a + ar = 28
ar^2 + ar^3 = 252
a(1+r) = 28
ar^2(1+r) = 252
ar^2(1+r)/a(1+r) = r^2 = 252/28 = 9
r = 3 or -3
a(1-3) = 28 ==> a = -14
a(1+3) = 28 ==> a = 7
So, now we need to know that
ar^6 = 5103
7*3^6 = 5103
Unfortunately, that works whether r is 3 or -3
Using r = -3
7+(-21) = -14
7+21 = 28
So, r=3
7r = 21
The bit about the 7th term is redundant and unhelpful.
the sum of the first two terms of a GP is 28,but the sum of the third and fourth terms is 252.find the second term of the gp if the seventh term is 5103. please help.
8 answers
Thanks for the solution
Thanks 4 d answer
Thanks for the reasonable solution
Nice one
How can I get the second term of the gp if the seventh term is 5103
Thanks for the help
thank for the help