Question

A sequence is defined recursively using the equation f(n + 1) = f(n) – 8. If f(1) = 100, what is f(6)?

Answer 1

If the sequence f is defined as

$\mathsf{f(n+1)=f(n)-8}$


then

$\mathsf{f(n+1)-f(n)=-8}$


As you can see above, the difference between two consecutive terms gives us a constant.


So f is actually an arithmetic progression (AP), and it has (– 8) as its common difference:

$\mathsf{f(n)=f(1)+(n-1)\cdot d}\\\\ \textsf{where }\mathsf{d=-8.}$


•   If f(1) = 100, then

$\mathsf{f(6)=f(1)+(6-1)\cdot (-8)}\\\\ \mathsf{f(6)=100+5\cdot (-8)}\\\\ \mathsf{f(6)=100-40}\\\\ \boxed{\begin{array}{l} \mathsf{f(6)=60} \end{array}}\qquad\checkmark$


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I hope this helps. :-)