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Question: Solve the following equation: \(-4+\left( -1 \right)+2+...+x=-437\). \[\]...

Solve the following equation: −4+(−1)+2+...+x=−437-4+\left( -1 \right)+2+...+x=-437. $$$$

Explanation

Solution

We recall the dentition of arithmetic progression (AP), the that the nth{{n}^{\text{th}}} term of an AP xn=a+(n−1)d{{x}_{n}}=a+\left( n-1 \right)d, the sum of the terms in an AP sequence up to nth{{n}^{\text{th}}} term {{S}_{n}}=\dfrac{n}{2}\left\\{ 2a+\left( n-1 \right)d \right\\} with common difference dd and the first term aa . We take xx to the right hand side and −437-437 to the left hand side and find xx as a sum of the terms in an AP. $$$$

Complete step-by-step answer:
Arithmetic sequence otherwise known as arithmetic progression, abbreviated as AP is a type mathematical sequence where the difference between any two consecutive numbers is constant. If (xn)=x1,x2,x3,...\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},... is an AP, then x2−x1=x3−x2...{{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}... . The difference between two terms is called common difference and denoted dd where d=x2−x1=x3−x2...d={{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}.... The first term x1{{x}_{1}} is conventionally denoted by aa.
We know that the nth{{n}^{\text{th}}} term of an AP with common difference dd and the first term aa is given by
xn=a+(n−1)d{{x}_{n}}=a+\left( n-1 \right)d
The sum of the terms in an AP sequence up to nth{{n}^{\text{th}}} term is given by;
{{S}_{n}}=\dfrac{n}{2}\left\\{ 2a+\left( n-1 \right)d \right\\}
We are given the equation −4+(−1)+2+...+x=−437-4+\left( -1 \right)+2+...+x=-437 and we are asked to solve it which means we have to find the value of xx. Let us take xx to the right hand side of the equation and 437 to the left hand side. We have;
−4+(−1)+2+...+437=−x-4+\left( -1 \right)+2+...+437=-x
We multiply −1-1 both sides of above equation to have;
4+1+(−2)+...−437=x4+1+\left( -2 \right)+...-437=x
We see that the left hand side of the above equation is a decreasing AP with first term 4 and common difference1−4=−2−1=−31-4=-2-1=-3. The value of xx is the sum of up to −437-437. We shall use the sum of the terms in an AP sequence up to nth{{n}^{\text{th}}} term but for that we need the value of nn. So let −437-437 be nth{{n}^{\text{th}}} term of the AP and then we use formula for the nth{{n}^{\text{th}}} term of an AP with common difference d=−3d=-3 and the firs term a=4a=4 and have;

& -437=4+\left( n-1 \right)\left( -3 \right) \\\ & \Rightarrow -441=\left( -3 \right)\left( n-1 \right) \\\ & \Rightarrow n-1=\dfrac{-441}{-3}=147 \\\ & \Rightarrow n=147+1=128 \\\ \end{aligned}$$ Now we use the formula for sum up to ${{n}^{\text{th}}}={{128}^{\text{th}}}$ term and have; $$\begin{aligned} & {{S}_{128}}=\dfrac{128}{2}\left\\{ 2\times 4+\left( 128-1 \right)\left( -3 \right) \right\\} \\\ & \Rightarrow {{S}_{128}}=64\left\\{ -373 \right\\} \\\ & \Rightarrow {{S}_{128}}=-23872 \\\ \end{aligned}$$ **Note:** We can alternatively solve using the sum of terms formula $S=\dfrac{n}{2}\left( a+l \right)$ where $l$ is the last term of the AP. We must be careful of the confusion for sum of first $n$ terms of an AP from GP which is given by $\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ where $r$ is the common ratio between two terms. Most mistakes happen here by finding the value of $-x$ instead of $x$.