The 1st,5th and 10th terms of a linear sequence are geometric progression. If the sum of the 2nd and 8th term of the linear sequence is 30,find
i. The 1st term
ii. The non- zero common difference of the linear sequence
According to your wording:
"the sum of the 2nd and 8th term of the linear sequence is 30"
a+d + a+7d = 30
2a + 8d = 30
a + 4d = 15 or a = 15-4d
"1st,5th and 10th terms of a linear sequence are geometric progression"
(a+4d)/a = (a+9d)/(a+4d)
(a+4d)^2 = a(a+9d)
a^2 + 8ad + 16d^2 = a^2 + 9ad
16d^2 = ad
16d^2 = d(15-4d)
16d^2 = 15d - 4d^2
20d^2 - 15d = 0
5d(4d - 3) = 0
d = 0 or d = 3/4
if d = 0 , then a = 15, and the linear sequence would be
15, 15, 15, .... , not very interesting, but it satisfies the conditon
if d= 3/4, a = 15 - 4(3/4) = 12
the sequence would be
12, 51/4, 27/2, ...
sum of 2nd and 8th
= 51/4 + 12+7(3/4) = 51/4 + 69/4 = 30 , as needed
term1 = 12
term5 = 12+4(3/4) = 15
term10 = 12 + 9(3/4) = 75/4
term5/term1 = 15/12 = 5/4
term10/term5 = (75/4) / (15) = 5/4 , as needed
all is fine here.