Question

Graph the six terms of a finite sequence where a1 = −6.25 and r = 1.25. Help pls i need this quick!!!!! graphed series showing point 1, 2.25, point 2, 3.75, point 3, 8.25, point 4, 21.75, point 5, 62.25, and point 6, 183.75 graphed series showing point 1, 1.5, point 2, 3.75, point 3, 10.5, point 4, 30.75, point 5, 91.5, and point 6, 273.75 graphed series showing point 1, negative 9, point 2, negative 12, point 3, negative 16.5, point 4, negative 23.25, point 5, negative 33.38, and point 6, negative 48.56 A coordinate graph is shown. The points are at ordered pairs, 1, negative 6.25, 2, negative 7.813, 3, negative 9.766, 4, negative 12.207, 5, negative 15.259, and 6, negative 19.073

Answer 1

Answer:

S = {-6.25, -7.813, -9.766, -12.207, -15.259 and -19.703}

Step-by-step explanation:

The first term (a₁) and the common ration (r) of a finite sequence are provided.

a₁ = -6.25

r = 1.25

The nth term of a geometric sequence is:

$T_{n}=a_{1}\cdot r^{n-1}$

Compute the first six terms as follows:

$T_{1}=a_{1}\cdot r^{1-1}=a_{1}=-6.25\\\\T_{2}=a_{1}\cdot r^{2-1}=-6.25\times 1.25=-7.8125\approx -7.813\\\\T_{3}=a_{1}\cdot r^{3-1}=-6.25\times (1.25)^{2}=-9.765625\approx -9.766\\\\T_{4}=a_{1}\cdot r^{4-1}=-6.25\times (1.25)^{3}=-12.20703\approx -12.207\\\\T_{5}=a_{1}\cdot r^{5-1}=-6.25\times (1.25)^{4}=-15.25879\approx -15.259\\\\T_{6}=a_{1}\cdot r^{6-1}=-6.25\times (1.25)^{5}=-19.07349\approx -19.073$

Thus, the six terms of a finite sequence are:

S = {-6.25, -7.813, -9.766, -12.207, -15.259 and -19.703}