Question 1: If the $n^{th}$ term $a_n$ of a sequence is given by $a_n = n^2-n+1$, write down the first five terms.

Given sequence is $a_n = n^2-n+1$

Let the first five terms be $a_1, a_2, a_3, a_4, a_5$

$a_1 = (1)^2 - 1 + 1 = 1$

$a_2 = (2)^2 - 2 + 1 = 3$

$a_3 = (3)^2 - 3 + 1 = 7$

$a_4 = (4)^2 - 4 + 1 = 13$

$a_5 = (5)^2 - 5 + 1 = 21$

Therefore the first five terms are $1, 3, 7, 13,$ and $21$

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Question 2: A sequence is defined by $a_n = n^3 - 6n^2 + 11n -6, n \in N$. Show that the first three terms of the sequence are zero and all other terms are positive.

The given sequence is $a_n = n^3 - 6n^2 + 11n -6, n \in N$

Let the first five terms be $a_1, a_2, a_3$

$a_1 = (1)^3 - 6(1)^2 + 11(1) - 6 = 0$

$a_1 = (2)^3 - 6(2)^2 + 11(2) - 6 = 0$

$a_1 = (3)^3 - 6(3)^2 + 11(3) - 6 = 0$

Therefore the first three terms are $0$

And $a_n = n^3 - 6n^2 + 11n -6 = (n-2)^3 - (n-2)$

Therefore $a_n$ will be positive for $n \geq 4$

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Question 3: Find the first four terms of the sequence defined by $a_1 =3$ and $a_n = 3a_{n-1} +2$ for all.

Given $a_1 =3$ and $a_n = 3a_{n-1} +2$

$a_2 = 3 a_{2-1}+ 2 = 3a_1 + 2 = 3(3) + 2 = 11$

$a_3 = 3 a_{3-1}+ 2 = 3a_2 + 2 = 3(11) + 2 = 35$

$a_4 = 3 a_{4-1}+ 2 = 3a_3 + 2 = 3(35) + 2 = 107$

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Question 4: Write the first five terms in each of the following sequences:

i) $a_1 = 1, a_n = a_{n-1} + 2, n > 1$

ii) $a_1 = 1 = a_2, a_n = a_{n-1}+ a_{n-2}, n > 2$

iii)  $a_1=a_2= 2, a_n = a_{n-1}-1, n > 2$

i)      $a_1 = 1, a_n = a_{n-1} + 2, n > 1$

$a_1 = 1$

$a_2 = a_{2-1} + 2 = a_1 + 2 = 1 + 2 = 3$

$a_3 = a_{3-1} + 2 = a_2 + 2 = 3 + 2 = 5$

$a_4 = a_{4-1} + 2 = a_3 + 2 = 5 + 2 = 7$

$a_5 = a_{5-1} + 2 = a_4 + 2 = 7 + 2 = 9$

Therefore the first five terms of the given sequence are $1, 3, 5, 7$ and $9$.

ii)      $a_1 = 1 = a_2, a_n = a_{n-1}+ a_{n-2}, n > 2$

$a_1 = 1$

$a_2 = 1$

$a_3 = a_{3-1} + a({3-2} = a_2 + a_1 = 1 + 1 = 2$

$a_4 = a_{4-1} + a({4-2} = a_3 + a_2 = 2 + 1 = 3$

$a_5 = a_{5-1} + a({5-2} = a_4 + a_3 = 3 + 2 = 5$

Therefore the first five terms of the given sequence are $1, 1, 2, 3$ and $5$.

iii)    $a_1=a_2= 2, a_n = a_{n-1}-1, n > 2$

$a_1 = 2$

$a_2 = 2$

$a_3 = a_{3-1} - 1 = a_2 -1 = 2 - 1 = 1$

$a_4 = a_{4-1} - 1 = a_3 -1 = 1 - 1 = 0$

$a_5 = a_{5-1} - 1 = a_4 -1 = 0 - 1 = -1$

Therefore the first five terms of the given sequence are $2, 2, 1, 0$ and $-1$.

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Question 5: The Fibonacci sequence is defined by $a_1 = 1 = a_2, a_n = a_{n-1} + a_{n-2}$ for $n > 2$. Find $\frac{a_{n + 1}}{a_n}$ for $n = 1, 2, 3, 4, 5$.

Given: $a_1= 1, \ \ a_2 = 1$

$a_3 = a_{3-1} + a_{3-2} = a_2 + a_1 = 1+1= 2$

$a_4 = a_{4-1} + a_{4-2} = a_3 + a_2 = 2+1= 3$

$a_5 = a_{5-1} + a_{5-2} = a_4 + a_3 = 3+2= 5$

$a_6 = a_{6-1} + a_{6-2} = a_5 + a_4 = 5+3= 8$

Therefore for $\frac{a_{n + 1}}{a_n}$

For $n = 1: \ \$ $\frac{a_{1 + 1}}{a_1} = \frac{a_2}{a_1} = \frac{1}{1}$ $= 1$

For $n = 2: \ \$ $\frac{a_{2 + 1}}{a_2} = \frac{a_3}{a_2} = \frac{2}{1}$ $= 2$

For $n = 3: \ \$ $\frac{a_{3 + 1}}{a_3} = \frac{a_4}{a_3} = \frac{3}{2}$

For $n = 4: \ \$ $\frac{a_{4 + 1}}{a_4} = \frac{a_5}{a_4} = \frac{5}{3}$

For $n = 5: \ \$ $\frac{a_{5 + 1}}{a_5} = \frac{a_6}{a_5} = \frac{8}{5}$

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Question 6: Show that each of the following sequences is in A.P. Also find the common difference and write 3 more terms in each case.

