+5 When a new school first opened, a students started at the school. Each year, the number of students increases by the same amount, d.
At the beginning of its 7th year, it had 449 students. Form an equation for a in terms of d?
At the end of the 13th year, the school had 750 students. Use this to form an equation relating a and d?
+37103 a = #students first year
a + d = #students SECOND year
for all of the following 'n' years
#students = a + (n-1) d
For 7 years #students = 449 = a + (6)d (1)
For 13 years #students = 750 = a + (12)d (2)
From (1) a = 449 - 6d
Substitute into (2): 750 = (449 - 6d) + 12d and solve for 'd'
750-449 = -6d + 12d
301 = 6 d
d=50 1/6
(not sure how you add 50 1/6 students per year...must be an AVERAGE per year)
Substitute this value of 'd' into one of the equations to find 'a' = ?
(1) 449 = a + 6 (50 1/6)
449 = a + 301
449 - 301 = a a = 148
The difference between the 13th term and the 6th term is:
13 - 6 =7 terms
The difference between the values of the two terms is:
750 - 449 =301
Therefore, the common difference between the two terms is:
301/7=43
a(n) =a(1) + (n-1)d...................(1)
449 =a(1) + (6-1)43
449=a(1) + 215
a(1) =449 -215
a(1) =234
a(n) =a(1) + (n - 1)d.................(2)
750=a(1) + (13- 1)43
a(1) =750 - 516
a(1) =234
+37103 I'll have to disagree with Guest #2.
Your equations do not work if d=43 and a = 234
a(n) =a(1) + (n-1)d results in a(7) = 492 the correct answer needs to be 449
Your mistake is here: 449 =a(1) + (6-1)43 where '6' should be '7' AND at year 7 the students are at 449 SIX (not SEVEN as you described)more terms later , at year 13, the total is 750.
Soooooo the 'common difference' as you describe it is (750-449)/6 = 50 1/6 (NOT 43).
Honest mistakes we all make (frequently!)
The young lady says "At the beginning of its 7th year, it had 449 students". Which is the same as saying "At the end of the 6th year, it had 449 students"!!!
+37103 Still disagree. At the beginning of the 7th year 'd' students have been added. At the beginning of the 2nd year 'd' students have been added. We'll just have to agree, as the English language is full of 'nuances' and unclear meaning, to disagree. G'Day! Thanx !
+129840 At the end of the 6th year [= the beginning of the 7th ], there were 449 students
At the end of the 12th year [ = the beginning of the 13th] , there were 750 students
So we have these two equations where a1 is the beginning enrollment [in the first year] and d is the yearly [constant ] increase
449 = a1 + d(6 -1) → 449 = a1 + 5d (1)
750 = a1 + d(13 - 1) → 750 = a1 + 12d (2)
Subtract (1) from (2)
301 = 7d divide both sides by 7
43 = d this is the yearly increase
And using (1) we can find the beginning enrollment, a1
449 = a1 + 5(43)
449 = a1 + 215 subtract 215 from each side
234 = a1
So.......the equation giving the number of students for any year, n =
an = 234 + 43 (n - 1)
