Math, asked by Venkatasaipalla1455, 9 months ago

Find the sum of whole numbers from 401 to 500

Answers

Answered by mysticd
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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