Find the indicated terms in each of the following sequences whose nth terms are:
(a) $a_{n}=5 n-4 ; a_{12}$ and $a_{15}$
(b) $a_{n}=\frac{3 n-2}{4 n+5} ; a_{7}$ and $a_{8}$
(c) $a_{n}=n(n-1)(n-2) ; a_{5}$ and $a_{8}$
(d) $a_{n}=(n-1)(2-n)(3+n) ; a_{1}, a_{2}, a_{3}$
(e) $a_{n}=(-1)^{n} n ; a_{3}, a_{5}, a_{8}$
Here, we are given the nth term for various sequences. We need to find the indicated terms of the A.P.
(i) $a_{n}=5 n-4$
We need to find and
Now, to find term we use, we get,
$a_{12}=5(12)-4$
$=60-4$
$=56$
Also, to find term we use, we get,
$a_{15}=5(15)-4$
$=75-4$
$=71$
Thus, $a_{12}=56$ and $a_{15}=71$
(ii) $a_{n}=\frac{3 n-2}{4 n+5}$
We need to find $a_{7}$ and $a_{8}$
Now, to find $a_{7}$ term we use $n=7$, we get,
$a_{7}=\frac{3(7)-2}{4(7)+5}$
$=\frac{21-2}{28+5}$
$=\frac{19}{33}$
Also, to find term we use, we get,
$a_{\mathrm{s}}=\frac{3(8)-2}{4(8)+5}$
$=\frac{24-2}{32+5}$
$=\frac{22}{37}$
Thus, $a_{7}=\frac{19}{33}$ and $a_{8}=\frac{22}{37}$
(iii) $a_{n}=n(n-1)(n-2)$
We need to find $a_{3}$ and $a_{5}$
Now, to find $a_{5}$ term we use $n=5$, we get,
$a_{5}=5(5-1)(5-2)$
$=5(4)(3)$
$=60$
Also, to find term we use, we get,
$a_{3}=8(8-1)(8-2)$
$=8(7)(6)$
$=336$
Thus, $a_{5}=60$ and $a_{8}=336$
(iv) $a_{n}=(n-1)(2-n)(3+n)$
We need to find $a_{1}, a_{2}$ and $a_{3}$
Now, to find $a_{1}$ term we use $n=1$, we get,
$a_{1}=(1-1)(2-1)(3+1)$
$=(0)(1)(4)$
$=0$
Also, to find term we use, we get,
$a_{2}=(2-1)(2-2)(3+2)$
$=(1)(0)(5)$
$=0$
Similarly, to find $a_{3}$ term we use $n=3$, we get,
$a_{3}=(3-1)(2-3)(3+3)$
$=(2)(-1)(6)$
$=-12$
Thus, $a_{1}=0, a_{2}=0$ and $a_{3}=-12$
(v) $a_{n}=(-1)^{n} n$
We need to find $a_{3}, a_{5}$ and $a_{8}$
Now, to find $a_{3}$ term we use $n=3$, we get,
$a_{3}=(-1)^{3} 3$
$=(-1) 3$
$=-3$
Also, to find $a_{5}$ term we use $n=5$, we get,
$a_{5}=(-1)^{5} 5$
$=(-1) 5$
$=-5$
Similarly, to find term we use, we get,
$a_{8}=(-1)^{5} 8$
$=(1) 8$
$=8$
Thus, $a_{3}=-3, a_{5}=-5$ and $a_{8}=8$