Question

determine the sum of all the numbers divisible by 9 in between 700 and 1100​

Answer 1

Answer: 40500

Step-by-step explanation:

• The first number which is divisible by 9 between 700 an 1100 is 702.
• So,it will form an A.P. with first term as a=702 and common difference d=9.
• The last term which will be divisible between 700 and 1100 is 1098.
• We have to calculate the number of terms.
• 1098=a+(n-1)d

    ⇒1098=702+(n-1)9

    ⇒1098-702=(n-1)9

    ⇒$\frac{396}{9} =(n-1)$

    ⇒n=44+1

    ∴ n=45

• So, sum of the A.P. with 45 terms where a=702 and d=9 will be

       =$\frac{45}{2}[2*(702)+(45-1)9]$

       =$\frac{45}{2}[1404+396]$

       =$\frac{45}{2}[1800]$

       =45×900

      =40500

• Hence,the required sum of all the numbers between 700 and 1100  divisible by 9 is 40500.