document.write( "Question 169946: The sum of even integers between 1 and 48 \n" ); document.write( "

Algebra.Com's Answer #125403 by chiefman(11) $\"\"$ $\"About$

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taking the formula n+1,n+3,n+5....where n=1,2,3,4,........to get even integers we have
\n" ); document.write( "2,4,6,8,.......48 this becomes an arithmetic series ie
\n" ); document.write( "2+4+6+8+......+48 where first term(a)=2,common diffrence(d)=2
\n" ); document.write( "we first find the number of terms in this seies
\n" ); document.write( "using the formula $\"L=%28a%2B%28n-1%29d%29\"$ where L is the last term.
\n" ); document.write( "48=2+(n-1)2
\n" ); document.write( "n=24 therefore there are 24 terms.
\n" ); document.write( "Now we find the the sum using the formula \r
\n" ); document.write( "\n" ); document.write( " $\"S=%28n%2F2%29%282a%2B%28n-1%29d%29%29\"$ where S=sum n=number of terms
\n" ); document.write( "S=[24/2(2*2+(24-1)2]
\n" ); document.write( " =12(4+46)
\n" ); document.write( " =600
\n" ); document.write( "therefore the sum is 600 \n" ); document.write( "

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