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2021-08-16

To find: The sum of the given arithmetic sequence.

Given: the sequence$\sum n=190(3-2n)$

Given: the sequence

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Skilled2021-08-17Added 95 answers

Formula used:

The formula for the summation of a polynomial with degree 1 is,$\sum k=1nk=n2(n+1)\dots .\left(1\right)$

The formula for the summation of a constant is,$\sum k=1nc=cn\dots ..\left(2\right)$

Calculation:

To find the sum of the arithmetic sequence, i.e, to find,

$Sn=\sum n=190(3-2n)$

Write$\sum n=1903-2n$ as follows,

$\sum n=190(3-2n)=\sum n=1903-2\sum n=190n$

Now to find$\sum n=1903$ ,

Substitute the values of n in equation (2)

We get,

$\sum n=1903$

$=3\left(90\right)$

$=270$

Now to find the$2\sum n=190n$

Substitute the values of n in equation (1).

We get,

$2\sum n=190n$

$=2\left(90(90+1)2\right)$

$=2(90\times 912)$

$=8190$

Then,

$\sum n=1903+2\sum n=190n=270-8190$

$=-7920$

Hence

$\sum n=190(3-2n)=-7920$ .

The formula for the summation of a polynomial with degree 1 is,

The formula for the summation of a constant is,

Calculation:

To find the sum of the arithmetic sequence, i.e, to find,

Write

Now to find

Substitute the values of n in equation (2)

We get,

Now to find the

Substitute the values of n in equation (1).

We get,

Then,

Hence

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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