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As per the latest CBSE Class 10 Mathematics Syllabus, Chapter 5 Arithmetic Progression is a part of Unit 2 Algebra. Unit 2 holds a weightage of 20 marks in CBSE Class 10 Maths Board Exams.
What is Arithmetic Progression?
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Arithmetic progression (AP) is an arithmetic sequence, a sequence of series or numbers with the common difference between two consecutive numbers in a sequence. A progression is arranged in an exceedingly particular order such that the relation between two consecutive terms of series or sequence is usually constant.
Progression comes across in our regular life just like the roll number of scholars in a class, days in a week or months in a very year. A progression may be a form of sequence from which it’s possible to get a formula for the nth term. Progression or AP could be a sequence where each new term after the primary is obtained by adding a continuing difference, called a common difference.
In simple language a progression could be a collection of integers within each term that is the result of adding a constant to the preceding term aside from the primary term.
E.g- 2,4,6,8,10,12
5,10,15,20,25,30,35,40
The video below explains this:
Arithmetic Progression Detailed Video Explanation:
Arithmetic Progression: Explanation
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The three styles of progressions are listed below:
- Arithmetic progression (AP)
- Geometric progression (GP)
- Harmonic progression (HP)
Arithmetic progression
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Arithmetic progression could be a pattern of sequence of a series which has a typical relationship between two consecutive terms. It’s a sequence of numbers, which each term of the sequence diverge from the subsequent term by a continuing amount.
Properties of Arithmetic Progression
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- If a relentless it’s added to every number of AP the resulting sequence is additionally an AP
- If each term of an AP is x a continuing the resulting sequence is additionally an AP.
- If each term of an AP is divided by a non-zero constant, then the resulting sequence is additionally an AP.
- If a constant is subtracted from every term of an AP the resulting sequence is additionally an AP.
Notation in Arithmetic Progression (A.P)
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a= First Term
l= Last Term
d= Common Difference
n= Total Number Of Terms
an= nth Term (where n=1,2,3,…,n)
Sn=Sum Of First n Terms
The General Term of an Arithmetic Progression
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Any sequence a1 a2 a3….and is termed a progression if an + 1=an+d, n€N
Let’s assume the primary term of an AP could be a common difference is d
i.e. a, a+d, a+2d, a+3d……
1st term = a = a+(1-1) d
2nd term= a+d = a+(2-1) d
3rd term = a+2d = a+(3-1) d.
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Nth term= an a+(n-1) d
Sum of n Terms of an Arithmetic Progression
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The sum of n terms of an AP is the addition of first n terms off the automated sequence. In simple words the sum is up to n divided by double of the sum of twice the primary term “a '' and also the product of the difference between 2nd and therefore the first common difference
| Sum of n terms in AP | n/2[2a+(n-1)d] |
| Sum of square of n natural numbers | [n(n+1)(2n+1)]/6 |
| Sum of Cube of a natural numbers | [n(n+1)/2]2 |
| Sum of natural numbers | n(n+1)/2 |
Example:- Find the addition of the first 22 terms of the AP: 8, 4, -2,......
Ans: a=8, d=4-8=-4, n=22
We know that, S= n/2[2a+(n-1)d]
Therefore, S= 22/2[16+21(-4)]
= 11(16-84)
= 11(-68)
= -748.
Arithmetic Mean
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If a,b,c are in A.P then b= (a+c)/2 and b is called the Arithmetic Mean.
Important Points
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- The general form of an AP is a, a+d, a+2d,…..
- If a, b, c are in AP, then b-a=c-b or 2b=a+c and b is called the arithmetic mean of a and c
- If 3 terms are in AP, then assume a-d, a, a+d
- If 4 terms are in AP, then assume a-3d, a-d, a+d, a+3d.
Sample Questions
Question: For what a value of K will K+9, 2k-1 and 2k+7 be consecutive terms of an A.P.? (2 marks)
Ans= (2k-1) - (k+9) = (2k+7) - (2k-1)
= k-10=8
= k=8
So, the value of K will be 8.
Question: Find the “d” of an A.P. in which a21-a7=84? (2017) (2 marks)
Ans= a+(21-1)d - [a+(7-1)d] =84 [formula: tn=a+(n-1)d]
= (a+20d) - (a+6d) =84
= 14d=84
= d= 84/14 = 6
So, the common difference of an A.P. is 6.
Question: Find the common difference of an A.P. whose first term is 4, the last term is 49 and the addition of all its terms is 265. (3 marks)
Ans:
a=4, l=49, S=265
We know that, Sn=n/2(a+l)
Therefore, 265=n/2(4+49)
265=n/2(53)
n/2=265/53
n/2=5
n=10
So, the total number of terms is 10.
We know that, l=a+(n-1)d
Therefore, 49=4+(10-1)d
45=9d
d=45/9
d= 5
So, the common difference is 5.
Question: The Addition of four consecutive numbers in an AP is 32 and the fraction of the multiple of the first and the last term to the multiple of two middle terms is 7:15. Find the numbers. (4 marks)
Ans: Let's suppose the four consecutive numbers are a-3d, a-d, a+d, a+3d
=The sum of four consecutive number is=32
=4a=32
=a=8
Furthermore, (a-3d)(a+2d)/(a-d)(a+d)= a2-9d2/a2-d2=7/15
15a2-135d2=7a2-7d2
8a2=128d2
d2=8/2
d2=4
d= ±2
Now let's find out four consecutive numbers ,
a-3d=8-3(2)=2
a-d=8-2=6
a+d=8+2=10
a+3d=8+3(2)=14
So, the four consecutive numbers are 2,6,10,and14.
