Amit has a packet of Candies .It consists of 288 candies .He arranges the candies in a way that first row contains 3 candies ,second row has 5 and third row has 7 and so on .
How many candies are there in the last row ?

Options

A. 33
B. 32
C. 31
D. 30

Answer

To solve the problem step by step, we will use the formula for the sum of an arithmetic progression (AP). Let's break it down:

### Step 1: Identify the parameters of the AP
- The first term \( A = 3 \) (the number of candies in the first row).
- The common difference \( D = 5 - 3 = 2 \) (the difference between the number of candies in consecutive rows).
- The total number of candies \( S_N = 288 \).

### Step 2: Use the formula for the sum of the first \( N \) terms of an AP
The formula for the sum of the first \( N \) terms of an arithmetic progression is given by:
\[ S_N = \frac{N}{2} \times (2A + (N - 1)D) \]
Substituting the known values:
\[ 288 = \frac{N}{2} \times (2 \times 3 + (N - 1) \times 2) \]

### Step 3: Simplify the equation
First, simplify the equation:
\[ 288 = \frac{N}{2} \times (6 + 2(N - 1)) \]
\[ 288 = \frac{N}{2} \times (6 + 2N - 2) \]
\[ 288 = \frac{N}{2} \times (2N + 4) \]
\[ 288 = N(N + 2) \]

### Step 4: Rearrange the equation
Rearranging gives us:
\[ N^2 + 2N - 288 = 0 \]

### Step 5: Solve the quadratic equation
We will use the quadratic formula \( N = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 2, c = -288 \):
\[ b^2 - 4ac = 2^2 - 4 \times 1 \times (-288) = 4 + 1152 = 1156 \]
Now, calculate \( N \):
\[ N = \frac{-2 \pm \sqrt{1156}}{2 \times 1} \]
Calculating \( \sqrt{1156} = 34 \):
\[ N = \frac{-2 \pm 34}{2} \]
This gives us two potential solutions:
1. \( N = \frac{32}{2} = 16 \)
2. \( N = \frac{-36}{2} = -18 \) (not valid since \( N \) cannot be negative)

Thus, \( N = 16 \).

### Step 6: Find the number of candies in the last row
The number of candies in the last row can be found using the formula for the \( N \)-th term of an AP:
\[ A_N = A + (N - 1)D \]
Substituting the values:
\[ A_{16} = 3 + (16 - 1) \times 2 \]
\[ A_{16} = 3 + 15 \times 2 \]
\[ A_{16} = 3 + 30 = 33 \]

### Final Answer
The number of candies in the last row is **33**.

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