SOLUTION: I have a maths c/w on Pythagorean triples (gcse level) and we are using the quadratic formula: . a + d (n-1) + 0.5 c (n-1) (n-2) . for the sequence of Pythagorean triples multi

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Question 117346: I have a maths c/w on Pythagorean triples (gcse level) and we are using the quadratic formula:
.
a + d (n-1) + 0.5 c (n-1) (n-2)
.
for the sequence of Pythagorean triples multiplied by 4.
where:
.
a = first term
d = first difference
c = second difference
.
length of middle side: 16, 48, 96, 160, 240
.
I have done so far:
.
a = 16
d = 32
c = 16
.
substitute:
.
16 + 32 (n-1) + 0.5 * 16 ( n-1) (n-2)
.
giving me:
.

.
simplified:
.

.
BUT when i am checking it against the third term (96) the 16 at the end of the equation is
left as extra and the answer I seem to get is 112 (an extra plus 16)
.
please help me urgently
Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website!
I am not completely sure of what you are doing in this problem, but it appears to me that you
have made a mistake as noted below. If this answer doesn't help you, please re-post the
problem and maybe another tutor will be able to get you through it.
.
You wrote:
.
I have a maths c/w on Pythagorean triples (gcse level) and we are using the quadratic formula:
.
a + d (n-1) + 0.5 c (n-1) (n-2)
.
for the sequence of Pythagorean triples multiplied by 4.
where:
.
a = first term
d = first difference
c = second difference
.
length of middle side: 16, 48, 96, 160, 240
.
I have done so far:
.
a = 16
d = 32
c = 16
.
substitute:
.
<==== here it appears that you forgot to use the 0.5 multiplier
.
giving me:
.
<=== the 0.5 multiplier would make this
.
simplified:
.
<=== this then should be
.
Hope this helps you complete the problem correctly