What is the sum of numbers between 250 and 350 which are divisible by 7?

Question

What is the sum of numbers between $250$ and $350$ which are divisible by $7$?

My attempted solution:

7|250|35
 |21 |
 -----
   40
   35
  ----
    5

The 1st number divisible by $7$ is: $(250-5)+7 = 252$.

7|350|50
 |35 |
 -----
    0

The last number divisible by $7$ is: $350$.

Total numbers between 250 and 350 divisible by $7$ is:

$((350-252)/7) + 1$

$ = (98/7)+1$

$ = 14+1$

$ = 15$

$ Sum = \frac{15}{2} [2 \cdot 252 + (15 - 1) \cdot 7 ]$

$ = \frac{15}{2} [2 \cdot 252 + 14 \cdot 7 ]$

$ = \frac{15}{2} [504 + 98 ]$

$ = \frac{15}{2} \cdot 602$

$ = 15 \cdot 301$

$ = 4515 $

(Ans.) $4515$

Is there a more efficient way to do this?

Answer 1

$$ 252 + 259 + 266 + 273+\cdots\cdots+336+343+350 $$ Since this is an arithmetic sequence, i.e. the difference between each number and the next is the same in all cases, the average of all the numbers in the list must be the same as the average of the first one and the last one: $$ \frac{252+350} 2 = 301 $$ Therefore the sum is $(301\times\text{the number of terms}).$

Since you have to add $7$ fourteen times, there are $15$ numbers, so the answer will be $301\times15.$

Answer 2

HINT

The first divisible number is $252$ the last $350$ then

$$252+\dots+350=7(36+\dots+50)=7\left(\sum_{k=1}^{50}k-\sum_{k=1}^{35}k\right)$$