# FIND THE SUM OF ARITHMETIC SERIES WITH GIVEN DESCRIPTION

We have to consider the given sum as Sn. Instead of Sn, we may use one of the formulas given below.

Sn  =  (n/2) [a + l] (or)

Sn  =  (n/2) [2a + (n - 1)d]

a = first term, d = common difference and n = number of terms.

Problem 1 :

Find the sum of all 3 digit natural numbers, which are divisible by 9.

Solution :

3 digit number starts from 100 and ends with 999. From this sequence we have to find number of terms which are divisible by 9 and also we have to find their sum.

108, 117, 126,........................999

The first number which is divisible by 9 from this sequence is 108, the second number will be 117 and the last number

will be 999.

108 + 117+ 126 + ........... + 999

a = 108, d = 117 - 108 = 9 and l = 999

to find number of terms, we have to use the formula for (n)

n  =  [(l-a)/d]+1

n  =  [(999-108)/9] + 1

n  =  [891/9] + 1

n  =  99 + 1

n  =  100

Sn  =  (n/2)[a+l]

= (100/2)[108+999]

=  50 (1107)

=  55350

Problem 2 :

Find the sum of first 20 terms of the arithmetic series in which 3rd term is 7 and 7th term is 2 more than three times its 3rd term.

Solution :

t3  =  7

t7  =  3t3+2

t7  =  3(7)+2

t7  =  23

a+2d  =  7   -----(1)

a+6d  =  23 -----(2)

(1) - (2)

-4d  =  -16

d  =  4

By applying the value of d in (1), we get

a = -1

Now we need to find sum of 20 terms

Sn  =  (n/2)[2a+(n-1)d]

S20  =  (20/2) [2(-1)+(20-1)(4) ]

=  10 [-2+19(4)]

=  10 (-2+76)

=  740

Problem 3 :

In an arithmetic series, the sum of first 11 terms is 44 and the that of the next 11 terms is 55.

Find the arithmetic series.

Solution :

Sum of first 11 terms  =  44

S11  =  44

(11/2)[2a+(11-1)d]  =  44

2a+10d  =  44 ⋅ (2/11)

2a+10d  =  8   ----- (1)

Sum of next 11 term  =  55

S22  =  S11 + 55

S22  =  44 + 55

S22  =  99

(22/2)[2a+(22-1)d]  =  99

2a+21d  =  99/11

2a+21d  =  9   ----- (2)

(1) - (2)

-11d  =  -1

d = 1/11

By applying the value of d = 1/11 in (1), we get

2a+10(1/11)  =  8

2a+10/11  =  8

2a  =  8-(10/11) ==> 78/11 ==> a = 39/11

Therefore the series is

(39/11) + (40/11) + (41/11) + ............

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