Question

sum of first ten even numbers is​

Answer 1

Step-by-step explanation:

step 1 Address the formula, input parameters & values.

Input parameters & values:

The number series 2, 4, 6, 8, 10, 12, .  .  .  .  , 20.

The first term a = 2

The common difference d = 2

Total number of terms n = 10

step 2 apply the input parameter values in the AP formula

Sum = n/2 x (a + Tn)

= 10/2 x (2 + 20)

= (10 x 22)/ 2

= 220/2

2 + 4 + 6 + 8 + 10 + 12 + .  .  .  .   + 20 = 110

Therefore, 110 is the sum of first 10 even numbers.

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Answer 2

Answer:

The sum of first 10 even numbers is 90.

Step-by-step explanation:

Every even number differ from the next even number by 2.

So the set of first 10 even numbers follow an arithmetic progression.

The first term being a = 0, the first even number and the common difference is d = 2.

The sum of first n terms of an AP is:

$S_{n}=\frac{n}{2}[2a+(n-1)d]$

Compute the sum of first 10 even numbers as follows:

$S_{n}=\frac{n}{2}[2a+(n-1)d]\\=\frac{10}{2}[(2\times0)+(10-1)\times 2]\\=5[0+18]\\=90$

Thus, the sum of first 10 even numbers is 90.