Question

What is the value of $ 1-2+3-4+5-...+101 $ ?
A. 51
B. 55
C. 110
D. 111

Answer 1

Hint: Here, we have been asked to evaluate the value of $ 1-2+3-4+5-...+101 $ . For this, we will first assume the series to be ‘S’ and then we will separate the series into two series- one where all the elements have the plus sign and the other where all the elements have a negative sign. Then we will assume these two separate series as $ {{S}_{1}} $ and $ {{S}_{2}} $ . Then we will solve both $ {{S}_{1}} $ and $ {{S}_{2}} $ by using different formulas of AP like $ {{a}_{n}}=a+\left( n-1 \right)d $ and $ Sum=\dfrac{n}{2}\left( a+{{a}_{n}} \right) $ . Hence, we will get the value of $ {{S}_{1}} $ and $ {{S}_{2}} $ . When we will get there, we will also obtain the value of S. Hence, we will have our answer.

Complete step by step answer:
We here have to evaluate the series $ 1-2+3-4+5-...+101 $ .
Now, in this series, we can see that all the elements in this series are consecutive natural numbers with alternate plus and minus signs.
Thus, we can separate the series into two series- one where all the elements have a plus sign and the other where all the elements have a negative sign.
Thus, on separating the series, we have:
 $ \begin{align}
  & 1-2+3-4+5-...+101 \\
 & \Rightarrow \left( 1+3+5+...+101 \right)+\left( -2-4-6-...-100 \right) \\
 & \Rightarrow \left( 1+3+5+...+101 \right)-\left( 2+4+6+...+100 \right) \\
\end{align} $
Now, if we assume the given series to be S, we will have:
 $ S=1-2+3-4+5-...+101 $
Now, let us assume the series $ 1+3+5+...+101 $ to be $ {{S}_{1}} $ and $ 2+4+6+...+100 $ to be $ {{S}_{2}} $ .
Thus, we will have the value of S as:
 $ S={{S}_{1}}-{{S}_{2}} $ …..(i)
Now, we will evaluate $ {{S}_{1}} $ .
We have:
 $ {{S}_{1}}=1+3+5+...+101 $
Now, we can see that this series is an AP with the first term as 1, the common difference is 2 and the last term as 101.
Now, in an AP, we know that:
 $ {{a}_{n}}=a+\left( n-1 \right)d $
Where, a is the first term, $ {{a}_{n}} $ is the last term, d is a common difference and n is the number of terms.
Thus, for $ {{S}_{1}} $ , the number of terms are:
 $ \begin{align}
  & {{a}_{n}}=a+\left( n-1 \right)d \\
 & \Rightarrow 101=1+\left( n-1 \right)2 \\
 & \Rightarrow 100=\left( n-1 \right)2 \\
 & \Rightarrow n=50+1 \\
 & \Rightarrow n=51 \\
\end{align} $
Now, we know that sum ‘n’ terms of an AP is given as:
 $ Sum=\dfrac{n}{2}\left( a+{{a}_{n}} \right) $
Thus, we get the value of $ {{S}_{1}} $ as:
 $ \begin{align}
  & {{S}_{1}}=\dfrac{51}{2}\left( 1+101 \right) \\
 & \Rightarrow {{S}_{1}}=\dfrac{51}{2}\left( 102 \right) \\
 & \Rightarrow {{S}_{1}}=51\left( 51 \right) \\
 & \Rightarrow {{S}_{1}}=2601 \\
\end{align} $
Hence, the value of $ {{S}_{1}} $ is 2601.
Now, we will find the value of $ {{S}_{2}} $ .
We have:
 $ {{S}_{2}}=2+4+6+...+100 $
Now, the number of terms for $ {{S}_{2}} $ are given as:
 $ \begin{align}
  & 100=2+\left( n-1 \right)2 \\
 & \Rightarrow 98=\left( n-1 \right)2 \\
 & \Rightarrow n=49+1 \\
 & \Rightarrow n=50 \\
\end{align} $
Thus, the value of $ {{S}_{2}} $ is given as:
 $ \begin{align}
  & {{S}_{2}}=\dfrac{50}{2}\left( 2+100 \right) \\
 & \Rightarrow {{S}_{2}}=\dfrac{50}{2}\left( 102 \right) \\
 & \Rightarrow {{S}_{2}}=50\left( 51 \right) \\
 & \Rightarrow {{S}_{2}}=2550 \\
\end{align} $
Hence, the value of $ {{S}_{2}} $ is 2550.
Now, putting the value of $ {{S}_{1}} $ and $ {{S}_{2}} $ in equation (i), we get:
 $ \begin{align}
  & S={{S}_{1}}-{{S}_{2}} \\
 & \Rightarrow S=2601-2550 \\
 & \therefore S=51 \\
\end{align} $
Thus, the value of S is 51.

Hence, option (A) is the first option.

Note:
Here, we have used the formula for the sum of the AP as $ Sum=\dfrac{n}{2}\left( a+{{a}_{n}} \right) $ . We also could have used the formula for the sum given by $ Sum=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right) $ according to our convenience. We just used this formula here because we know the last terms of the series and hence this formula seemed easier.