Asked by Juma
In geometric progression sum of 2nd term and 3rd term is 12 and sum of 4th term and 5th term is 300 find first term and common ratio
MATHEMATICS
MATHEMATICS
Answers
R_scott
a r + a r^2 = 12 ... a r (1 + r) = 12
a r^3 + a r^4 = 300 ... a r^3 (1 + r) = 300
dividing equations ... r^2 = 25 ... r = 5
substitute back to find a
a r^3 + a r^4 = 300 ... a r^3 (1 + r) = 300
dividing equations ... r^2 = 25 ... r = 5
substitute back to find a
mathhelper
ar +ar^2 = 12 ---> ar(1+r) = 12
ar^3 + ar^4 = 300 --> ar^3(1+r) = 300
divide the 2nd equation by the first:
r^2 = 300/12 = 25
r = ± 5
sub into ar(1+r) = 12
<b>if r = 5, 5a(6) = 12 ----> a = 2/5</b>
<b>if r = -5, -5a(-4) = 12 ----> a = 3/5</b>
check:
case 1: a = 2/5, r = 5, we have 2/5, 2, 10, 50, 250, ...
sum of 2nd and 3rd = 12, sum of 4th and 5th = 300, check!
case 2. a = 3/5, r = -5, we have 3/5, -3, 15, -75, 375
sum of 2nd and 3rd = -3+15 = 12, sum of 4th and 5th = -75+375 = 300,
my answers are correct
ar^3 + ar^4 = 300 --> ar^3(1+r) = 300
divide the 2nd equation by the first:
r^2 = 300/12 = 25
r = ± 5
sub into ar(1+r) = 12
<b>if r = 5, 5a(6) = 12 ----> a = 2/5</b>
<b>if r = -5, -5a(-4) = 12 ----> a = 3/5</b>
check:
case 1: a = 2/5, r = 5, we have 2/5, 2, 10, 50, 250, ...
sum of 2nd and 3rd = 12, sum of 4th and 5th = 300, check!
case 2. a = 3/5, r = -5, we have 3/5, -3, 15, -75, 375
sum of 2nd and 3rd = -3+15 = 12, sum of 4th and 5th = -75+375 = 300,
my answers are correct