If sum of first n terms of a series is `n/(n+1)` find `1/(T_8)`where `T_8` is the eighth term of the series (A) `64` (B) `80` (C) `75` (D) `72`

Options

A. 64
B. 80
C. 75
D. 72

Answer

To solve the problem, we need to find \( \frac{1}{T_8} \) where \( T_8 \) is the eighth term of the series. We are given that the sum of the first \( n \) terms of the series, denoted as \( S_n \), is:

\[ S_n = \frac{n}{n+1} \]

### Step 1: Find \( S_8 \) and \( S_7 \)

First, we calculate \( S_8 \):

\[ S_8 = \frac{8}{8+1} = \frac{8}{9} \]

Next, we calculate \( S_7 \):

\[ S_7 = \frac{7}{7+1} = \frac{7}{8} \]

### Step 2: Find \( T_8 \)

The eighth term \( T_8 \) can be found using the relationship:

\[ T_8 = S_8 - S_7 \]

Substituting the values we found:

\[ T_8 = \frac{8}{9} - \frac{7}{8} \]

To perform this subtraction, we need a common denominator. The least common multiple of 9 and 8 is 72. Therefore, we convert both fractions:

\[ T_8 = \frac{8 \times 8}{9 \times 8} - \frac{7 \times 9}{8 \times 9} = \frac{64}{72} - \frac{63}{72} \]

Now, we can subtract the two fractions:

\[ T_8 = \frac{64 - 63}{72} = \frac{1}{72} \]

### Step 3: Find \( \frac{1}{T_8} \)

Now, we need to find \( \frac{1}{T_8} \):

\[ \frac{1}{T_8} = \frac{1}{\frac{1}{72}} = 72 \]

### Conclusion

Thus, the value of \( \frac{1}{T_8} \) is:

\[ \boxed{72} \]