EVALUATING ARITHMETIC SERIES

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[103]

  =  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

S_n = 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.

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