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The sum of all natural numbers lying between 100 and 300 which are multiplies of 5:
1. 7800
2. 7900
3. 8000
4. 7500

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Correct Answer - Option 1 : 7800

Concept:

For an AP, a= First term , a= Last term and d = Common difference

The sum of n terms  = \(\rm \dfrac n 2 (a_1 + a_n)\)

nth terms of AP = an = a1 + (n - 1)d

Calculations:

To Fnd: The sum of all natural numbers lying between 100 and 300 which are multiplies of 5

First term = a1  = 105

Last term = an = 295

Common difference = d = 5

nth terms of AP = an = a1 + (n - 1)d

no. of terms  = n = \(\rm \dfrac {a_n -a_1}{d}+ 1\)

⇒ n = \(\rm \dfrac {295 -105}{5}+ 1\)

⇒ n = 38 + 1

⇒ n = 39

Now, The sum of all natural numbers lying between 100 and 300 which are multiplies of 5 = \(\rm \dfrac n 2 (a_1 + a_n)\)

=  \(\rm \dfrac {39}{2}(105+ 295)\) = 39 × 200

= 7800

The sum of all natural numbers lying between 100 and 300 which are multiplies of 5 is 7800

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