Question

Sum of all even integers between 100 and 800 which are divisible by 6

Answer 1

Answer: 5265

Step-by-step explanation:

First number between 100 and 800 divisible by 6 = 102

Similarly, last number = 798

So, an = 798; a = 102; d = 6; n = ?

an = a + (n-1)d

=> 798 = 102 + 6(n-1)

=> 696 = 6(n-1)

=> n-1 = 696/6 = 116

=> n = 117

Now, sum of n integers in an AP,

Sn = (n/2)[2a + (n-1)d]

=> (117/2)[2(102) + 6(117-1)]

=> (117/2)(2)(102 + 3 x 116)

=> 117(102 + 348)

=> 117 x 450

=> 52650

NOTE: All multiples of 6 are even integers.