Each term in a sequence of numbers, except for the 1st term, is 2 less than the square root of the previous term.If the 3rd term of this sequence is 1, what is the first term?
2 answers
121
according to your wording of the problem,
term(n) = √term(n-1) - 2
or
√term(n-1) = term(n) + 2
square both sides
term(n-1) = term(n) ^2 + 4term(n) + 4
for n = 3, term(3) = 1
term(2) = 1^2 + 4(1) + 4
= 9
term(4) = √term(3) - 2 = -1
so we can't go any higher than term(4)
since term(5) = √-1 - 2 , which is undefined in the set of real numbers.
if we follow the rule,
term(1) = term(2) ^2 + 4term(2) + 4
= 81 + 4(9) + 4 = 121
But you said it does not follow the rule for term(1), so I guess term(1) shall remain a mystery since you state no rule for term(1)
Check your typing
term(n) = √term(n-1) - 2
or
√term(n-1) = term(n) + 2
square both sides
term(n-1) = term(n) ^2 + 4term(n) + 4
for n = 3, term(3) = 1
term(2) = 1^2 + 4(1) + 4
= 9
term(4) = √term(3) - 2 = -1
so we can't go any higher than term(4)
since term(5) = √-1 - 2 , which is undefined in the set of real numbers.
if we follow the rule,
term(1) = term(2) ^2 + 4term(2) + 4
= 81 + 4(9) + 4 = 121
But you said it does not follow the rule for term(1), so I guess term(1) shall remain a mystery since you state no rule for term(1)
Check your typing
1 answer