document.write( "Question 502147: Isabella sends money to each niece and nephew on his or her b'day. She gives each child $10.00 on his or her 1st b'day. On each b'day thereafter, you and your cousins get $20.00 more than the b'day before. How old will you be when you have rec'd a total of $4,000.00 form Aunt Isabella? \n" ); document.write( "
Algebra.Com's Answer #338820 by Maths68(1474)\"\" \"About 
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1st Birthday I received = $10
\n" ); document.write( "2nd Birthday I received = $30
\n" ); document.write( "3rd Birthday I received = $50
\n" ); document.write( "4th Birthday I received = $70
\n" ); document.write( "And so on
\n" ); document.write( "10+30+50+70+…………. To sum of 4000
\n" ); document.write( "This is an arithmetic progression so we will use the following formula
\n" ); document.write( "s= n/2[2a+(n-1)d]
\n" ); document.write( "where
\n" ); document.write( "s = sum of progression= 4000
\n" ); document.write( "n = number of terms = ?
\n" ); document.write( "a = value of first term = 10
\n" ); document.write( "d = difference between two terms = 20
\n" ); document.write( "Plug in the values in the formula
\n" ); document.write( "s= n/2[2a+(n-1)d]
\n" ); document.write( "4000=n/2[(2*10)+(n-1)20]
\n" ); document.write( "4000=n/2[20+(20n-20)]
\n" ); document.write( "4000=(n/2)(20+20n-20)
\n" ); document.write( "4000=(n/2)(20n)
\n" ); document.write( "4000=20n^2/2
\n" ); document.write( "4000=10n^2
\n" ); document.write( "Divide by 10 both sides of the above equation
\n" ); document.write( "4000/10=10n^2/10
\n" ); document.write( "400=n^2
\n" ); document.write( "Taking square root both sides of the equation
\n" ); document.write( "√400=√n^2
\n" ); document.write( "20=n
\n" ); document.write( "n=20\r
\n" ); document.write( "\n" ); document.write( "On my 20th birthday I will have received total amount of $4000.00 from Aunt Isabella.
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