Exercise 15.6

1. Find the sum of integers which are divisible by 5 from 1 to 100.

Solution:

The integers divisible by 5 between 1 and 100 are 5, 10, 15, ..., 100.

This is an arithmetic progression with first term a = , common difference d = , and last term l = .

To find the number of terms (n), we use the formula l = + (n-1)(n+1)d.

= + (n-1)

= (n-1)5

= n-1

n =

The sum of an arithmetic progression is given by S = n2a+l

S = (202)(5 + 100)

S = ×

S =

2. Find the sum of integers which are divisible by 2 from 11 to 50.

Solution:

The integers divisible by 2 between 11 and 50 are 12, 14, 16, ..., 50.

This is an arithmetic progression with first term a = , common difference d = , and last term l = .

To find the number of terms (n), we use the formula l = a + (n-1)d.

= + (n-1)

= (n-1)2

= n-1

n =

The sum of an arithmetic progression is given by S = n2a+l

S = 20212+50

S = ×

S =

3. Find the sum of integers which are divisible by 2 and 3 from 1 to 50.

Solution:

Integers divisible by both 2 and 3 are divisible by their least common multiple, which is .

The integers divisible by 6 between 1 and 50 are , , , ..., .

This is an arithmetic progression with first term a = , common difference d = , and last term l = .

To find the number of terms (n), we use the formula l = a + (n-1)d.

48 = 6 + (n-1)6

= (n-1)6

= n-1

n =

The sum of an arithmetic progression is given by S = n2a+l

S = 826+48

S = ×

S =

4. (n3 - n) is divisible by 3. Explain the reason.

Solution:

n3 - n can be factored as (n2 - 1) = n(n - 1)(n+1)(n-1).

This represents the product of three consecutivesequential integers.

In any three consecutive integers, at least one of them must be divisible by .

Therefore, the product n(n - 1)(n + 1) must be divisible by .

5. Sum of ‘n’ odd number of consecutive numbers is divisible by ‘n’. Explain the reason.

Solution:

Let the n consecutive odd numbers be represented as:

a, a+2, a+4, ..., a + 2(n-1)

The sum of these n odd numbers is given by:

S = n2(2a + 2(n-1))

S = (n/2)(2a + - )

S = n2(2(a + n - 1))

S = nn2(a + n - 1)

Since 'a' and 'n' are integers, (a + n - 1) will be an integer. Thus, the sum S is a multipledivisible of n, hence divisiblemultiple by n.

6. Is 111 + 211 + 311 + 411 divisible by 5? Explain.

Solution:

To determine if the sum is divisible by 5, we can examine the remainders of each term when by 5.

111 leaves a remainder of when divided by 5.

211 leaves a remainder of when divided by 5.

311 leaves a remainder of when divided by 5.

411 leaves a remainder of when divided by 5.

Therefore, the sum 111 + 211 + 311 + 411 leaves a remainder of + + + = when divided by 5.

Since the remainder is 4 (not 0), the sum 111 + 211 + 311 + 411 is notis divisible by 5.