Question 844488: Find five consecutive terms of an a.p whose sum is 45 and sum of cubes of 2nd and 4th term is 1944
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! S = ½(a + an)n
45 = ½(2a + (5-1)d)5
an = a1 + (n - 1)d
9=a + 2d,
b^3+c^3=1944,
a2=b = a + d,
a4=c = a + 3d,
two answers
a1 = 3, a2= b = 6, a4=c = 12, d = 3
3,6,9,12,15
S = ½(a + an)n
s=1/2(3+15)*5
s=1/2(18)*5
s=45
6^3+12^3
216+1728=1944
and
a1 = 15, a2= b = 12, a4= c = 6, d = -3
15,12,9,6,3
same numbers in reverse
a2 and a4 are again 12 and 6
same results
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