Which shows a recursive and an explicit formula for a sequence whose initial term is 14 and whose common difference is −4? A. A(1) = 14; A(n) = (n − 1) −4; A(n) = 14 + (n − 1)(−4) B. A(1) = 14; A(n) = (n + 1) −4; A(n) = 14 + (n − 1)(−4) C. A(1) = 14; A(n) = (n − 1) − 14; A(n) = 14 + (n − 1)(−14) D. A(1) = 14; A(n) = (n + 1) − 14; A(n) = 14 + (n + 1)(14)

A. A(1) = 14; A(n) = (n − 1) −4; A(n) = 14 + (n − 1)(−4)

Step-by-step explanation:

Arithmetic sequence is a sequence that is identified by their common difference. Let a be the first term, n be the number of terms and d be the common difference.

For an arithmetic sequence, common difference 'd' is added to the preceding term to get its succeeding term. For example if a is the first term of a sequence, second term will be a+d, third term will give a+d+d and so on to generate sequence of the form,

a, a+d, a+3d, a+4d...

Notice that each new term keep increasing by a common difference 'd'

The nth term of the sequence Tn will therefore give Tn = a+(n-1)d

If the initial (first) term is 14 and common difference is -4, the nth of the sequence will be gotten by substituting a = 14 and d = -4 in the general formula to give;

Tn = 14+(n-1)-4 (which gives the required answer)

Tn = 14-4n+4

Tn = 18-4n

72 fudge and 75 toffee

step-by-step explanation:

i hope this : )