The sum of n terms of a series is `3n^2 + 5n`. Find the value of n if nth term is 152.

Options

A. A. 15
B. B. 18
C. C. 20
D. D. 25

Answer

To solve the problem, we need to find the value of \( n \) such that the \( n \)-th term of the series is 152, given that the sum of the first \( n \) terms is \( S_n = 3n^2 + 5n \).

### Step 1: Find the first few sums of the series
1. **Calculate \( S_1 \)**:
\[ S_1 = 3(1^2) + 5(1) = 3 + 5 = 8 \]
Thus, \( a_1 = S_1 = 8 \).

2. **Calculate \( S_2 \)**:
\[ S_2 = 3(2^2) + 5(2) = 3(4) + 10 = 12 + 10 = 22 \]
Thus, \( a_2 = S_2 - S_1 = 22 - 8 = 14 \).

3. **Calculate \( S_3 \)**:
\[ S_3 = 3(3^2) + 5(3) = 3(9) + 15 = 27 + 15 = 42 \]
Thus, \( a_3 = S_3 - S_2 = 42 - 22 = 20 \).

### Step 2: Identify the pattern in the terms
Now we have the first three terms:
- \( a_1 = 8 \)
- \( a_2 = 14 \)
- \( a_3 = 20 \)

We can see that the terms are increasing. To find the common difference \( d \):
\[ d = a_2 - a_1 = 14 - 8 = 6 \]
\[ d = a_3 - a_2 = 20 - 14 = 6 \]
So, the common difference \( d = 6 \).

### Step 3: Write the formula for the \( n \)-th term
The \( n \)-th term of an arithmetic progression can be expressed as:
\[ a_n = a_1 + (n - 1)d \]
Substituting the known values:
\[ a_n = 8 + (n - 1) \cdot 6 \]

### Step 4: Set the \( n \)-th term equal to 152
We need to find \( n \) such that:
\[ 8 + (n - 1) \cdot 6 = 152 \]

### Step 5: Solve for \( n \)
1. Subtract 8 from both sides:
\[ (n - 1) \cdot 6 = 152 - 8 \]
\[ (n - 1) \cdot 6 = 144 \]

2. Divide both sides by 6:
\[ n - 1 = \frac{144}{6} = 24 \]

3. Add 1 to both sides:
\[ n = 24 + 1 = 25 \]

### Conclusion
The value of \( n \) for which the \( n \)-th term is 152 is \( n = 25 \).

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