Arithmetic Progression

1. Arithmetic Progression SUBMITTED BY- PRIYARANJAN KUMAR CLASS-10 ‘A’ ROLL NO.-17

2. Arithmetic Sequence • ArithmeticSequence is a sequence of numbers such that the difference between the consecutive terms is constant. • For instance, the sequence 5, 7, 9, 11, 13, 15 … is an arithmetic progression with common difference of 2. • 2,6,18,54(next term to the term is to be obtained by multiplying by 3. Arithmetic Sequence

3. Arithmeticseries • A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of a finite arithmetic progression is called an Arithmetic series. • The behavior of the arithmetic progression depends on the common difference d. If the common difference is: • Positive, the members (terms) will grow towards positive infinity. • Negative, the members (terms) will grow towards negative infinity.

4. Common Difference • If we take first term of an AP as a andCommon Difference asd. Then-- • nth term of that AP will be An = a + (n-1)d. • For instance--- 3, 7, 11, 15, 19 … d =4 a =3 • Notice in this sequence that if we find the difference between any term and the term before it we always get 4. • 4 is then called the common difference and is denoted with the letter d. • To get to the next term in the sequence we would add 4 so a recursive formula for this sequence is: • The first term in the sequence would be a1 which is sometimes just written as a.

6. ArithmeticProgression • If the initial term of an arithmetic progression is a1 and the common difference of successive members is d, then the nth term of the sequence (an) is given by: • An= a1+(n-1)d • And in general An= a1+ (n-m)d

7. To find the nth Term of an A.P. Let us consider an A.P. with first term ‘a’ and common difference ‘d’ ,then The first term = a1 =a +0 d = a + (1-1)d The second term = a2 = a + d = a + (2-1)d The third term = a3 = a + 2d = a + (3-1)d The fourth term = a4 = a + 3d = a + (4-1)d ------------------------------------------- ------------------------------------------- The nth term = an = a + (n-1)d

8. Example Let a=2, d=2, n=12,find An An=a+(n-1)d =2+(12-1)2 =2+(11)2 =2+22 Therefore, An=24 Hence solved.

9. To check that a given term is in A.P. or not. • 2, 6, 10, 14…. • Here first term a = 2, • find differences in the next terms • a2-a1 = 6 – 2 = 4 • a3-a2 = 10 –6 = 4 • a4-a3 = 14 – 10 = 4 • Since the differences are common. • Hence the given terms are in A.P.

10. Problem1. Find 10th term of A.P. 12, 18, 24, 30.. … Solution. Given A.P. is 12, 18, 24, 30.. First term is a = 12 Common difference is d = 18- 12 = 6 nth term is an = a + (n-1)d Put n = 10, a10= 12 + (10-1)6 = 12 + 9 x 6 = 12 + 54 a10 = 66

11. Arithmetic Mean To find the Arithmetic Mean Between Two Numbers. • Let A be the AM between a and b. • A, A, b are in AP (A-a)=(b-A) • A=1/2(a+b) AM between a and b A=1/2(a+b) Practice Find the AM between 13 & 19 AM= ½(13+ 19) = 16

12. Some convenient methods to determine the AP • It is always convenient to make a choice of • 3 numbers in an AP as (a-d), a, (a+d); • 4 numbers in an AP as (a-3d), (a-d), a, (a+d),(a+3d); • 5 numbers in an AP as (a-2d), (a-d), a, (a+d), (a+2d).

13. Practice Q---- The sum of three numbers in an AP is 21 & their product is 231. Find the numbers. • Let the required Numbers be (a-d), a, (a+d) • Then (a-d)+ a+ (a+d)=21 • A=7 • And (a-d) a (a+d)=231 • A(a2-d2) =231 • Substituting the value of ‘a’ • 7(49-d2)=231 • 7d2= (343-231)= 112 • D2= 16 • D= 4

14. Sum of n-term of an ap The sumofnterms,we find as, Sum = n X [(first term + last term) / 2] Now last term will be = a + (n-1) d Therefore, • Sn= ½ n [ 2a + (n - 1)d ] • It can also be written as • Sn = ½ n [ a + an ]

15. Problem 1. Find the sum of 30 terms of given A.P. 12 + 20 + 28 + 36……… Solution : Given A.P. is 12 , 20, 28 , 36 Its first term is a = 12 Common difference is d = 20 – 12 = 8 • The sum to n terms of an arithmetic progression • Sn = ½ n [ 2a + (n - 1)d ] • = ½ x 30 [ 2x 12 + (30-1)x 8] • = 15 [ 24 + 29 x8] • = 15[24 + 232] • = 15 x 246 • = 3690

16. Practice Q1: Find the sum of the following APs: (i) 2, 7, 12, ……, to 10 terms Q3: Given a = 2, d = 8, Sn = 90, find n and an. Question: 2. Find the sums given below: (i) 7+10.5+14+…..+84

17. Problem-- Find number of terms of A.P. 100, 105, 110, 115……500 • Solution. • First term is a = 100 , an = 500 • Common difference is d = 105 -100 = 5 • nth term is an = a + (n-1)d • 500 = 100 + (n-1)5 • 500 - 100 = 5(n – 1) • 400 = 5(n – 1) • 5(n – 1) = 400 • n – 1 = 400/5 • n - 1 = 80 • n = 80 + 1 • n = 81 • Hence the no. of terms are --81.

18. Practice Question: 1. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149. Question: 2. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

19. General Formulas of AP • The general forms of an AP is a,(a+d), (a+2d),. .. , a + ( m - 1)d. • Nth term of the AP is Tn =a+(n-1)d. • Nth term form the end ={l-(n-1)d}, where l is the last term of the word. • Sum of 1st n term of an AP is Sn=N/2{2a=(n-1)d}. • Also Sn=n/2 (a+l)

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