Question

Find the sum of whole numbers from 401 to 500

Answer 1

$Whole \: numbers \:from \: 401 \:to \: 500$

$401, 402, 403, \cdot\cdot\cdot ,500 \: is \\an\: A .P$

$First \:term (a) = 401$

$Common \: difference (d) = a_{3} - a_{2} \\= 403 - 402 = 1$

$\boxed { \pink { n^{th} \: term (a_{n}) = a + (n-1)d }}$

$\implies i )a_{n} = 500 \: ( given )$

$\implies a + (n-1)d = 500$

$\implies 401 + ( n - 1 )1= 500$

$\implies 401 + n - 1 = 500$

$\implies 400 + n = 500$

$\implies n = 500 - 400$

$\implies n = 100$

$Here , a = 401 , n = 100 \:and \: a_{n} = 500$

$\boxed { \pink { Sum \:of \: n \:terms (S_{n} ) = \frac{n}{2} ( a + a_{n} ) }}$

$\implies S_{100} = \frac{100}{2}[ 401 + 500 ] \\= 50 \times 901 \\= 45050$

Therefore.,

$\red{ sum\: of \:whole\: numbers\: from\: 401}\\\red{ to\: 500} \green { = 45050 }$

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