Arithmatic progression : Arithmatic progression Some basics :
Any series with a common difference is known as Arithmetic Progression.
By adding the common difference to the last term we get the next term of the series.
Similarly by subtracting the common difference from the first term we get the previous term.
If we know the first term and the common difference we can frame the series.
Let a = the first termd = the common differencethen,to find the nth term ,nth term = a + (n – 1) d : Let a = the first termd = the common differencethen,to find the nth term ,nth term = a + (n – 1) d
Eg.Find the 10th and nth term of the series,3, 7, 11, 15……?ans : Here a = 3 d = 7 – 3=410th term = 3 + (10-1) 4=39nth term = 3 + (n-1)4= 4n – 1 : Eg.Find the 10th and nth term of the series,3, 7, 11, 15……?ans : Here a = 3 d = 7 – 3=410th term = 3 + (10-1) 4=39nth term = 3 + (n-1)4= 4n – 1
To find the no. of terms in an AP:n = last term – first term + 1common difference : To find the no. of terms in an AP:n = last term – first term + 1common difference
How many terms are there in the AP,13,17,21,25,29,……65 ? : How many terms are there in the AP,13,17,21,25,29,……65 ? Ans : n = tn - a + 1
d
= 65 – 13 + 1
4
= 14
Sum of ‘ n ‘ natural nos. : Sum of ‘ n ‘ natural nos. 1 + 2 + 3 + ………..(n-2) + (n-1) + n (n+1) (n+1) (n+1) Like wise we have “ n / 2 ” times of (n+1)’s .
Hence, Sum of ‘n’ natural nos = n(n+1)
2
Sum of ‘n’ even nos. : Sum of ‘n’ even nos. 2 + 4 + 6 +………..2n
Taking 2 out, we get
= 2 ( 1 + 2 + 3 + …….n)
= 2 n(n+1)
2
Similarly we can find the sum of the series which is the multiple of any no.
Sum of ‘n’ odd nos. : Sum of ‘n’ odd nos. 1 + 3 + 5 +………2n -1
We can write this as,
1 + (1+2) + (1+4)……….(1+2(n-1))
Adding all 1’s together, we wud have n 1’s
ie., (1+1+1+…..n) +(2+4+6….2(n-1))
= (1*n) + 2 [ 1+2+3+….(n-1)]
= n + 2 (n-1)n
2
= n + n2 – n
= n2
Sum of an AP of ‘n’ nos. : Sum of an AP of ‘n’ nos. a + (a+d) + (a+2d)……….+(a+(n-1)d)
Adding all ‘a’ s together, we get
= (a+a+a…n times) +( d+2d+3d…..+(n-1)d
= (a * n) + d[ 1+2+3+…..+(n-1)]
= na + d (n-1)n
2
= 2na +dn(n-1)
2
= n 2a + (n-1)d
2
Hence Sum of an AP = n 2a + (n-1)d 2 : Hence Sum of an AP = n 2a + (n-1)d 2 From the above formula, we can derive another formula for the
Sum of an AP = n [ a + a + (n-1)d]
2
= n first term + nth term
2