Zarrolou wrote:
Here is my question about Patterns/Series
Given \(a_1=-81\) and \(a_n=a_{n-1}+3\) for \(n>1\), what is the value of the sum of the first 54 elements: \(a_1+a_2+...+a_{53}+a_{54}\) ?
A)0
B)-78
C)-81
D)81
E)78
OA: C)-81
the sequence is as follow:
\(a_1=-81\) \(a_2=-78\) ... and this will reach 0 is 27 passages \(a_{28}=0\) and then will become positive and every term will balance its negative correspondent.
\(a_{27}=-3\) will be balanced by \(a_{29}=3\) => sum=0 and so on...
This process continues for all terms, if the question were "What is the sum of the first 55 terms the balance will be perfect and the sum would be 0.
But here we are asked the sum of the 54 terms: the very first term will not be balanced!
\(a_{1}=-81\), \(a_{2}=-78\), ..., \(a_{54}=78\)
The sum is -81.
Given \(a_1=-81\) , and \(a_n=a_{n-1}+3\) for \(n>1\),
Therefore, \(a_2=a_{1}+3\)
& \(a_3=a_{2}+3\)
This means the Common increment of 3, & as we know that there are 54 terms in total but if we remove \(a_1\) , then we will have 53 terms each increasing with the common increment of 3. Therefore we have \(a_54\) as 3*53-81.
\(a_54\) =159 - 81 \(\Rightarrow\) \(a_54\) =78.
Now, we have first term & the last term & the common difference. so as per the properties.....
--> \(\frac{Last Term + First Term}{2} * Total # of Terms\)
---> \(\frac{78-81}{2}*54\)
-----> -27*3 \(\Rightarrow\) Hence, -81.