Onnly the sequences where each term is found by multiplying the previous term by the same common ratio. Example, 5, 10, 20 works because each term is the previous term multiplied by 2.

7 x 3 = 21, 21 x 3 = 63, 63 x 3 = 189

Each term is 1.5 times bigger than the previous term

3 x 5⁵ = 9375

2.5 x (-2)⁹

The general expression for the nth term of a geometric sequence is U_n = U₁ x r^(n-1), where U₁ is the first term, r is the common ratio and n is the postion/number of terms. We can see that the common ratio is 4 and the first term is 2.

6, 12, 24 are the first 3 terms. If you divide the terms 192, 384, 768 by 3 (the first term) you get 64, 128, 256, which are successive powers of 2 and so each of these terms are 3 x a power of 2 and so they must be part of the sequence.

Substitute n = 8, U1 = 4.5 and r = 2 in to the sum formula.

\({ S }_{ n }={ u }_{ 1 }\left( \frac { { r }^{ n }-1 }{ r-1 } \right) \\ in\quad this\quad case,\quad { u }_{ 1 }=4.5\quad and\quad r=2\\ SO,\\ { S }_{ 8 }=4.5\left( \frac { { 2 }^{ 8 }-1 }{ 2-1 } \right) \\ { S }_{ 8 }=4.5\left( \frac { 255 }{ 1 } \right) \\ { S }_{ 8 }=1147.5\)

Substitute n = 20, U₁ = 15 and r = 1.2 in to the sum formula. Then round carefully.

\({ S }_{ n }={ u }_{ 1 }\left( \frac { { r }^{ n }-1 }{ r-1 } \right) \\ in\quad this\quad case,\quad { u }_{ 1 }=15\quad and\quad r=1.2\\ SO,\\ { S }_{ 20 }=15\left( \frac { { 1.2 }^{ 20 }-1 }{ 1.2-1 } \right) \\ { S }_{ 20 }=2800.319994\\ { S }_{ 20 }=2800\quad (3sf)\)

\({ u }_{ 3 }\times { r }\times { r }\times { r }={ u }_{ 6 }\\ { u }_{ 3 }\times { r }^{ 3 }={ u }_{ 6 }\\ 9\times { r }^{ 3 }=-243\\ { r }^{ 3 }=\frac { -243 }{ 9 } =-27\\ r=\sqrt [ 3 ]{ -27 } =-3\)