Understanding the Problem

We are asked to find the maximum possible sum of the series: 51 + $\frac{101}{2}$ + 50 + ...

This appears to be an arithmetic series. Let's analyze the pattern.

Step 1: Identify the Pattern

The given terms are: 51, $\frac{101}{2}$ (which is 50.5), 50, ...

Let's write them as fractions for clarity:

First term, a₁ = 51 = $\frac{102}{2}$

Second term, a₂ = $\frac{101}{2}$

Third term, a₃ = 50 = 1002

We can see a clear pattern. The numerators are decreasing by 1: 102, 101, 100,...

This is an arithmetic progression (AP) with a common difference, d.

d = a₂ - a₁ = $\frac{101}{2}$ - $\frac{102}{2}$ = $\frac{-1}{2}$

So, the common difference is $\frac{-1}{2}$.

Step 2: Find the Last Term

The series will have a maximum sum if we sum all terms until the last positive term. The sum decreases after we add negative terms.

We need to find the term number (n) for which the nth term, aₙ, is just greater than 0.

The general formula for the nth term of an AP is:

$a_n=a_1+(n-1)d$

Let's set aₙ > 0 to find the last positive term.

$\frac{102}{2}+(n-1)\left(\frac{-1}{2}\right)>0$

Multiply both sides by 2:

$102-(n-1)>0$

$102-n+1>0$

$103-n>0$

$n<103$

Since n must be an integer, the largest value of n for which aₙ is positive is n = 102.

Let's find the 102nd term, a₁₀₂:

$a_{102}=\frac{102}{2}+(102-1)\left(\frac{-1}{2}\right)$

$a_{102}=51+\left(101\right)\left(\frac{-1}{2}\right)$

$a_{102}=51-\frac{101}{2}$

$a_{102}=\frac{102}{2}-\frac{101}{2}=\frac{1}{2}$

The 103rd term would be $\frac{1}{2}+\left(\frac{-1}{2}\right)=0$. After this, terms become negative.

Therefore, to get the maximum sum, we sum the first 102 terms (from n=1 to n=102).

Step 3: Apply the Sum Formula

The formula for the sum of the first n terms of an AP is:

$S_n=\frac{n}{2}[2a_1+(n-1\left)d\right]$

We have:
n = 102
a₁ = $\frac{102}{2}$
d = $\frac{-1}{2}$

Let's substitute these values into the formula:

$S_{102}=\frac{102}{2}[2\times \frac{102}{2}+(102-1)\times (\frac{-1}{2}\left)\right]$

Simplify inside the brackets:

$S_{102}=51[102+(101)\times (\frac{-1}{2}\left)\right]$

$S_{102}=51[102-\frac{101}{2}]$

Write 102 as $\frac{204}{2}$ to combine the terms:

$S_{102}=51[\frac{204}{2}-\frac{101}{2}]$

$S_{102}=51\left[\frac{103}{2}\right]$

Now, multiply:

$S_{102}=\frac{51}{1}\times \frac{103}{2}=\frac{5253}{2}$

Final Answer

The maximum possible sum is $\frac{5253}{2}$.

Related Topics & Formulae

Arithmetic Progression (AP): A sequence of numbers in which the difference between consecutive terms is constant. This difference is called the common difference (d).

n-th Term of an AP: