Question

17.
(1) Find the sum of all two digit natural numbers which are divisible by​

Answer 1

The sum of all two-digit natural numbers which are divisible by 4 = 1188

QUESTION:-

Find the sum of all two-digit natural numbers which are divisible by 4

GIVEN:- All two-digit natural numbers which are divisible by 4

TO FIND:- The sum of the arithmetic progression given.

SOLUTION:-

• An arithmetic progression (A.P) is a series of numbers in which the difference between its consecutive terms is constant.

The two-digit natural numbers are { 10, 11, 12, 13,.....99}

The two-digit natural numbers divisible by 4 =

• A.P = {4, 8, 12, 16, 20,.....96}

The first term of the A.P is

• a = 12

The common difference between the consecutive number

• d = 4

The last term of the A.P is

• L = 96

The number of terms in the A.P is

• n =?

As we know,

Last term = a + (n - 1)d

By putting values in the formula

$96 = 12 + (n - 1)4 \\ 96 = 12 + 4n - 4 \\ 4n = 96 - 12 + 4 \\ n = \frac{84}{4} \\ n = 22$

n= 22

Then,

Sum of the AP = $\frac{n}{2} (2a + (n - 1)d)$

By putting the values in the formula

$= \frac{22}{2} (2 \times 12 + (22 - 1) \times 4\\ = 11 \times 24 +( 21 \times 4) \\ = 11 \times 24 + 84 \\ = 11 \times 108 \\ = 1188$

Hence, the sum of all two-digit natural numbers which are divisible by 4 = 1188

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Answer 2

Correct question: Find sum of all two digit natural numbers which are divisible by 4

The sum of all two digit natural numbers which are divisible by​ 4 is 1188.

Given:

All two digit numbers divided by 4.

To find:

Sum of all two digit numbers divisible by 4.

Solution:

First two digit number divisible by 4 = 12

Last two digit number divisible by 4 = 96.

A.P. = {12, 16, 20, 24, ….. 96}

Last term of an A.P = a + (n-1)d

a = first term = 12

d = difference between terms

96 = 12 + (n-1)4

96-12 = (n-1)4

84/4 = n-1

21 = n-1

n = 22

Sum of A.P. = n/2(a+l)

= 22/2 (12 + 96)

= 11 * 108

= 1188

So, the sum of all two digit natural numbers which are divisible by​ 4 is 1188.

#SPJ2