Arithmetic Progression

Arithmetic Progression Arithmetical series structure An arithmetical series starts with the first term, usually given the letter 'a'. For each subsequent term of the series another term is added. This is a multiple of the letter 'd' called 'the common difference'. So the series has the structure: Sn=a+(a+d)+(a+2d)+(a+3d)+(a+4d)+(a+5d)+………+l where Snis the sum to 'n' terms, the letter 'l' is the last term. The common difference 'd' is calculated by subtracting any term from the subsequent term. The nth term (sometimes called the 'general term')is given by: An=a+(n-1)d Arithmetic Progression Problem 1) Is the row 1,11,21,31... arithemtic progression?

Solution: Yes it is arithmetic progression with first term 1 and common differnece 10.

Proof of the sum of an arithmetical series The sum of an arithmetical series is found by adding two identical series together, but putting the second series in reverse order.

Sn = a+(a+d)+(a+2d)+…….+(l-2d)+(l-d)+l…………………………………(i) Sn= l+(l-d)+(l-2d)+…………..+(a+2d)+(a+d)+a……………………………(ii) 2Sn=(a+l)+( a+l)+( a+l)+………………+( a+l)+( a+l)+( a+l) Now,2Sn=n(a+l) Then, Sn=n/2(a+l). Therefore Sum of n terms of AP=NO.OF TERMS(1ST TERM+LAST TERM)

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Example #2 In an arithmetical progression the 8th term is 23 and the 11th term is 4 times the 3rd term. Find the 1st term, the common difference and the sum of the first 10 terms.

Solution:

8th term,a+7d=23……………….(i) a+10d=4(a+2d) =>a+10d=4a+8d =>3a=2d

a =2d/3 Now putting this value of a in (i) we get d =3 and then a=2 Therefore,1st term=a=2 c.d.=d=3 Sum=n[2a+(n-1)d]/2 =155

Example #3 The sum of terms of an arithmetic progression is 48. If the first term is 3 and the common difference is 2, find the number of terms.

Solve by yourself Answer->n=6

,n=-8(neglected bcoz n cannt be -ve)

Arithmetic Mean This is a method of finding a term sandwiched between two other terms. The required term is calculated by taking an average of the term before and the term after. So if we have a sequence of terms: a b c and a and c are known. Then term b is given by: b= a+c 2

a, A1,A2,A3,………………….An,b………………………………(i) HERE, A1,A2,A3…………An ARE THE N ARITHMETIC MEANS NOW IF WE HAVE TO FIND IT THEN WE CAN PROCEED AS Here In eqn.i we see that there are (n+2)terms Then (n+2)th term,b=a+(n+2-1)d =>b=a+(n+1)d =>b-a=(n+1)d =>d=(b-a)/(n+1) Then , A1=a+d A2=a+2d and so on

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