Question

what is the sum of the consecutive number from 11 to 20 ?​

Answer 1

The consecutive numbers from 11 to 20 are:

11, 12, 13, ........, 20

Here, first term (a) = 11 and common difference (d) = 12 - 11 = 1

Last term(l) = 20

The given sequence are in AP.

Let n be the nth number of terms.

We have to find, the sum of the consecutive number from 11 to 20.

Solution:

We know that:

The nth term of an A.P.

$t_{n}$ = a + (n - 1)d

∴ 11 + (n - 1)1 = 20

⇒ n - 1 = 20 - 11 = 9

⇒ n = 10

The sum of nth term of an AP

$S_{n} =\dfrac{n}{2}(a+l)$

The sum of the consecutive number from 11 to 20

∴ $S_{10} =\dfrac{10}{2}(11+20)$

⇒ $S_{10}$ = 5(31)

⇒ $S_{10}$ = 155

Thus, the sum of the consecutive number from 11 to 20 is 155.

Answer 2

Given:

The consecutive number from 11 to 20

To find:

The sum of the consecutive number from 11 to 20

Solution:

We have given the series of the consecutive number from 11 to 20

which can be written as:

11, 12, 13..................20

As the common difference is same for all so the given series is an AP

a = 11, d = 1

The nth term of an A.P.

• aₙ = a + (n - 1)d
• 20 =  11 + (n - 1)1
• n - 1 = 20 - 11
• n = 9+1
• n = 10

The sum of nth term of an AP

• sₙ = n/2(2a+(n-1)d)
• s₁₀ = 10/2(22+9×1)
• s₁₀ = 5(22+9)
• s₁₀ =  5(31)
• s₁₀ = 155

Hence,

The sum of the consecutive number from 11 to 20 is 155.