Correct Answer - Option 1 : 7800
Concept:
For an AP, a1 = First term , an = 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