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Oksana_A [137]
3 years ago
9

Given the sequence 8, 12, 16, 20, 24, what is the sum of the 31st and 19th terms?

Mathematics
1 answer:
erica [24]3 years ago
7 0

Answer:

First term (a) =8

Common difference (d)= t2-t1

=12-8

=4

Now, sum of first 31th term (tn31) =n/2{2a+(n-1)d}

= 31/2{2×8+(31-1)4}

=31/2{16+(30×4)

=31/2(16+120)

=31/2×126

=31×63

Step-by-step explanation:

Similarly use 19 as (n) for the 19th term

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Answer:

\frac{4x+2}{x^2 - 9x + 8} = \frac{4x}{(x-8)(x-1)} + \frac{2}{(x-8)(x-1)}

Step-by-step explanation:

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\frac{4x+2}{x^{2}-9+8 } = \frac{A}{()(x-1)} + \frac{B}{()(x-8)}

Required

Fill in the gaps

Going by the given parameters, we have that

\frac{4x+2}{x^{2}-9+8 } = \frac{A}{()(x-1)} + \frac{B}{()(x-8)}

x^2 - 9x + 8, when factorized is (x-1)(x-8)

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B = 2

Hence, the complete expression is

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