Find the first four terms of the sequence given by the following.
aₙ=43-3(n-1) n=1,2,3. . .

Given :-

To Find :-

Solution :-

The given expression is

→ aₙ=43-3(n-1)

where n > 0

Finding the first term :

Substituting n = 1 , we have ,

→ T1 = 43 - 3(1-1)

→ T1 = 43 - 3*0

→ T1 = 43 - 0 = 43

Finding the second term :

Substituting n = 2 , we have,

→ T2 = 43 -3(2-1)

→ T2 = 43 -3*1

→ T2 = 43 -3 = 40

Finding the third term :

Substituting n = 3 , we have,

→ T3 = 43 -3(3-1)

→ T3 = 43 -3*2

→ T3 = 43 -6 = 37

Finding the fourth term :

→ T4 = 43 -3(4-1)

→ T4 = 43 -3*3

→ T4 = 43-9 = 34

Hence the first four terms of the sequence are 43 , 40 , 37 and 34 .

I hope this helps . Let me know if you need further clarification .

step-by-step explanation:

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.