Answer to Question #234837 in Algebra for Bikoye

Question

The first 1st,5th and 10th term of a linear sequence are geometric progression. If the 2nd and 8th term of the linear sequence is 30, find the non-zero common difference of the linear sequence

Answer 1

The 2nd and 8th term of the linear sequence is 30

$a+d+a+7d = 30 \\ 2a + 8d = 30 \\ a + 4d = 15 \\ a = 15-4d$

1st, 5th and 10th terms of a linear sequence are geometric progression

$\frac{(a+4d)}{a} = \frac{(a+9d)}{(a+4d)} \\ (a+4d)^2 = a(a+9d) \\ a^2 + 8ad + 16d^2 = a^2 + 9ad \\ 16d^2 = ad \\ 16d^2 = d(15-4d) \\ 16d^2 = 15d – 4d^2 \\ 20d^2 - 15d = 0 \\ 5d(4d - 3) = 0$

d = 0 or $d = \frac{3}{4}$

if d = 0, then a = 15, and the linear sequence would be

15, 15, 15

if $d= \frac{3}{4}$ , $a = 15 - 4(\frac{3}{4}) = 12$

the sequence would be

$12, \frac{51}{4}, \frac{27}{2}, ...$

sum of 2nd and 8th

$= \frac{51}{4} + 12+7(\frac{3}{4}) = \frac{51}{4} + \frac{69}{4} = 30$

term1 = 12

term5 $= 12+4(\frac{3}{4}) = 15$

term10 $= 12 + 9(\frac{3}{4}) = \frac{75}{4}$

term5/term1 $= \frac{15}{12} = \frac{5}{4}$

term10/term5 $= \frac{(\frac{75}{4}) }{(15)} = \frac{5}{4}$