Question

What is the average of the sum of first 20 odd natural numbers and
sum of first twenty even natural numbers and sum of squares of
the natural numbers upto twenty?​

Answer 1

Answer:

hey here is ur answer

pls mark it as brainliest

Explanation:

1.for sum of first 20 odd natural numbers

so there are 10 odd natural numbers in category of first 20 odd natural nos

so now we have to use arithmetic progression (AP) concept here

so where as the common difference ie value of d would be constant through out

series of first 20 odd ntrl numbers forms an ap

so for above ap

a=t1 ie first term is 1 tn=last term is 19

n=15

so use formula sn=n/2 (t1+tn)

so put n=10

so we get

solving and substituting values in formula

s15=100

ii) now similarly for second apply ap concept

so here n=10

because as there are 10 odd nos remaining 10 would be even

use same procedure same formula

where t1=2 tn=20 ie last term

so then s10=110

iii.npw for third you have add squares of each number up to 20

ie (1) whole square +(2) square+up to +(20)square

which gives you

2870 as answer

now average of all these would be average of 1st qnswer+2nd+3rd upon total possible elements of each of three

ie I mean to say 10+10+20

=40

so 2870+110+100/40

=3080/40

ie 770

hence 770 is ur answer

pls mark it as brainliest

Answer 2

Answer:

The average of the sum of first 20 odd natural numbers is 400 .

The sum of first twenty even natural numbers is 420 .

The sum of squares of the natural numbers upto twenty is 2870 .

Explanation:

Given :

According to the question the numbers will be of first $20$.

To Solve :

The concept of sequence and series will help us to solve this question.

(i) The formula of the sum of first $n$ terms is A.P : $\frac{\mathrm{n}}{2}(2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d})$

The value of the first term $a=1$

Common difference $d=2$

and the value of $n=20$

Put the values in the above A.P

$\frac{20}{2}(2(1)+(20-1) 2)\\=10\times(2+19\timess 2)=10\times40\\=400$

Therefore, the average of the sum of first $20$ odd natural numbers is $400$ .

(ii) We need to find out the sum of $n$ terms of the A.P $2,4,6,...$ .

The formula of required sum is $\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$

Here $a=2$ , $d=4-2=2$ and $n=20$

Put the values in the above formula

$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \times 2+(\mathrm{n}-1) \times 2]\\\\=\mathrm{n}[2+\mathrm{n}-1]\\=n(n+1)\\\mathrm{S}_{\mathrm{20}}=20(20+1)=420$

So , the sum of first twenty even natural numbers is $420$ .

(ii) The formula of the sum of squares of first n natural numbers is

$\sum n^2=1^2+2^2+3^2+\ldots+n^2$

Here the value of $n=20$ .

Now the sum will be

$\Sigma 20^2=[n(n+1)(2 n+1)] / 6\\= [20(20+1)(40+1)] / 6 \\= [20(21)(41)] / 6 \\= 2870$

So the sum of squares of the natural numbers upto twenty is $2870$ .

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