Question

Find the first term and the common ratio in a geometric progression, given that \(T_4-T_2=240\) and \(T_6-T_4=6000.\)

Answer 1

since it is a geometric progression then
\[ \large T_n=r^{n-1}T_1\qquad\text{for }n\geq2 \].

Answer 2

then
\[ \large 240=T_4-T_2=r^3T_1-rT_1=(r^3-r)T_1 \]
and
\[ \large 6000=T_6-T_4=r^5T_1-r^3T_1=(r^5-r^3)T_1 \]

Answer 3

then
\[ \large T_1=\frac{240}{r^3-r}\qquad\text{and}\qquad
T_1=\frac{6000}{r^5-r^3} \]

Answer 4

can u finish?

Answer 5

Isn't \(T_n=ar^{n-1}?\)