## Arithmetic Series Worksheet Solution4

In the page arithmetic series worksheet solution4 you are going to see solution of each questions from the arithmetic series worksheet.

(10) Find the sum of all natural numbers between 300 and 500 which are divisible by 11.

Solution:

From this sequence we have to find number of terms which are divisible by 11 and also we have to find their sum.

308,319,330,........................495

The first number which is divisible by 11 from this sequence is 308,the second number will be 319 and the last number will be 495.

308 + 319 + 330 + ........... + 495

a = 308    d = 319 - 308     and L = 495

d = 11

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

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

n = [(495-308)/11] + 1

n = [187/11] + 1

n = 17 + 1

n = 18

S n = (n/2) [a+ L]

= (18/2) [308 + 495]

= 9 [803]

= 7227

(11) Solve 1 + 6 + 11 + 16 + ..........  + x = 148

Solution:

In this problem 148 represents sum of x number of terms starting from 1.

Sn = (n/2) [ 2 a + (n-1) d ]

a = 1  d = 6 - 1    n = [(l-a)/d] + 1

d = 5            = [(x - 1)/5] + 1

= (x - 1 + 5)/5

= (x + 4)/5

148 = [(x + 4)/5] /2 [1+x]

148 = (x+4)/10 [1+x]

148 x 10 = (x + 4) (1 + x)

1480 = x + x² + 4 + 4x

1480 = x² + 4 + 5 x

x² + 5 x + 4 - 1480 = 0

x² + 5 x - 1476 = 0

(x - 36) (x + 41) = 0

x = 36,-41

Therefore the last term of the series is 36.

(12) Find the  sum of all natural numbers between 100  and 200 which are not divisible by 5.

Solution:

To find sum of numbers which are not divisible by 5 first we have to find the sum all numbers between 100 and 200 and we have to subtract it by sum of numbers which are divisible by 5.

= (100 + 101 + 102 + 103 + ......... + 200)-(100+105+110+..........+200)

a = 100   d = 101- 100                    a = 100   d = 105- 100

= 1                                              = 5

n = [(200-100)/1] + 1                           n = [(200 -100)/5] + 1

n = 100 + 1                                         n = (100/5) + 1

n = 101                                              n = 20 + 1

n = 21

No of terms of the series (100 + 101 + 102 + 103 + ......... + 200) is 101 and no of terms of the series (100+105+110+..........+200) is 21.

Sn = (n/2) [a+L]

= (101/2)[100 + 200] - (21/2)[100 + 200]

= (101/2)[300] - (21/2) [300]

= (101)[150] - (21) [150]

= 15150 - 3150

= 12000

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arithmetic series worksheet solution4