Find the indicated terms in each of the following sequences whose nth terms are:
Question:

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) $a_{n}=(-1)^{n} n ; a_{3}, a_{5}, a_{8}$

Solution:

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 $a_{8}=8$

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