Question:: Find the addition of all factors of 7 lying between 500 and 900. (2 marks)
Ans= Our first term will be 504 and the last term will be 896.
We know that, l=a+(n-1)d
Therefore, 896=504+(n-1)7
392=(n-1)7
n-1=392/7=56
n=57
We know that, Sn=n/2(a+l)
Therefore, Sn=57/2(504+896)
=57/2×1400
=39900
Question: If the ratio of the addition of the first n terms of two A.P’s is (7n+1):(4n+27), then find the fraction of their 9th terms. (3 marks)
Ans: Let's assume first term of (7n+1)= a1 and common difference= d1
Let's assume first term of (4n+27)= a2 and common difference=d2
So, our ratio is 24:19.
Question: The sum of the first n terms of an A.P is given by Sn=3n2+2n. Determine the A.P and its 15th term. (4 marks)
Ans: we know that, Sn= 3n2+2n
First, let's find out S1, S2 and a2,
S1=a1=3(1)2+2(1)
=5
S2=3(2)2+2(2)
=12+4
=16
a2=S2-S1=16-5=11
So, d=a2-a1=11-5= 6
Now we will find out AP,
AP= a, a+d, a+2d,…….= 5, 11, 17,…..
And now we will see 15th term,
a15=a+14d
=5+14(6)
=5+84=89
So, the common difference is 6 and its 15th term is 89.
Question:: An operator of TV sets produced 600 sets in the 3rd year and 700 sets in the 7th year. Assuming that the production increases constantly by a fixed number every year, find the production in the 1st year, 10th year and the total production in the first 7 years. (5 marks)
Ans: Since the production increases constantly by a fixed number every year, the no. of TV sets operated in 1st,2nd,3rd,….., years will form an AP.
Let us denote the number of TV sets operator in the nth year by an.
Then, a3= 600 and a7= 700
or, a+2d=600
And a+6d=700
Solving these equations, we get d=25 and a=550.
Therefore production of TV sets in the 1st year is 550.
Now, a10=a+9d
=550+9×25
=775
So, production of TV sets in the year is 775.
Also, s7=7/2[2×550+(7-1)×25]
=7/2(1100+150)
=4375
Thus, the total production of TV sets in the first 7 years is 4375.
Question: A man manages to pay off a loan of Rs 3600 by annual installments which form an arithmetic series. When 30 of the installments are paid, he dies leaving one-third of the loan unpaid. Find the value of the first installment. (5 marks)
Ans: A man manages to pay off a loan of Rs 3600 by 40 annual installments which form an A.P i.e., sum of all 40 installments= Rs 3600
S40= 3600
We know that, Sn= n/2 [2a+(n-1)d]
40/x[2a+(40-1)d]=3600
2a+39d=3600/20=180….. (i)
He died by leaving one third of the loan unpaid. That means he paid the remaining money in 30 installments.
Therefore, the money he paid in 30 installments= 3600-3600/3=3600-1200
So, S30=2400
S40=2400=30/2[2a+(30-1)d]=2400
Therefore, Sn=n/2[2a+(n-1)d]
2a+2ad=2400/15=160…. (ii)
(i) – (ii) = 2a+39d=180
2a+29d=160/0+10d=20
d=20/10=2
Put d= 2 in (ii) 2a+29(2)=160
2a=102
A=102/2=51
Therefore, the value of his first installment = 51.
Question: A man secured Rs 32 during the 1st year, Rs 36 in the 2nd year and in this way he raised his savings by Rs 4 every year. Find in what time his secured money will be Rs 200. (4 marks)
Ans: Saving in 1st year (a1)=Rs 32
Saving in 2nd year (a2)=Rs 36
Increase in salary every year (d)=Rs 4
Let in n years his saving will be Rs 200
= Sn=200
= n/2[2a+(n-1)d=200
= n/2(64+4n-4)=200
N/2(4n+60)=200
=2n2+30n=200
=n2+15-100=0 [when divided by 2]
=n2+20n-5n-100=0
=n(n+20)-5(n+20)=0
(n+20)(n-5)=0
If n+20=0 or n-5=0
N=-20 or n=5 [Rejected as n can’t be negative]
Therefore, in 5 years his savings will be Rs 200.
Question: A man secured 16500 in 10 years. In each year after the first he secured Rs 100 more than he did in the previous year. How much did he secure in the 1st year? (4 marks)
Ans: Let “a” be the money he secured in 1st year
= First year he secured the money= Rs a
He saved Rs 100 more than he did in the previous year.
=Second year he secured the money=Rs (a+100)
= Third year he saved the money= Rs [a+2(100)]
So, the sequence is a, a+100, a+2(100), ……, This is an AP with common difference (d)=100
=Sum of money he secured in ten years S10=16,500 rupees
We know that, Sn=n/2[2a+(n-1)d]
S10=10/2[2a+(10-1)100]
16500=5(2a+9×100)
2a +900=16500/5=3300
2a=2400
A=2400/2=1200
Therefore, he saved the money in the first year (a)= Rs 1200.
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