Question

5. Find the sum of the integers between 1000 and 2000 that are not divisible by 9.​

Answer 1

$\large \sf Solution :$

To find the sum of the integers between 1000 and 2000 that are not divisible by 9

firstly we need to compute ,

(i) Sum of the integers that are divisible by 9 between 1000 and 2000

(ii) Sum of the integers between 1000 and 2000

(i)

AP :

1008 , 1017 , 1026 , ............... ,1998

Given ,

First term = 1008

Common difference = 9

Last Term = 1998

We know that , The nth term of an AP given by :

$\large \sf\fbox{ \fbox{a_{n} = a + (n - 1)d}}$

$\sf \implies 1998 = 1008 + (n - 1)9 \\ \\ \sf \implies 1998 - 1008 = (n - 1)9 \\ \\ \sf \implies 990 = (n - 1)9 \\ \\ \sf \implies n - 1 = \frac{990}{9} \\ \\ \sf \implies n - 1= 110 \\ \\ \sf \implies n = 111$

The sum of the first n terms of an AP is given by :

$\large \sf \fbox{ \fbox{s_{n} = \frac{n}{2} (a + l) }}$

$\sf \implies s_{111} = \frac{111}{2} (1008 + 1998) \\ \\\sf \implies s_{111} = \frac{111}{2} \times 3006 \\ \\\sf \implies s_{111} = 111 \times 1503 \\ \\ \sf \implies s_{111} = 166833$

(ii)

AP : 1001 , 1002 , 1003 , ...... , 1999

Given ,

First term = 1001

Last term = 1999

Common difference = 1

Number of term = 999

We know that ,

$\large \sf \fbox{ \fbox{s_{n} = \frac{n}{2} (a + l) }}$

$\sf \implies s_{999} = \frac{999}{2} (1001 + 1999) \\ \\\sf \implies s_{999} = \frac{999}{2} \times 1500 \\ \\ \sf \implies s_{999} =999 \times 1500 \\ \\\sf \implies s_{999} =1098500$

Now ,

The sum of the integers that are not divisible by 9 b/w 1000 and 2000 is equals to the difference between the sum of the integers that are divisible by 9 and sum of integers b/w 1000 and 2000

$\sf \implies The \: required \: number =1098500 - 166833 \\ \\ \sf \implies The \: required \: number = 931667$

Hence , 931667 is the sum of the integers between 1000 and 2000 that are not divisible by 9