Identify the following series as arithmetic, geometric, both, or neither. 3a + 3a² + 3a³ + . . . + 3an geometric arithmetic neither both

solution

given=3a+3a^2+3a^3++...+...3an

first term(t1)=3a

second term(t2)=3a^2

third term(t3)=3a^3

for arithmetic mean ,d1=d2

d1=t2_t1=3a^2_3a=3a(a_1)

d2=3a^3_3a^2=3a^2(a_1)

here d1 is not equal to d2.

so it's not an arithmetic series.

Again,for geometric mean,r1=r2

r1=d2/d1=3a^2/3a=a

r2=d3/d2=3a^3/3a^2=a

here ,r1 is equal to r2.

so ,it's an geometric series.

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1