1 Marks Question

Question 11 Mark

$11^{th}$ term of an AP: – 3, $-\frac{1}{2}$ , 2, . . . is

Answer

We have given the AP: $-3,-\frac{1}{2}, 2, \ldots$
Here, $a =-3, d=a_2-a_1 d=\frac{-1}{2}-(-3)=\frac{-1}{2}+3=\frac{-1+6}{2}=\frac{5}{2}$ and $n =11$
We know that, $a_n=a+(n-1) d$
Or, $a_{11}=-3+(11-1) \times \frac{5}{2}$
$=-3+10 \times \frac{5}{2}$
$=-3+5 \times 5$
$=-3+25$
$a_{11}=22$

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Question 21 Mark

$30^{\text {th }}$ term of an AP: 10, 7, 4, .......... is

Answer

We have AP: $10,7,4, \ldots$
Here, $a=10, d=7-10=-3$ and $n =30$
We know:
$a_n = a + (n - 1)d$
$or, a_{30} = 10 + (30 - 1)(-3)$
$= 10 + 29(-3)$
$= 10 - 87$
$a_{30} = -77$

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Question 31 Mark

For the AP $\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{{13}}{3},.....$, write the first term and the common difference.

Answer

$\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{{13}}{3},.....$
First term (a) $ = \frac{1}{3}$
Common difference (d) $ = \frac{5}{3} - \frac{1}{3} = \frac{4}{3}$

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Question 41 Mark

For the AP –5, –1, 3, 7,... write the first term and the common difference.

Answer

Given A.P is: -5, - 1, 3, 7, ...
First term (a) = -5
Common difference (d) = -1 - (-5) = -1 + 5 = 4

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Question 51 Mark

For the AP 3, 1, -1, -3 ......, write the first term and the common difference.

Answer

3, 1, -1, -3...
First term = a= 3,
Common difference (d) = Second term - first term = Third term - second term and so on
Therefore, Common difference (d) = 1 - 3 = -2

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Question 61 Mark

Find the $10^{\text {th }}$ term of the AP $2,7,12, \ldots$

Answer

Here $a = 2$;
$d = 7 - 2 = 5$.
So, $a_{10}=a+9 d=2+9 \times 5=47$

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Question 71 Mark

Find the sum of the first 1000 positive integers.

Answer

According to question we are given that a spiral is made up of successive semi-circles, with centres alternately at $A$ and $B$, starting with centre at $A$, of radii $0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm$, ....as shown in Fig.Let $I _1, I _2, I _3, I _4, \ldots I _{13}$ be the lengths (circumferences) of semi-circles of radii $r_1=0.5 cm, r_2=1.0 cm, r_3=1.5 cm, r_4=2.0 cm, r_5=2.5 cm, \ldots$ respectively.

Now, Semi-perimeter of circle = $\pi\cdot r $
Therefore,
$l _ { 1 } = \pi r _ { 1 } = \pi \times 0.5 = \frac { \pi } { 2 } \mathrm { cm }$
$l _ { 2 } = \pi r _ { 2 } = \pi \times 1 = 2 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
$l _ { 3 } = \pi r _ { 3 } = \pi \times \frac { 3 } { 2 } = 3 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
$l _ { 4 } = \pi r _ { 4 } = \pi \times 2 = 4 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
and
$l _ { 13 } = \pi r _ { 13 } = \pi \times \frac { 13 } { 2 } \mathrm { cm } = 13 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
Therefore total length of the spiral $= I _1+ I _2+ I _3+\ldots+ I _{13}$
$\bf= \left\{ \frac { \pi } { 2 } + 2 \left( \frac { \pi } { 2 } \right) + 3 \left( \frac { \pi } { 2 } \right) + \dots + 13 \left( \frac { \pi } { 2 } \right) \right\} $
$\bf= \frac { \pi } { 2 } ( 1 + 2 + 3 + \cdots + 13 ) $
$\bf= \frac { \pi } { 2 } \times \frac { 13 } { 2 } ( 1 + 13 ) \quad \left[ \text { Using } S _ { n } = \frac { n } { 2 } ( a + l ) \right]$
$\bf= \frac { \pi } { 2 } \times \frac { 13 } { 2 } \times 14$ = $\bf\frac { 1 } { 2 } \times \frac { 22 } { 7 } \times 13 \times 7$ = $\bf {143 cm}$
which is required length of the spiral made up of thirteen consecutive semi-circles.

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Question 81 Mark

Find the sum of first 22 terms of the AP 8,3, - 2,...

Answer

Here $a=8, d=3-8=-5$.
So, $S _{22}=\frac{22}{2}[2 a +(22-1) d ]$
$\Rightarrow S_{22}=11(16-105)=-979$

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Question 91 Mark

For the AP: $\frac { 3 } { 2 } , \frac { 1 } { 2 } , -\frac {1 } { 2 } , -\frac {3} {2}$, ... write the first term a and the common difference d.

Answer

Given AP = $\frac { 3 } { 2 } , \frac { 1 } { 2 } , \frac { - 1 } { 2 } , \frac { - 3 } { 2 }$...$1^{st}$ term = $\frac{3}{2}$
Common difference = $\frac { 1 } { 2 } - \frac { 3 } { 2 }$ = -1

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