The 12th term of a sequence is 41 and the 15th term is 140. What is the nth term?
1 answer:
An= mth term.
an=a₁+(n-1)*d
a₁₂=41
a₁₅=140
a₁₂=41
41=a₁+(12-1)*d
41=a₁+11d
a₁+11d=41 (1)
a₁₅=140
140=a₁+(15-1)*d
140=a₁+14d
a₁+14d=140 (2)
With the equiations (1) and (2) build a system of equations
a₁+11d=41
a₁+14d=140
we solve it.
-(a₁+11d=41)
a₁+14d=140
--------------------
3d=99 ⇒d=99/3=33
a₁+11d=41
a₁+(11*33)=41
a₁+363=41
a₁=41-363=-322
an=a₁+(n-1)*d
an=-322+(n-1)*33
an=-322+33n-33
an=-355+33n
an=-355+33n
To check:
a₁₂=-355+33*12=-355+396=41
a₁₅=-355+33*15=-355+495=140.
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Read the question, this problem is not that hard to solve!
Answer:
y = 40 / x^1/3
Step-by-step explanation:
Given that :
y α x^1/3
y = k * 1 / x^1/3
y = k / x^1/3
When x = 125 ; y = 8
8 = k / 125^1/3
Cube root of 125 = 5
8 = k / 5
8 * 5 = k ; k = 40
Hence, expression becomes :
y = 40 / x^1/3
Answer:
c(3c - 1)
Step-by-step explanation:
3c² - c
Common factor = c
Factored:
c(3c - 1)
Rule of trigonometric functions:-
sin a = cos(90 - a)
Here a = 90°.
sin 90 = cos (90 - 90)
1 = cos 0.
cos 0 = 1.
So the cos 0 = 1.
Answer:
Step-by-step explanation:
a+√b
a-√b