Question

Find the sum of – 64, – 66, – 68, ...., – 100.
(a) – 1458
(b) – 1558
(c) – 1568
(d) – 1664

Answer 1

Hint: Here, first we have to determine which type of series we have, the given is arithmetic progression series, Use the summation formula for arithmetic progression, $ {{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right] $ to find the required result.

Complete step-by-step answer:
We have a sequence, – 64, – 66, – 68, ...., – 100 which is in the form of arithmetic progression because it is gradually increasing with a constant difference. We need to use the formula for the summation of numbers in a series, $ {{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right] $ , where $ {{S}_{n}} $ is the sum of all the numbers in the series, $ a $ is the first number in the series, $ n $ is the total numbers in a series, $ d $ is the constant difference between all the consecutive numbers in the series.
First let us count how many numbers we have from – 64 to – 100.
– 64, – 66, – 68, – 70, – 72, – 74, – 76, – 78, – 80, – 82,
 – 84, – 86, – 88, – 90, – 92, – 94, – 96, – 98, –100
So, after counting, we found that we have about 19 numbers, therefore $ n=19 $ .
Now, let us find which is the first number in the series, $ a= $ – 64.
Let us also find the constant difference, $ d= $ – 66 – (– 64)
                                                                             = – 66 + 64
                                                                             = – 2
Hence, the common difference is – 2.
Now, we have all the required values, now we will find the summation of the series.
 $ \begin{align}
  & {{S}_{n}}=\dfrac{19}{2}\left[ 2\left( -64 \right)+\left( 19-1 \right)\left( -2 \right) \right] \\
 & =\dfrac{19}{2}\left[ -128+\left( 18 \right)\left( -2 \right) \right]
\end{align} $
Now, open the brackets and solve further with required operations and then find the sum of these numbers, etc.
 $ \begin{align}
  & {{S}_{n}}=\dfrac{19}{2}\left[ -128-36 \right] \\
 & =\dfrac{19}{2}\left( -164 \right) \\
 & =19\left( -82 \right) \\
 & =-1558
\end{align} $
Hence, the summation of all these numbers in the series is – 1558.


Note: In this particular question, we can either use the formula which is used in the solution or you can use, $ {{S}_{n}}=n\left( \dfrac{{{a}_{1}}+{{a}_{n}}}{2} \right) $ , where $ {{a}_{1}} $ is the first term in the series and $ {{a}_{n}} $ is the last term in the series and $ n $ is the total number of terms in the series.