Question

find the sum of all the odd numbers from 1 to 100​

Answer 1

Answer:

he guess here your answer:--

Step-by-step explanation:

can we write this series ......

1+3+5+7+....+99.

late here...

a=1,d= 2,l= 99,n=50

by using this formula

s= n/2[(a +l]

s= 99/2(1+99)

s= 99/2*100

s= 4950.

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Answer 2

Given:-

• An A.P 1, 3, 5, . . . . . , 99

To Find:-

• Sum of the A.P

Formula used:-

• $a_n= a+(n-1)d$

• $S_n=\dfrac{n}{2}(a+l)$

where,

• a = First term

• n = No. of terms

• d = Common Difference

• l = last term

• $a_n=\:n^{th}\:term$

• $S_n=\:Sum\:of\:n^{th}\:term$

Solution:-

Here,

• a = 1

• d = 3 - 1 = 2

• $a_n\:or\:l\:=99$

Now,

$\implies\:a_n= a+(n-1)d$

$\implies\:99= 1+(n-1)2$

$\implies\:99= 1+2n-2$

$\implies\:2n=99+1$

$\implies\:n=\dfrac{100}{2}$

$\implies\:n=50$

Now using,

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

$\implies\:S_n=\dfrac{50}{2}(1+99)$

$\implies\:S_n=25×100$

$\implies\:S_n=2500$

Hence, The sum of all even numbers from 1 to 100 is 2500.

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More Formulas:-

• $S_n=\dfrac{n}{2}[2a+(n-1)d]$

• $S_n=\dfrac{n}{2}[a+a+(n-1)d]$

• $S_n=\dfrac{n}{2}(a+a_n)$

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