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Ugo [173]
3 years ago
12

Which rule can you use to find the nTH term of an arithmetic sequence in which the common difference is 5 and a12 = 63?

Mathematics
1 answer:
snow_tiger [21]3 years ago
5 0
Arithmetic,  nth =  a + (n - 1)d,   common difference, d = 5

a₁₂ = a + (12 -1)d = 63

<span>a + 11d = 63
</span>
a + 11*5 = 63

a = 63 - 11*5

a = 63  - 55

a = 8

So the nth term = a + (n - 1)d = 8<span> + (n - 1)5

8 + 5*(n - 1)

8 + 5n - 5

8 - 5 + 5n

3 + 5n

So nth term = 3 + 5n</span>
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I have 8 questions witch means 80 points answer correctly!
kolezko [41]
<h3>Answer : </h3>

\Large \pink{\tt \dfrac{35}{36}}

<h3>Solution :</h3>

\tt \dfrac{3}{4} + \dfrac{2}{9}

By taking LCM = 4 × 9 = 36

\tt = \dfrac{3 \times 9}{4 \times 9} + \dfrac{2 \times 4}{9 \times 4}

\tt = \dfrac{27}{36} + \dfrac{8}{36}

\tt = \dfrac{27 + 8}{36}

\tt = \dfrac{35}{36}

\pink{\tt \therefore \: Answer \: is \: \dfrac{35}{36}}

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2 years ago
Please help!
erik [133]
For the answer to the question above, 
1 + nx + [n(n-1)/(2-factorial)](x)^2 + [n(n-1)(n-2)/3-factorial] (x)^3 

<span>1 + nx + [n(n-1)/(2 x 1)](x)^2 + [n(n-1)(n-2)/3 x 2 x 1] (x)^3 </span>

<span>1 + nx + [n(n-1)/2](x)^2 + [n(n-1)(n-2)/6] (x)^3 </span>

<span>1 + 9x + 36x^2 + 84x^3 </span>

<span>In my experience, up to the x^3 is often adequate to approximate a route. </span>

<span>(1+x) = 0.98 </span>

<span>x = 0.98 - 1 = -0.02 </span>

<span>Substituting: </span>

<span>1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3 </span>

<span>approximation = 0.834 </span>

<span>Checking the real value in your calculator: </span>

<span>(0.98)^9 = 0.834 </span>

<span>So you have approximated correctly. </span>

<span>If you want to know how accurate your approximation is, write out the result of each in full: </span>

<span>1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3 = 0.833728 </span>

<span> (0.98)^9 = 0.8337477621 </span>

<span>So it is correct to 4</span>
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