Question

how to find the sum of 1/100 1/101 1/102 1/103 ..... 1/200

Answer 1

To find sum=1/100,1/101,1/102,1/103,------,1/200
so,103-102=1 and 101-1=100
so,1/100+0/100=1/200

Answer 2

The sum of $\dfrac{1}{100}, \dfrac{1}{101},\dfrac{1}{102},\dfrac{1}{103} .....,\dfrac{1}{200}$ = 0.75

Step-by-step explanation:

The given series are:

$\dfrac{1}{100}, \dfrac{1}{101},\dfrac{1}{102},\dfrac{1}{103} .....,\dfrac{1}{200}$

Here, first term(a) $=\dfrac{1}{100}$, last term(l) $=\dfrac{1}{200}$ and

number of terms(n) = 100

To find, the sum of $\dfrac{1}{100}, \dfrac{1}{101},\dfrac{1}{102},\dfrac{1}{103} .....,\dfrac{1}{200}$ = ?

We know that,

The sum of nth term of an AP

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

$=\dfrac{100}{2}(\dfrac{1}{100}+\dfrac{1}{200})$

$=\dfrac{100}{2}\times \dfrac{1}{100}(1+\dfrac{1}{2})$

$=\dfrac{1}{2}\times \dfrac{3}{2}$

$=\dfrac{3}{4}$

= 0.75

Hence, the sum of $\dfrac{1}{100}, \dfrac{1}{101},\dfrac{1}{102},\dfrac{1}{103} .....,\dfrac{1}{200}$ = 0.75