Find the sum
$1 + 3 + 5 + 7 + . . . . . . + 199$ .

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Solution

Sum of $1 + 3 + 5 + 7 + . . . . . + 199$

$a = 1$ $d = 2$

Finding $n^{t h}$ term in A.P; $a_{n} = a + (n - 1) d$

If $a_{n} = 199$ then $199 = 1 + (n - 1) 2$

$\Rightarrow n = 100$

Therefore, no. of terms $= n = 100$

sum of 100 terms = $\frac{n}{2} [a + a_{n}] = \frac{100}{2} [1 + 199]$

$= \frac{100}{2} [200]$

$= 10000$

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Similar questions

Find the sum of the terms of the finite AP :
1, 3, 5, 7,…,199

199 \cdot 1 + 197 \cdot 3 + 195 \cdot 5 + . . . . . + 103 \cdot 197 + 101 \cdot 99

N = \frac{1}{3} + \frac{1}{15} + \frac{1}{35} . . . . . . . . .100

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