Since, each row has one more log than the row above it
so, this is arithematic sequence
We are given that
First row is
[tex]=14[/tex]
so,
[tex]a_1=14[/tex]
Last row is
[tex]=20[/tex]
so,
[tex]a_n=14[/tex]
Each row has one more log than the row above it
so,
[tex]d=1[/tex]
now, we can find number of rows
[tex]a_n=a_1+(n-1)d[/tex]
we can plug values
[tex]20=14+(n-1)1[/tex]
we can solve for n
[tex]6=(n-1)1[/tex]
[tex]n=7[/tex]
now, we can find total number of logs
[tex]S=\frac{n(a_1+a_n)}{2}[/tex]
now, we can plug values
[tex]S=\frac{7(14+20)}{2}[/tex]
[tex]S=\frac{7(14+20)}{2}[/tex]
[tex]S=119[/tex]
So,
Number of logs in the pile are 119........Answer
The log in the pile is an illustration of arithmetic progression.
The number of logs in the pile is 119.
The first term of the progression is:
[tex]\mathbf{a =14}[/tex]
The last term is:
[tex]\mathbf{L =20}[/tex]
The common difference is:
[tex]\mathbf{d =1}[/tex]
First, we calculate the number of terms using:
[tex]\mathbf{L = a +(n - 1)d}[/tex]
So, we have:
[tex]\mathbf{20= 14 +(n - 1) \times 1}[/tex]
[tex]\mathbf{20 = 14 +(n - 1) }[/tex]
Subtract 14 from both sides
[tex]\mathbf{n - 1 = 6}[/tex]
Add 1 to both sides
[tex]\mathbf{n = 7}[/tex]
The number of logs in the pile is calculated using the sum of n terms of an AP formula:
[tex]\mathbf{S_n = \frac{n}{2}(a + L)}[/tex]
So, we have:
[tex]\mathbf{S_n = \frac{7}{2}(14 + 20)}[/tex]
[tex]\mathbf{S_n = 3.5(34)}[/tex]
[tex]\mathbf{S_n = 119}[/tex]
Hence, the number of logs in the pile is 119.
Read more about arithmetic progressions at:
https://brainly.com/question/13989292
