# Find the 12th partial sum of the summation of negative 2i minus 10, from i equals 1 to infinity.

A partial sum is basically the sum of part of a sequence. Sometimes partial sums are called "Finite Series".

## Answer: The 12th partial sum of the summation of negative 2i minus 10, from i equals 1 to infinity is -276.

Let's see the type of series given in the question.

**Explanation: **

Given: the summation is: \(x(i) = \sum_{i=1}^{\infty} (-2i-10)\)

Putting i = 1, 2, 3, 4, ...,12 in equation \(x(i) = \sum_{i=1}^{\infty} (-2i-10)\), we get

- x(1) = -2 - 10 = -12
- x(2) = -4 - 10 = -14
- x(3) = -6 - 10 = -16
- ....
- x(12) = -24 - 10 = -34

We need to determine the sum of the first 12 terms of the series \(\sum_{i=1}^{\infty} (-2i-10)\)

The given series is an arithmetic series as the difference between any two consecutive terms is constant, that is, -14-(-12) = -16-(-14) = -2

The sum of an arithmetic sequence when the last term is given is: (n/2) (a+l), where n is the number of terms, a = first term, l = last term

We have n = 12, a = x(1) = -12, l = x(12) = -34

The 12^{th} partial sum of an arithmetic series is given by

S(12) = (12/2) (-12 + (-34)) = 6 × -46 = -276

### Therefore, the 12th partial sum of the summation of negative 2i minus 10, from i equals 1 to infinity is -276.

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