+128731 So...we need to find "d"
The sum of the first n terms of an arithmetic series is given by:
S = (n/2)[a1 + a1 + d(n-1) ] .... so we have
(8/2)[a1 + a1 +d(8 - 1)] = 3* (4/2) [ a1 + a1 + d(4 - 1) ]
4 [2a1 + d(7)] = 6 [ 2a1 + d(3) ]
8a1 + 28d = 12a1 + 18d
10d = 4a1
d = (4/10)a1 = (2/5)a1
I'm not overly sure about this one....but I'll submit it for review/criticism....!!!
+128731 So...we need to find "d"
The sum of the first n terms of an arithmetic series is given by:
S = (n/2)[a1 + a1 + d(n-1) ] .... so we have
(8/2)[a1 + a1 +d(8 - 1)] = 3* (4/2) [ a1 + a1 + d(4 - 1) ]
4 [2a1 + d(7)] = 6 [ 2a1 + d(3) ]
8a1 + 28d = 12a1 + 18d
10d = 4a1
d = (4/10)a1 = (2/5)a1
I'm not overly sure about this one....but I'll submit it for review/criticism....!!!
