In a flower bed there are 23 rose plants in the first row, 21 in the second, 19 in the third and so on. There are 5 rose plants in the last row. Then the number of rows in the flower bed is:

Question

Let $a_1,a_2,a_3,a_4$ be an A.P. of four terms such that each term of the A.P. and its common difference are integers. If $a_1+a_2+a_3+a_4=48$ and $a_1^2a_2a_3a_4+1^4=361$, then the largest term of the A.P. is equal to

Answer 1

Step 1: Concept Identification:
The number of plants in each row forms an arithmetic progression (AP) due to a constant difference between successive terms.

Step 2: Applicable Formula/Method:
The formula for the nth term of an AP is:
$a_n = a_1 + (n-1)d$
Here, $a_n$ represents the nth term, $a_1$ is the first term, n is the total number of terms, and d is the common difference. The objective is to determine 'n', which signifies the number of rows.

Step 3: Detailed Calculation:
The sequence representing the number of plants per row is 23, 21, 19, ..., 5.
The first term, $a_1 = 23$.
The last term, $a_n = 5$.
The common difference, $d = 21 - 23 = -2$.
Substituting these values into the nth term formula:
$5 = 23 + (n-1)(-2)$
Subtract 23 from both sides:
$5 - 23 = (n-1)(-2)$
$-18 = (n-1)(-2)$
Divide both sides by -2:
$\frac{-18}{-2} = n-1$
$9 = n-1$
Add 1 to both sides:
$n = 10$

Step 4: Conclusion:
There are 10 rows of plants in the flower bed.

In a flower bed there are 23 rose plants in the first row, 21 in the second, 19 in the third and so on. There are 5 rose plants in the last row. Then the number of rows in the flower bed is:

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The Correct Option is B

Solution and Explanation

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