Question

1-2+3-4+5-6+........+99-100

Answer 1

Answer:

Step-by-step explanation:

By grouping in pairs (1–2) + (3–4) + (5–6) + …, + (97–98) + (99-100) each of which equals -1, you will have a total of -50.

Answer 2

Answer:

Step-by-step explanation:

Concept:

Arithmetic Progression:

The difference between any two consecutive integers in an arithmetic progression (AP) sequence of numbers is always the same amount. It also goes by the name Arithmetic Sequence. For instance, the natural number sequence 1, 2, 3, 4, 5, 6,... is an example of an arithmetic progression. It has a common difference of 1 between two succeeding terms (let's say 1 and 2). (2 -1). We can see that the common difference between two subsequent words will be equal to 2 in both the case of odd and even numbers.

A progression of numbers known as an arithmetic sequence is one in which, for every pair of consecutive terms, the second number is derived by adding a predetermined number to the first one.

The common difference of the AP is the constant amount that must be multiplied by any phrase in order to obtain the subsequent term.

Given: The series $1-2+3-4+5-6+........+99-100$

Find: To solve $1-2+3-4+5-6+........+99-100$

Solution:

Let  $S =$ $1-2+3-4+5-6+........+99-100$

The above series can be split into $2$ series as:

$S = (1+3+5+7+.....+99) - (2+4+6+8+.....+100)$

Let first split be $S1$ and second split be $S2$

$S1$ be an AP with $A= 1 , D = 2 , N = 50 , L = 99$

$S1 = N/2 * [A+L] = 50/2 * [1+99] = 2500$

$S2$ is an AP with $A= 2 , D = 2 , N = 50 , L = 100$

$S2 = N/2*[A+L] = 50/2*[2+100] = 2550$

$S = S1 -S2$ $= 2500 - 2550 = -50$

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