What is a1 of the arithmetic sequence for which a3=10 and a51=−182?

a₁ = 18

Step-by-step explanation:

The n th term of an arithmetic sequence is

= a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Given a₃ = 10 and a₅₁ = - 182, then

a₁ + 2d = 10 → (1)

a₁ + 50d = - 182 → (2)

Subtract (1) from (2) term by term to eliminate a₁

48d = - 192 ( divide both sides by 48 )

d = - 4

Substitute d = - 4 into (1) and evaluate for a₁

a₁ + 2(- 4) = 10

a₁ - 8 = 10 ( add 8 to both sides )

a₁ = 18

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

24 units

step-by-step explanation: