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.

add all the numbers and check the ones place.

the answer is 486. as the ones place is more than 4, the estimation will be the next 10 ie, 490

answer: i think it's a

step-by-step explanation: