term of a finite arithmetic progression from the end
Let there be an arithmetic progression with first term a and common difference d. If there are n terms in the arithmetic progression, then rth term from the end =a+(n−r)d
- If ℓ is the last term of an arithmetic progression then rth term from the end is the rth term of an arithmetic progression whose first term is ℓ and common difference is -d .
rth term from the end =ℓ+(r−1)(−d)
- Whenever we find number of terms(n) for any arithmetic progression, then n should be natural number only.
Numerical Ability 3
Find the 20th term of the sequence : 7, 3, −1,−5,….
Solution
t2−t1=3−7=−4
t3−t2=−1−3=−4
t4−t3=−5−(−1)=−4
The given sequence is an AP in which the first term a=7 and the common difference d=−4.
t20=a+(20−1)d{∵tn=a+(n+1)d}
=7+(20−1)×(−4)
=7−4×19=7−76=−69
Hence, t20=−69.
Numerical Ability 4
Find the 6th term from the end of the AP 17, 14, 11, ...., −40.
Solution:
The given AP : 17, 14, 11, ..., -40
Here, a=17,d=14−17=−3,ℓ=−40
Let there be n terms in the given AP.
Then, nth term =−40
⇒a+(n−1)d=−40(∵an=a+(n−1)d)
⇒17+(n−1)(−3)=−40
⇒(n−1)(−3)=−40−17
⇒(n−1)(−3)=−57
⇒n−1=−3−57
⇒n−1=19
⇒n=19+1
⇒n=20
Hence, there are 20 terms in the given AP. Now, 6th term from the end
=a+(20−6)d(∵rth term from the end =a+(n−r)d)
=a+14 d
=17+14(−3)
=17−42=−25
Hence, the 6th term from the end of the given AP is -25 .
Alternate solution
6th term from end =ℓ+(n−1)(−d)
=−40+(6−1)(−(−3))
=−40+5(3)
=−40+15
=−25
A sequence is an AP if the sum of its first n terms is of the form An2+Bn, where A, B are constants independent of n. In such a case, the common difference is 2A.
4.0Selection of terms in an AP
Sometimes we require certain number of terms in AP whose sum is given. The following ways of selecting terms are generally very convenient.
It should be noted that in case of an odd number of terms, the middle term is a and the common difference is d while in case of an even number of terms the middle terms are a−d,a+d and the common difference is 2 d .
- Selection of terms is useful when sum of certain terms are given. If the sum of terms is not given, then select terms as a,a+d,a+2 d……..
- If three numbers a,b,c in order are in AP. Then
b−a= Common difference =c−b
⇒b−a=c−b
⇒2b=a+c
Thus, a,b,c are in AP if and only if 2b=a+c.
- If a,b,c are in AP, then b is known as the arithmetic mean (AM) between a and c.
- If a,x,b are in AP then, 2x=a+b
⇒x=2a+b. Thus, AM between a and b is 2a+b.
Numerical Ability 5
The sum of three numbers in AP is −3, and their product is 8 . Find the numbers.
Solution:
Let the three numbers in AP be (a−d),a,(a+d). Then,
⇒ Sum =−3
⇒(a−d)+a+(a+d)=−3
⇒3a=−3
⇒a=−1
Now, product =8
⇒(a−d)(a)(a+d)=8
⇒a(a2−d2)=8
⇒(−1)(1−d2)=8(∵a=−1)
⇒d2=9
⇒d=±3
If d=3, the numbers are −4,−1,2. If d=−3, the numbers are 2,−1,−4.
Thus, the numbers are −4,−1,2 or 2,−1,−4.
Numerical Ability 6
If 2x,x+10,3x+2 are in AP, find the value of x.
Solution:
As, 2x,x+10,3x+2 are in AP,
∴2(x+10)=2x+(3x+2)
⇒2x+20=5x+2
⇒3x=18
⇒x=6
5.0Sum of first \mathbf{n} terms of an arithmetic progression
The sum ' Sn ' of n terms of an arithmetic progression with first term 'a' and common difference ' d ' is
Sn=2n[2a+(n−1)d] or Sn=2n[a+ℓ]; Where ℓ= last term.
- In the formula Sn=2n[2a+(n−1)d], there are four quantities viz. Sn,a,n and d. If any three of these are known, the fourth can be determined. Sometimes, two of these quantities are given. In such a case, remaining two quantities are provided by some other relation.
- If the sum Sn of n terms of a sequence is given, then nth term an of the sequence can be determined by using the following formula an=Sn−Sn−1 i.e., the nth term of an AP is the difference of the sum of first n terms and the sum of first ( n−1 ) terms of it.
Numerical Ability 7
Find the sum of the first 20 terms of the AP : 5, 8, 11, 14, ..... .
Solution
First term of the AP=5, i.e., a=5
Common difference =3, i.e., d=3 and n=20
S20=220(2a+19d)=10×(10+19×3)=10×67=670
Numerical Ability 8
Find the sum : 34+32+30+….+10
Solution:
As the given AP is 34+32+30+….+10
Here, a=34
d=32−34=−2ℓ=10
Let the number of terms of the AP be n.
We know that
an=a+(n−1)d⇒10=34+(n−1)(−2)⇒(n−1)(−2)=−24⇒n−1=−2−24=12⇒n=13 Again, we know that, Sn=2n(a+ℓ)⇒S13=213(34+10)⇒S13=286
Hence, the required sum is 286.
Numerical Ability 9
Find the number of terms of the AP 54, 51, 48,...so that their sum is 513.
Solution:
The given AP is 54,51,48,…
Here, a=54,d=51−54=−3
Let the sum of n terms of this AP be Sn.
We know that
Sn=2n[2a+(n−1)d]
⇒513=2n[2(54)+(n−1)(−3)]
⇒513=2n[108−3n+3]
⇒513=2n[111−3n]
⇒1026=n[111−3n]
⇒1026=111n−3n2
⇒3n2−111n+1026=0
⇒n2−37n+342=0
(Dividing throughout by 3 )
⇒n2−18n−19n+342=0
⇒n(n−18)−19(n−18)=0
⇒(n−18)(n−19)=0
⇒n−18=0 or n−19=0
⇒n=18,19
Hence, the sum of 18 terms or 19 terms of the given AP is 513.
So, n=18 or 19
Note: Here 19th term =a19
a19=a+(19−1)d
(∵an=a+(n−1)d)
a19=a+18 d
a19=54+18(−3)
a19=54−54=0
Numerical Ability 10
Rakesh has to buy a TV. He can buy TV either making cash down payment of ₹ 15000 at once or by making 14 monthly instalments as below :
₹ 1500 ( 1st month), ₹ 1450 ( 2nd month), ₹ 1400 ( 3 rd month), ₹ 1350 ( 4 th month), .....
Each instalment except the first is ₹ 50 less than the previous one.
Find:
(i) Amount of the instalment paid in the 9th month.
(ii) Total amount paid in 14 instalments.
(iii) How much extra he has to pay in addition to the amount of cash down payment?
Solution:
t1=1500,t2=1450,t3=1400,t4=1350,…...
Now, t2−t1=t3−t2=…=−50.
So, the fourteen monthly instalments form an AP.
Here, a=1500 and d=−50
(i) t9=a+8 d=1500−8×50=1500−400
Thus, 9th instalment =₹1100
(ii) S14=214(2a+13 d)=7×(2×1500+13×−50)
=7×(3000−650)=7×2350=16,450
Here, total amount paid = ₹ 16,450
(iii) Extra paid amount (i.e., in addition to cash down payment)
=₹16,450−₹15,000=₹1450