Answer :
Let's tackle each part of the question step by step.
Find the 8th term from the end of the AP 7, 10, 13, ..., 184.
Given an arithmetic progression (AP) with first term [tex]a = 7[/tex] and last term [tex]l = 184[/tex], and common difference [tex]d = 3[/tex], we know the general formula for the [tex]n[/tex]-th term of an AP is:
[tex]a_n = a + (n-1) \, d[/tex]
The total number of terms [tex]N[/tex] can be found as:
[tex]l = a + (N-1) \, d[/tex]
[tex]184 = 7 + (N-1) \, 3[/tex]
[tex]184 - 7 = 3(N-1)[/tex]
[tex]177 = 3N - 3[/tex]
[tex]180 = 3N[/tex]
[tex]N = 60[/tex]
The 8th term from the end is the [tex](N - 8 + 1)[/tex]-th term:
[tex]a_{53} = 7 + (53-1) \, 3 = 7 + 156 = 163[/tex]
Thus, the 8th term from the end is 163.Check whether -150 is a term of the AP: 11, 8, 5, 2, ...
This AP has a first term [tex]a = 11[/tex] and a common difference [tex]d = -3[/tex]. We use the formula for the [tex]n[/tex]-th term to check if [tex]-150[/tex] is a term:
[tex]a_n = 11 + (n-1)(-3) = -150[/tex]
[tex]11 - 3n + 3 = -150[/tex]
[tex]14 - 3n = -150[/tex]
[tex]-3n = -150 - 14[/tex]
[tex]-3n = -164[/tex]
[tex]n = \frac{164}{3} \approx 54.67[/tex]
Since [tex]n[/tex] is not an integer, [tex]-150[/tex] is not a term of the AP.Which term of the AP 3, 15, 27, 39, .... will be 132 more than its 54th term?
The given AP has [tex]a = 3[/tex] and [tex]d = 12[/tex]. The 54th term is:
[tex]a_{54} = 3 + (54-1) \, 12 = 3 + 636 = 639[/tex]
Suppose [tex]a_n = a_{54} + 132[/tex]:
[tex]a_n = 639 + 132 = 771[/tex]
We find [tex]n[/tex] using the formula for the [tex]n[/tex]-th term:
[tex]3 + (n-1) \, 12 = 771[/tex]
[tex]12(n-1) = 771 - 3[/tex]
[tex]12(n-1) = 768[/tex]
[tex]n-1 = \frac{768}{12} = 64[/tex]
[tex]n = 65[/tex]
Thus, the 65th term will be 132 more than the 54th term.
