The minimum number of terms from the beginning of the series `20+22(2)/(3)+25(1)/(3)+ . . . .`, so that the sum may exceed 1568, is

Options

A. 25
B. 27
C. 28
D. 29

Answer

To find the minimum number of terms from the beginning of the series \( 20 + \frac{22 \cdot 2}{3} + \frac{25 \cdot 1}{3} + \ldots \) such that the sum exceeds 1568, we can follow these steps:

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

Next, we need to find the common difference \( d \). The second term is \( \frac{22 \cdot 2}{3} = \frac{44}{3} \) and the third term is \( \frac{25 \cdot 1}{3} = \frac{25}{3} \).

Calculating the common difference:
\[ d = \frac{44}{3} - 20 = \frac{44}{3} - \frac{60}{3} = \frac{-16}{3} \]
\[ d = \frac{25}{3} - \frac{44}{3} = \frac{-19}{3} \]

From the calculations, it appears that the terms are not consistent, so let's check the series again. The series appears to be an arithmetic progression (AP) with a first term of 20 and a common difference of \( \frac{8}{3} \).

### Step 2: Use the formula for the sum of the first \( n \) terms of an AP
The formula for the sum \( S_n \) of the first \( n \) terms of an arithmetic series is given by:
\[ S_n = \frac{n}{2} \left( 2a + (n-1)d \right) \]

Substituting \( a = 20 \) and \( d = \frac{8}{3} \):
\[ S_n = \frac{n}{2} \left( 2 \cdot 20 + (n-1) \cdot \frac{8}{3} \right) \]
\[ S_n = \frac{n}{2} \left( 40 + \frac{8(n-1)}{3} \right) \]
\[ S_n = \frac{n}{2} \left( 40 + \frac{8n - 8}{3} \right) \]
\[ S_n = \frac{n}{2} \left( \frac{120 + 8n - 8}{3} \right) \]
\[ S_n = \frac{n}{2} \left( \frac{8n + 112}{3} \right) \]
\[ S_n = \frac{4n(8n + 112)}{3} \]

### Step 3: Set up the inequality to find \( n \)
We need \( S_n \) to exceed 1568:
\[ \frac{4n(8n + 112)}{3} > 1568 \]
Multiplying both sides by 3:
\[ 4n(8n + 112) > 4704 \]
Dividing by 4:
\[ n(8n + 112) > 1176 \]
Expanding:
\[ 8n^2 + 112n - 1176 > 0 \]

### Step 4: Solve the quadratic equation
To find the roots of the equation \( 8n^2 + 112n - 1176 = 0 \), we can use the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 8, b = 112, c = -1176 \):
\[ b^2 - 4ac = 112^2 - 4 \cdot 8 \cdot (-1176) \]
\[ = 12544 + 37632 = 50176 \]
\[ n = \frac{-112 \pm \sqrt{50176}}{16} \]
Calculating \( \sqrt{50176} = 224 \):
\[ n = \frac{-112 \pm 224}{16} \]
Calculating the two possible values for \( n \):
1. \( n = \frac{112}{16} = 7 \)
2. \( n = \frac{-336}{16} = -21 \) (not valid)

### Step 5: Determine the minimum \( n \)
Since \( n = 7 \) is the root, we check the inequality:
\[ 8(7)^2 + 112(7) - 1176 = 392 + 784 - 1176 = 0 \]
To exceed 1568, we need \( n \) to be greater than 7. Thus, the minimum number of terms \( n \) that exceeds 1568 is:
\[ n = 8 \]

### Conclusion
The minimum number of terms from the beginning of the series so that the sum may exceed 1568 is:
\[ \boxed{8} \]