Question

Given a sequence of integers between 20 and 300 which are divisible by 9.Find their sum.

Answer 1

Answer:

Required sequence=27,36,45........396.

Let the no of terms be 'n'.

27+(n-1)9=396

(n-1)9=396-27

n-1=369/9

n=41+1=42

Sum= 42/2(27+396)

=21×423=8883

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Answer 2

Concept

Arithmetic Progression is a sequence of numbers where the common difference between the numbers is constant. Division is one out of the four basic mathematical operations used In arithmetic.

Given

Integers between 20 and 300

Find

Sum of integers between 20 and 300 which are divisible by 9

Solution

Numbers are between 20 and 300 divisible by 9

The first number is 27

Last number is 297

Common Difference = 9

Series : 27,36,45………..297

Calculating total number of terms

An = a+ (n-1)d

297 = 27 + (n-1)9

270 = (n-1)9

(n-1) = 30

n = 31

Calculating sum

Sn = n2(a+an)

Sn = 31/2(297+27)

= 31/2 * 324

= 31*162

= 5022

Sum of integers between 20 and 300 which are divisible by 9 is 5022

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