Question
Find the sum:1 + 3 + 5 + 7 + ..... + 199.

Answer

Given,
A.P. 1 + 3 + 5 + 7 + ..... + 199.
Here,
First term, a = 1
Difference, d = 3 - 1 = 2
and Last term $a_n = 199$
$We know, a_n = a + (n - 1)d$
$\Rightarrow 199 = 1 + (n - 1)2$
$\Rightarrow 199 = 1 + 2n - 2$
$\Rightarrow 200 = 2n$
$\Rightarrow n = 100$
We know, Sum of n terms,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ \text{S}_{100}=\frac{100}{2}[2(1)+(100-1)2]$
$\Rightarrow S_{100} = 50[2 + 99 \times 2]$
$\Rightarrow S_{100} = 50[2 + 198]$
$\Rightarrow S_{100} = 50[200]$
$\Rightarrow S_{100} = 10000.$
Hecne, Sum of given A.P. is 10000.

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