# EVALUATING ARITHMETIC SERIES

A series whose terms are in arithmetic sequence is called arithmetic series.

To find sum of terms in arithmetic sequence, we can use one of the formulas given below.

Sn  =  (n/2) [a + l]

Sn  =  (n/2) [2a + (n - 1)d]

a = first term,

n = number of terms of the series,

d = common difference and l = last term

Question 1 :

Find the sum of the following.

(i) 3, 7, 11,… up to 40 terms.

Solution :

Number of terms (n) = 40

First term (a) = 3

Common difference (d)  =  7 - 3  =  4

Sn  =  (n/2) [2a + (n - 1)d]

=  (40/2) [2(3) + (40 - 1)4]

=  20 [6 + 156]

=  3240

Hence the sum of 40 terms of the given series is 3240.

(ii) 102, 97, 92,… up to 27 terms.

Solution :

Number of terms (n) = 27

First term (a) = 102

Common difference (d)  =  97 - 102  =  -5

Sn  =  (n/2) [2a + (n - 1)d]

=  (27/2) [2(102) + (27 - 1)(-5)]

=  (27/2) [204 + 26(-5)]

=  (27/2) [204 - 130]

=  (27/2) (74)

=  27 (37)

=  999

Hence the sum of 27 terms of the given series is 999.

(iii) 6 + 13 + 20 +...........+ 97

Solution :

l = 97, a = 6, d = 13 - 6  =  7

From this, we have to find the number of terms.

n = [(l - a)/d] + 1

n = [(97 - 6)/7] + 1

=  (91/7) + 1

n  =  13 + 1

n = 14

Sn  =  (n/2) [a + l]

=  (14/2)[6 + 97]

=  7

=  721

Hence the sum of given series is 721.

Question 2 :

How many consecutive odd integers beginning with 5 will sum to 480?

Solution :

Consecutive integers starting from 5 are 5, 7, 9, ............

5 + 7 + 9 + 11 + .............  =  480

Sn = 480

(n/2)[2a + (n - 1)d]  =  480

a  =  5, d = 7 - 5  =  2

(n/2)[2(5) + (n - 1)(2)]  =  480

(n/2)[10 + 2n - 2]  =  480

(n/2)[2n + 8]  =  480

2n2 + 8n  =  960

Dividing by 2, we get

n2 + 4n  - 480  =  0

(n + 24) (n - 20)  =  0

n = -24 or n = 20

Hence by finding sum of 20 terms we will get 480. Apart from the stuff given aboveif you need any other stuff in math, please use our google custom search here.

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