The sum of n terms of a progression is `3n^2 + 5` . The number of terms which equals 123 is :

Options

A. a)12
B. b)18
C. c)21
D. d)23

Answer

To find the number of terms in the progression that equals 123, we start with the given information about the sum of the first \( n \) terms of the progression, which is:

\[ S_n = 3n^2 + 5 \]

### Step 1: Find the \( n \)-th term \( T_n \)

The \( n \)-th term \( T_n \) can be found using the formula:

\[ T_n = S_n - S_{n-1} \]

Where \( S_{n-1} \) is the sum of the first \( n-1 \) terms.

### Step 2: Calculate \( S_{n-1} \)

To find \( S_{n-1} \), we substitute \( n-1 \) into the sum formula:

\[ S_{n-1} = 3(n-1)^2 + 5 \]

Expanding this gives:

\[ S_{n-1} = 3(n^2 - 2n + 1) + 5 \]
\[ S_{n-1} = 3n^2 - 6n + 3 + 5 \]
\[ S_{n-1} = 3n^2 - 6n + 8 \]

### Step 3: Substitute \( S_n \) and \( S_{n-1} \) into \( T_n \)

Now we can substitute \( S_n \) and \( S_{n-1} \) into the formula for \( T_n \):

\[ T_n = S_n - S_{n-1} \]
\[ T_n = (3n^2 + 5) - (3n^2 - 6n + 8) \]

### Step 4: Simplify \( T_n \)

Now simplify the expression:

\[ T_n = 3n^2 + 5 - 3n^2 + 6n - 8 \]
\[ T_n = 6n - 3 \]

### Step 5: Set \( T_n \) equal to 123

We want to find \( n \) such that \( T_n = 123 \):

\[ 6n - 3 = 123 \]

### Step 6: Solve for \( n \)

Add 3 to both sides:

\[ 6n = 126 \]

Now divide by 6:

\[ n = 21 \]

### Conclusion

The number of terms which equals 123 is:

\[ \boxed{21} \]

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