i) $3, -1, -5, -9, \ldots$

ii) $-1,$ $\frac{1}{4}$ $,$ $\frac{3}{2}$ $,$ $\frac{11}{4}$ $, \ldots$

iii) $\sqrt{2}, 3\sqrt{2}, 5\sqrt{2}, 7\sqrt{2}, \ldots$

iv)  $9, 7 , 5, 3 \ldots$

i)       Given sequence: $3, -1, -5, -9, \ldots$

$a_1 = 3, \ \ a_2 = -1, \ \ a_3 = -5, \ \ a_4 = -9$

$a_2 - a_1 = -1-3 = - 4$

$a_3 - a_2 = -5 - (-1) = - 4$

$a_4 - a_3 = -9 - (-5) = - 4$

Therefore the common difference $(d) = -4$

$a_5 = a_4 + d = -9 +(-4) = - 13$

$a_6 = a_5 + d = -13 +(-4) = -17$

$a_7 = a_6 + d = -17 +(-4) = -21$

ii)      Given sequence: $-1,$ $\frac{1}{4}$ $,$ $\frac{3}{2}$ $,$ $\frac{11}{4}$ $, \ldots$

$a_1 = -1, \ \ a_2 =$ $\frac{1}{4}$ $, \ \ a_3 =$ $\frac{3}{2}$ $, \ \ a_4 =$ $\frac{11}{4}$

$a_2 - a_1 =$ $\frac{1}{4}$ $- (-1) =$ $\frac{5}{4}$

$a_3 - a_2 =$ $\frac{3}{2}$ $- ($ $\frac{1}{4}$ $) =$ $\frac{5}{4}$

$a_4 - a_3 =$ $\frac{11}{4}$ $- ($ $\frac{3}{2}$ $) =$ $\frac{5}{4}$

Therefore the common difference $(d) =$ $\frac{5}{4}$

$a_5 = a_4 + d =$ $\frac{11}{4}$ $+$ $\frac{5}{4}$ $= 4$

$a_6 = a_5 + d = 4 +$ $\frac{5}{4}$ $=$ $\frac{21}{4}$

$a_7 = a_6 + d =$ $\frac{21}{4}$ $+$ $\frac{5}{4}$ $=$ $\frac{26}{4}$ $=$ $\frac{13}{2}$

iii)    Given sequence: $\sqrt{2}, 3\sqrt{2}, 5\sqrt{2}, 7\sqrt{2}, \ldots$

$a_1 = \sqrt{2}, \ \ a_2 = 3\sqrt{2}, \ \ a_3 = 5\sqrt{2}, \ \ a_4 = 7\sqrt{2}$

$a_2 - a_1 = 3\sqrt{2}-\sqrt{2} = 2\sqrt{2}$

$a_3 - a_2 = 5\sqrt{2} - 3\sqrt{2} = 2\sqrt{2}$

$a_4 - a_3 = 7\sqrt{2} - 5\sqrt{2} = 2\sqrt{2}$

Therefore the common difference $(d) = 2\sqrt{2}$

$a_5 = a_4 + d = 7\sqrt{2} +2\sqrt{2} = 9\sqrt{2}$

$a_6 = a_5 + d = 9\sqrt{2} +2\sqrt{2} = 11\sqrt{2}$

$a_7 = a_6 + d = 11\sqrt{2} +2\sqrt{2} = 13\sqrt{2}$

iv)     Given sequence: $9, 7 , 5, 3 \ldots$

$a_1 = 9, \ \ a_2 = 7, \ \ a_3 = 5, \ \ a_4 = 3$

$a_2 - a_1 = 7-9 = - 2$

$a_3 - a_2 = 5-7 = - 2$

$a_4 - a_3 = 3-5 = - 2$

Therefore the common difference $(d) = -2$

$a_5 = a_4 + d = 3 +(-2) = 1$

$a_6 = a_5 + d = 1 +(-2) = -1$

$a_7 = a_6 + d = -1 +(-2) = -3$

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Question 7: The $n^{th}$ terms of a sequence is given by $a_n = 2n+ 7$. Show that it is an A.P. Also, find its $7^{th}$ terms.

Given sequence $a_n = 2n+ 7$

$a_1 = 2 ( 1) + 7 = 9$

$a_2 = 2(2) + 7 = 11$

$a_3 = 2(3) + 7 = 13$

$a_2 - a_1 = a_3 - a_2 = 2$

Therefore the common difference $(d) = 2$. Therefore the given sequence is an A.P.

$a_7 = 2(7) + 7 = 21$

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Question 8: The $n^{th}$ terms of a sequence is given by $a_n = 2n^2 +n + 1$. Show that it is not an A.P.

Given sequence $a_n = 2n^2 +n + 1$
$a_1 = 2(1)^2 + 1 + 1 = 4$
$a_2 = 2(2)^2 + 2 + 1 = 11$
$a_3 = 2(3)^2 + 3 + 1 = 21$
Therefore $a_3-a_2 \neq a_2-a_1$