Answer :
To solve these problems related to Arithmetic Progressions (AP), we will use the definition of the nth term of an AP.
An Arithmetic Progression is a sequence of numbers in which the difference of any two successive members is a constant. This constant is known as the common difference, denoted as [tex]d[/tex]. The general formula for the nth term [tex]a_n[/tex] of an AP is:
[tex]a_n = a + (n-1)\cdot d[/tex]
where [tex]a[/tex] is the first term and [tex]d[/tex] is the common difference.
Let's break down the given problems:
If the 5th term of an AP = 10 and the 10th term = 5, find:
a) The first term
Given: [tex]a_5 = 10[/tex] and [tex]a_{10} = 5[/tex].
For the 5th term: [tex]a + 4d = 10[/tex] (Equation 1)
For the 10th term: [tex]a + 9d = 5[/tex] (Equation 2)
We have two equations:
[tex]a + 4d = 10[/tex]
[tex]a + 9d = 5[/tex]By subtracting Equation 1 from Equation 2:
[tex](a + 9d) - (a + 4d) = 5 - 10[/tex]
[tex]5d = -5[/tex]
Solving for [tex]d[/tex]:
[tex]d = -1[/tex]
Substituting [tex]d = -1[/tex] into Equation 1:
[tex]a + 4(-1) = 10[/tex]
[tex]a - 4 = 10[/tex]
[tex]a = 14[/tex]
So, the first term [tex]a = 14[/tex] and the common difference [tex]d = -1[/tex].
If the 12th term of an AP is double the 5th term, find the common difference given that the first term is 7.
Given: [tex]a_1 = 7[/tex], [tex]a_{12} = 2a_5[/tex].
The formula for the 12th term is:
[tex]a + 11d = a_{12}[/tex]The formula for the 5th term is:
[tex]a + 4d = a_5[/tex]Since [tex]a_{12} = 2a_5[/tex]:
[tex]a + 11d = 2(a + 4d)[/tex]Substituting [tex]a = 7[/tex]:
[tex]7 + 11d = 2(7 + 4d)[/tex]Simplifying:
[tex]7 + 11d = 14 + 8d[/tex][tex]11d - 8d = 14 - 7[/tex]
[tex]3d = 7[/tex]
Solving for [tex]d[/tex]:
[tex]d = \frac{7}{3}[/tex]
Thus, the common difference [tex]d[/tex] is [tex]\frac{7}{3}[/tex].
