Answer: 5220 Step-by-step explanation: Lets make an ap as follows: 102, 108, 114… 246. This will have all numbers divisible by 6 between 100 and 250. Now, a = 102 d = 6 tn = 246. [tex]tn = a + (n – 1)d \\ 246 = 102 + (n – 1) \times 6 \\ 6n – 6 = 144 \\ 6n = 150 \\ n = 30[/tex] Thus there are 40 terms. Now let’s find sum. [tex]sn = \frac{n}{2} (a + l) \\ = \frac{30}{2} \times (102 + 246) \\ = 15 \times 348 \\ = 5220[/tex] Thus, the required sum is 5220 Reply

Answer:5220

Step-by-step explanation:Lets make an ap as follows: 102, 108, 114… 246.

This will have all numbers divisible by 6 between 100 and 250.

Now,

a = 102

d = 6

tn = 246.

[tex]tn = a + (n – 1)d \\ 246 = 102 + (n – 1) \times 6 \\ 6n – 6 = 144 \\ 6n = 150 \\ n = 30[/tex]

Thus there are 40 terms. Now let’s find sum.

[tex]sn = \frac{n}{2} (a + l) \\ = \frac{30}{2} \times (102 + 246) \\ = 15 \times 348 \\ = 5220[/tex]

Thus, the required sum is 5220