If the 4th and 7th terms of a GP are 250 and 31250 respectively. Find the two possible values of a and r

a = 2, r = 5

Step-by-step explanation:

Nth term of a GP = a×r^(n-1)

Where 'a' is the first term and 'r' is the common ratio

4th term = a×r^3 = 250

r^3 = 250/a

7th term = a×r^6 = 31250

a×r^6 = 31250

a×(r^3)^2 = 31250

a×(250/a)^2 = 31250

a×(62500/a^2) = 31250

62500/a = 31250

a = 62500/31250 = 2

a = 2

since r^3 = 250/a,

r^3 = 250/2 = 125

r = (125)^(1/3)

r = 5

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option

idc

step-by-step explanation:

idk