Question

find the sum of the first 100 natural numbers.​

Answer 1

Answer:

5050

Step-by-step explanation:

A. p. = 1,2,3,...,100

A=First term of the A.P.

D=distance between two numbers of A.P. (D=T2-T1)

n=the number of terms.

A=1, D=1 ,n=100

S100=n/2×[2a+(n-1)d]

S100=100/2×[2×1+(100-1)1)

S100=50[2+99×1]

S100=50×101

S100=5050

∴ The sum of first 100 natural numbers are 5050

Answer 2

$\large\bf\red{To\:Find}$

the sum of the first 100 natural numbers.

____________________________

$\large\bf\blue{Natural\:Numbers➠}$

Natural numbers are those Integer numbers which are greater than 0 or sometime start from 0 as well.

Example:-

1,2,3,4,5........................etc

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$\large\bf\green{Arithmetic\: Progression(A.P)➛}$

When the difference between two consecutive terms of series always come same,Then the following given series is in A.P.

$\large\bf\blue{Main\:terms\:of\:A.p:-}$

a=stands for 1st term

t stands for last term

d=difference between two consecutive terms

n= nth term of series.

_____________________________

$\large\bf\purple{Formula➽}$

Sum of nth term in A.P:-

$\frac{n}{2} \times (2a + (n - 1)d)$

_____________________________

$\large\bf\orange{SOLUTION➛}$

Series:-1,2,3,4,5......................100

a=1

n=100

d=(2-1)=(3-2)=(4-3)......

=1

Now,

$sum = \frac{n}{2} \times (2a + (n - 1)d)$

$sum = \frac{100}{2} \times (2(1) + (100 - 1)(1))$

$sum = 50\times (2 + 99)$

$sum = 50\times 101 \\ = 5050$

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$\large\bf\red{ANSWER➠}$

Sum of the first 100 natural nnumber:-

15050