Express the terms of the following geometric sequence recursively. 7 , 14 , 28 , 56 , 112 , ...
1 .t1 = 7 and tn = 2(tn-1) - 7, for n > 2
2. t1 = 7 and tn = tn-1 + 7, for n > 2
3. t1 = 7 and tn = tn-1 + 2n, for n > 2
4. t1 = 7 and tn = 2(tn-1), for n > 2

The most correct option for the recursive expression of the geometric sequence is;

4. t₁ = 7 and tₙ = 2·tₙ₋₁, for n > 2

Step-by-step explanation:

The general form for the nth term of a geometric sequence, aₙ is given as follows;

aₙ = a₁·r⁽ⁿ⁻¹⁾

Where;

a₁ = The first term

r = The common ratio

n = The number of terms

The given geometric sequence is 7, 14, 28, 56, 112

The common ratio, r = 14/7 = 25/14 = 56/58 = 112/56 = 2

r = 2

Let, 't₁', represent the first term of the geometric sequence

Therefore, the nth term of the geometric sequence is presented as follows;

tₙ = t₁·r⁽ⁿ⁻¹⁾ = t₁·2⁽ⁿ⁻¹⁾

tₙ =  t₁·2⁽ⁿ⁻¹⁾ = 2·t₁2⁽ⁿ⁻²⁾ = 2·tₙ₋₁

∴ tₙ = 2·tₙ₋₁, for n ≥ 2

Therefore, we have;

t₁ = 7 and tₙ = 2·tₙ₋₁, for n ≥ 2.

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:

my acc plz

step-by-step explanation: