Question

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

Answer 1

Hint: We recall the dentition of arithmetic progression (AP), the that the ${{n}^{\text{th}}}$ term of an AP ${{x}_{n}}=a+\left( n-1 \right)d$, the sum of the terms in an AP sequence up to ${{n}^{\text{th}}}$ term ${{S}_{n}}=\dfrac{n}{2}\left\{ 2a+\left( n-1 \right)d \right\}$ with common difference $d$ and the first term $a$ . We take $x$ to the right hand side and $-437$ to the left hand side and find $x$ 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 $\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...$ is an AP, then ${{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}...$ . The difference between two terms is called common difference and denoted $d$ where $d={{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}...$. The first term ${{x}_{1}}$ is conventionally denoted by $a$.
We know that the ${{n}^{\text{th}}}$ term of an AP with common difference $d$ and the first term $a$ is given by
\[{{x}_{n}}=a+\left( n-1 \right)d\]
The sum of the terms in an AP sequence up to ${{n}^{\text{th}}}$ term is given by;
\[{{S}_{n}}=\dfrac{n}{2}\left\{ 2a+\left( n-1 \right)d \right\}\]
We are given the equation $-4+\left( -1 \right)+2+...+x=-437$ and we are asked to solve it which means we have to find the value of $x$. Let us take $x$ to the right hand side of the equation and 437 to the left hand side. We have;
\[-4+\left( -1 \right)+2+...+437=-x\]
We multiply $-1$ both sides of above equation to have;
\[4+1+\left( -2 \right)+...-437=x\]
We see that the left hand side of the above equation is a decreasing AP with first term 4 and common difference$1-4=-2-1=-3$. The value of $x$ is the sum of up to $-437$. We shall use the sum of the terms in an AP sequence up to ${{n}^{\text{th}}}$ term but for that we need the value of $n$. So let $-437$ be ${{n}^{\text{th}}}$ term of the AP and then we use formula for the ${{n}^{\text{th}}}$ term of an AP with common difference $d=-3$ and the firs term $a=4$ and have;
\[\begin{align}
  & -437=4+\left( n-1 \right)\left( -3 \right) \\
 & \Rightarrow -441=\left( -3 \right)\left( n-1 \right) \\
 & \Rightarrow n-1=\dfrac{-441}{-3}=147 \\
 & \Rightarrow n=147+1=128 \\
\end{align}\]
Now we use the formula for sum up to ${{n}^{\text{th}}}={{128}^{\text{th}}}$ term and have;
\[\begin{align}
  & {{S}_{128}}=\dfrac{128}{2}\left\{ 2\times 4+\left( 128-1 \right)\left( -3 \right) \right\} \\
 & \Rightarrow {{S}_{128}}=64\left\{ -373 \right\} \\
 & \Rightarrow {{S}_{128}}=-23872 \\
\end{align}\]

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 $x$.