[1 Mark Questions]

Question 11 Mark

Find the sum of the following series
$6^2+7^2+8^2+\ldots+21^2$

Answer

$\begin{aligned} & 6^2+7^2+8^2+\ldots+21^2 \\ & \left(1^2+2^2+\ldots+21^2\right)-\left(1^2+2^2+\ldots+5^2\right) \\ & =\sum_1^{21} n ^2-\sum_1^5 n ^2 \\ & =\left(\frac{ n ( n +1)(2 n +1)}{6}\right)_{ n =21}-\left(\frac{ n ( n +1)(2 n +1)}{6}\right)_{ n =5} \\ & =\left(\frac{21 \times 22 \times 43}{6}\right)-\left(\frac{5 \times 6 \times 11}{6}\right) \\ & =3311-55 \\ & =3256\end{aligned}$

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Question 21 Mark

Find the sum of the following series
$10^3+11^3+12^3+\ldots+20^3$

Answer

$\begin{aligned} & 10^3+11^3+12^3+\ldots+20^3 \\ & =\left(1^3+2^3+3^3+\ldots+20^3\right)-\left(1^3+2^3+3^3+\ldots+9^3\right) \\ & =\sum_1^{20} n ^3-\sum_1^9 n ^3 \\ & =\left(\frac{ n ( n +1)}{2}\right)_{ n =20}^2-\left(\frac{ n ( n +1)}{2}\right)_{ n =9}^2 \\ & =\left(\frac{20 \times 21}{2}\right)^2-\left(\frac{9 \times 10}{2}\right)^2 \\ & =210^2-45^2 \\ & =44100-2025 \\ & =42075\end{aligned}$

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Question 31 Mark

Find the sum of the following series
1 + 3 + 5 + ... + 71

Answer

$\begin{aligned} & 1+3+5+\ldots+71 \\ & = n ^2 \\ & n =\frac{1- a }{ d }+1 \\ & \Rightarrow\left(\frac{71-1}{2}\right)+1 \\ & =\frac{70}{2}+1 \\ & =36 \\ & \therefore 1+3+5+\ldots+71=(36)^2 \\ & =1296\end{aligned}$

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Question 41 Mark

Find the sum of the following series
1 + 4 + 9 + 16 + ... + 225

Answer

$\begin{aligned} & 1+4+9+16+\ldots+225 \\ & =1^2+2^2+3^2+4^2+\ldots+15^2 \\ & \sum_1^{ n } n ^2=\frac{ n ( n +1)(2 n +1)}{6} \\ & \sum_1^{15} n ^2=\frac{15(15+1)(2 \times 15+1)}{6} \\ & =\frac{15 \times 16 \times 31}{6} \\ & =1240\end{aligned}$

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Question 51 Mark

Find the sum of the following series
1 + 2 + 3 + ... + 60

Answer

$\begin{aligned} & 1+2+3+\ldots+60 \\ & \sum_1^{ n } n =\frac{ n ( n +1)}{2} \\ & 1+2+3+\ldots+60=\sum_1^{60} n \\ & =\frac{60(60+1)}{2} \\ & =30 \times 61 \\ & =1830\end{aligned}$

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Question 61 Mark

Find the sum of the following series
3 + 6 + 9 + ... + 96

Answer

$\begin{aligned} & 3+6+9+\ldots+96 \\ & =3(1+2+3+\ldots+32) \\ & =3\left[\sum_1^{32} n \right] \\ & =3\left[\frac{ n ( n +1)}{2}\right]_{ n =32} \\ & =3\left[\frac{32(33)}{2}\right] \\ & =1584\end{aligned}$

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Question 71 Mark

Find the sum of the following series
51 + 52 + 53 + ... + 92

Answer

$\begin{aligned} & 51+52+53+\ldots+92 \\ & =(1+2+3+\ldots+92)-(1+2+3+\ldots+50) \\ & =\sum_1^{92} n -\sum_1^{50} n \\ & =\left(\frac{ n ( n +1)}{2}\right)_{ n =92}-\left(\frac{ n ( n +1)}{2}\right)_{ n =50} \\ & =\frac{92 \times 93}{2}-\frac{50 \times 51}{2} \\ & =4278-1275 \\ & =3003\end{aligned}$

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Question 81 Mark

Find the first term and common difference of the Arithmetic Progressions whose $n ^{\text {th }}$ term is given below
$t_n=-3+2 n$

Answer

a $= t _1=-3+2(1)$
$=-3+2=-1$
$d=t_2-t_1$
$\text { Here } t_2=-3+2(2)$
$=-3+4=1$
$\therefore d=t_2-t_1$
$=1-(-1)=2$

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Question 91 Mark

Find the first term and common difference of the Arithmetic Progressions whose $n ^{\text {th }}$ term is given below $t _{ n }=4-7 n$

Answer

$a=t_1=4-7(1)$
$=4-7=-3$
$d=t_2-t_1$
$\text { Here } t_2=4-7(2)$
$=4-14-10$
$\therefore d=t_2-t_1$
$=10-(-3)=-7$

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Question 101 Mark

Find the next three terms of the following sequence.
$8, 24, 72, …$

Answer

$8, 24, 72, …$
In an arithmetic sequence $a = 8,$
$d = t2 – t1 = t3 – t2$
$= 24 – 8 = 72 – 24$
$= 16 \neq 48$
So,it is not an arithmetic sequence. In a geometric sequence,
$r=\frac{t_2}{t_1}=\frac{t_3}{t_2}$
$\Rightarrow \frac{24}{8}=\frac{72}{24}$
$\Rightarrow 3=3$
$\therefore$ It is a geometric sequence
$\therefore$ The $n ^{\text {th }}$ term of a G.P is $t_n=\operatorname{ar}^{n-1}$
$\therefore t _4=8 \times 3^{4-1}$
$=8 \times 3^3$
$=8 \times 27$
$=216$
$t _5=8 \times 3^{5-1}$
$=8 \times 3^4$
$=8 \times 8$
$=648$
$t_6=8 \times 3^{6-1}$
$=8 \times 3^5$
$=8 \times 243$
$=1944$
The next 3 terms are $8,24,72,216,648,1944$.

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Question 111 Mark

Find the next three terms of the following sequence.
5, 1, – 3, …

Answer

$5,1,-3, \ldots$
$d=t_2-t_1=t_3-t_2$
$\Rightarrow 1-5=-3-1$
$-4=-4$
$\therefore$ It is an A.P.
$t_n=a+(n-1) d$
$t_4=5+3 \times-4$
$=5-12$
$=-7$
$15=a+4 d$
$=5+4 \times-4$
$=5-16$
$=-11$
$t_6=a+5 d$
$=5+5 \times-4$
$=5-20$
$=-15$
$\therefore$ The next three terms are $5,1,-3,-7,-11,-15$.

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Question 121 Mark

Find the next three terms of the following sequence
$\frac{1}{4}, \frac{2}{9}, \frac{3}{16}, \ldots$

Answer

$\frac{1}{4}, \frac{2}{9}, \frac{3}{16}, \ldots \ldots \ldots$
Here $a_n=$ Numerators are natural numbers and denominators are squares of the next numbers
$
\frac{1}{4}, \frac{2}{9}, \frac{3}{16}, \frac{4}{25}, \frac{5}{36}, \frac{6}{49} \ldots \ldots \ldots
$

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