How many terms of the A.P. `20, 19 1/3, 18 2/3,`……… must be taken so that their sum is 300?

Answer

To solve the problem of how many terms of the A.P. \(20, 19 \frac{1}{3}, 18 \frac{2}{3}, \ldots\) must be taken so that their sum is 300, we can follow these steps:

### Step 1: Identify the first term and the common difference
The first term \(a\) of the A.P. is:
\[ a = 20 \]

To find the common difference \(d\), we subtract the first term from the second term:
\[ d = 19 \frac{1}{3} - 20 = \frac{58}{3} - 20 = \frac{58}{3} - \frac{60}{3} = -\frac{2}{3} \]

### Step 2: Use the formula for the sum of the first \(n\) terms of an A.P.
The formula for the sum of the first \(n\) terms \(S_n\) of an A.P. is given by:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
We know \(S_n = 300\), so we set up the equation:
\[ 300 = \frac{n}{2} \left(2 \times 20 + (n-1) \left(-\frac{2}{3}\right)\right) \]

### Step 3: Simplify the equation
Substituting \(a\) and \(d\) into the equation:
\[ 300 = \frac{n}{2} \left(40 - \frac{2}{3}(n-1)\right) \]
Now, simplifying the expression inside the parentheses:
\[ 300 = \frac{n}{2} \left(40 - \frac{2n - 2}{3}\right) \]
To combine the terms, we need a common denominator:
\[ 40 = \frac{120}{3} \]
Thus,
\[ 300 = \frac{n}{2} \left(\frac{120 - 2n + 2}{3}\right) = \frac{n}{2} \left(\frac{122 - 2n}{3}\right) \]

### Step 4: Clear the fraction
Multiply both sides by 6 to eliminate the fractions:
\[ 1800 = n(122 - 2n) \]

### Step 5: Rearrange the equation
Rearranging gives us:
\[ 2n^2 - 122n + 1800 = 0 \]

### Step 6: Solve the quadratic equation
We can factor the quadratic equation:
\[ 2n^2 - 50n - 72n + 1800 = 0 \]
Factoring gives:
\[ (2n - 72)(n - 25) = 0 \]

### Step 7: Find the values of \(n\)
Setting each factor to zero:
1. \(2n - 72 = 0 \Rightarrow n = 36\)
2. \(n - 25 = 0 \Rightarrow n = 25\)

Thus, the number of terms that must be taken to achieve a sum of 300 is either \(n = 25\) or \(n = 36\).

### Final Answer:
The number of terms of the A.P. that must be taken to achieve a sum of 300 is \(25\) or \(36\).

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