The sum of the first n terms is given by S_n = (n)/(2)[2a + (n - 1)d]. Substituting n = 18, a = 12, and d = 5, we get 9 × (24 + 85) = 9 × 109 = 981 total tiles.

School decorate a rectangular were using tiles arranged in rows the First row contain 12 tiles the second row contains 17 times the drought 22 tiles and the pattern continuous find common difference how many times are in the 18th of find total number of tile used in the first 18

Answer: Common difference $d = 5$ , number of tiles in the 18th row = 97, and total tiles in the first 18 rows = 981.

Step-by-step solution

Step 1: Identify the Arithmetic Progression and Common Difference

The number of tiles in successive rows are $12, 17, 22, \dots$ . Since the difference between consecutive terms is $17 - 12 = 22 - 17 = 5$ , the terms form an Arithmetic Progression with first term $a = 12$ and common difference $d = 5$ .

Step 2: Find the Number of Tiles in the 18th Row

We use the $n^{\text{th}}$ term formula $a_n = a + (n - 1)d$ with $n = 18$ , $a = 12$ , and $d = 5$ . Multiplying $17$ by $5$ gives $85$ , and adding $12$ results in $97$ tiles in the 18th row.

Step 3: Find the Total Number of Tiles in the First 18 Rows

The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$ . Substituting $n = 18$ , $a = 12$ , and $d = 5$ , we get $9 \times (24 + 85) = 9 \times 109 = 981$ total tiles.

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