Alex's Simple Addition

We have 9 numbers. The average of these numbers is 15.Thus the sum 15*9=135

This is a more easier solution.

Sum of consecutive numbers $=\frac{n(n+1)}{2}$

Sum of numbers (11 to 19) $=\frac{19(19+1)}{2}-\frac{10(10+1)}{2}$

$=\frac{19\times20}{2}-\frac{10\times11}{2}$

$=\frac{380}{2}-\frac{110}{2}$

$=190-55=135$

Thus, the answer is: $11+12+13+14+15+16+17+18+19=\boxed{135}$

numbers 1 to 9 in sequence will always add upto 45. so thats is one side. the remaining values are just 10s place values which can be multiplied if they are the same 10 s value. In this case it is 90 . Hence 90 + 45 = 135

Simple addition gives the answer as $\color{#69047E}{\boxed{135}}$ .

Just add all the digits in the tens position and carry it on to all the ones. U get 135

here a=11, d=12-11=1 so S=(n/2){2a+(n-1) d}=(9/2) {2 11+(9-1) 1}=135

All you need to do is add the numbers starting from the outside and work your way in: (11+19) + (12+18) + (13+17) + (14+16) + 15 = 30 + 30 + 30 + 30 + 15 since all of the number in brackets add up to 30, so you can just do 30*4 + 15 = 120 + 15 = 135

Step 1: Count the number. How many numbers are there? (There is 9 numbers) Step 2: Multiply 11 and 9. (99 is the product) Step 3: Count all the numbers except 11. (There is 8 numbers if you do not consider 11) Step 4: Add numbers 1 to 8. (36 is the answer) Step 5: Add the product of 11 and 9 and the sum of 1 to 8. (The overall answer is 135)

Using A.P. n=9 So, Sn = n/2 (first term+last term) S9 = 9/2 (11+19) = 9/2 *30 = 135

11+19,12+18,13+17,14+16,15 ->30.4+15=135!

you can get it by this equation n(n+k/2 k n is the last number of the sequence and the beginning of the sequence is 1 ,k is the difference between any two numbers of the sequence and if the result is a fraction you must round it ,and because you began with 11 you must subtract the sum of 1-10 from the sum of 1-19 (19 (19+1)-2)-(10 (10+1)/2)=135 or you can get it by middle number multiplied by the position of the last number 15 9=135

This can be done easily using A.P.