Question

Which of the following functions describes the sequence 4, –2, 1, –12, 14, . . .? A. f(1) = 4, f(n + 1) = –2f(n) for n ≥ 1 B. f(1) = 4, f(n + 1) = f(n) – 2 for n ≥ 1 C. f(1) = 4, f(n) = 12f(n + 1) for n > 1 D. f(1) = 4, f(n) = –12f(n – 1) for n > 1

Answer 1

Answer: D. $f(1) = 4,\ \ f(n+1)=\dfrac{-1}{2}f(n)$

Step-by-step explanation:

The given sequence:  $4, -2,1,\dfrac{-1}{2},\dfrac14,....$

Here, first term: $f(1)=4$

Second term: $f(2)=-2$

Third term : $f(3)=\dfrac{-1}{2}$

It can be observed that it is neither increasing nor decreasing sequence but having the common ratio.

Common ratio: $r=\dfrac{f(2)}{f(1)}=\dfrac{-2}{4}=\dfrac{-1}{2}$

So, $f(n+1)=\dfrac{-1}{2}f(n)$  [as in G.P. nth term= $a_{n+1}=ar^n$]

Hence, correct option is D. $f(1) = 4,\ \ f(n+1)=\dfrac{-1}{2}f(n)$