The 5th,9th and 16th terms of a linear sequence (A.P) are the consecutive terms of an exponential sequence (G.P)
(i) find the common difference of the linear sequence in terms of the first term
(ii)show that the 21st,37th and 65th terms of the linear sequence are also in an exponential sequence and hence find the common ratio
3 answers
Pls help me solve
just use your formulas.
(a+8d)/(a+4d) = (a+15d)/(a+8d)
3a = 4d
d = 3/4 a
Now show that the other three terms also have a common ratio
(a+8d)/(a+4d) = (a+15d)/(a+8d)
3a = 4d
d = 3/4 a
Now show that the other three terms also have a common ratio
for the AP,
term5 = a+4d
term9 = a+8d
term16 = a + 15d
but these are now in a GP
(a+8d)/(a+4d) = (a+15d)/(a+8d)
(a+8d)^2 = (a+4d)(a+15d)
a^2 + 16ad + 64d^2 = a^2 + 19ad + 60d^2
3ad - 4d^2 = 0
d(3a - 4d) = 0
d = 0 which does not give an interesting AP
or
d = 3a/4
repeat with:
(a+36d)^2 = (a+20d)(a+64d)
a^2 + 72ad +1296d^2 = a^2 + 84ad + 1280d^2
16d^2 - 12ad = 0
4d^2 - 3ad = 0
d = 3a/4 , same result
r = term9/term5 = (a+8d)/(a+4d)
= (a + 8(3a/4))/(a + 4(3a/4) = 7a/4a = 7/4
term5 = a+4d
term9 = a+8d
term16 = a + 15d
but these are now in a GP
(a+8d)/(a+4d) = (a+15d)/(a+8d)
(a+8d)^2 = (a+4d)(a+15d)
a^2 + 16ad + 64d^2 = a^2 + 19ad + 60d^2
3ad - 4d^2 = 0
d(3a - 4d) = 0
d = 0 which does not give an interesting AP
or
d = 3a/4
repeat with:
(a+36d)^2 = (a+20d)(a+64d)
a^2 + 72ad +1296d^2 = a^2 + 84ad + 1280d^2
16d^2 - 12ad = 0
4d^2 - 3ad = 0
d = 3a/4 , same result
r = term9/term5 = (a+8d)/(a+4d)
= (a + 8(3a/4))/(a + 4(3a/4) = 7a/4a = 7/4
4 answers