[2 Mark Question Answer]

Question 12 Marks

The $n^{th}$ term of a sequence is $8 - 5$n. Show that the sequence is an $A.P.$

Answer

$t_n = 8 - 5n$
Replacing n by $(n + 1)$, we get
$t_{n+1} = 8 - 5(n + 1) = 8 - 5n - 5 = 3 - 5n$
Now,
$t_{n+1} - t_n = (3 - 5n) - (8 - 5n) = -5$
Since, $(t_{n+1} - t_n)$ is independent of n and is therefore a constant.
Hence, the given sequence is an $A.P.$

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Question 22 Marks

Find the sum of first $20$ terms of an A.P. whose first term is $3$ and the last term is $57.$

Answer

Here,
First term, $a = 3$
Last term, $l = 57$
$n = 20$
$S=\frac{n}{2}(a+I)$
$=\frac{20}{2}(3+57) $
$=10 \times 60$
$ =600$

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Question 32 Marks

Find the sum of first 10 terms of the A.P. $4 + 6 + 8 + .......$

Answer

Here,
First term, $a = 4$
Common difference, $d = 6 - 4 = 2$
$n = 10$
$S=\frac{n}{2}(2 a+(n-1) d) $
$ =\frac{10}{2}(2(4)+9(2)) $
$ =5(8+18) $
$=5 \times 26$
$=130$

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Question 42 Marks

Find the arithmetic mean of:
$(m + n)^2$ and $(m - n)^2$

Answer

Arithmetic mean of $(m+n)^2$ and $(m-n)^2=\frac{(m+n)^2+(m-n)^2}{2}$
$=\frac{m^2+n^2+2 m n+m^2+n^2-2 m n}{2}$
$=\frac{2\left(m^2+n^2\right)}{2}$
$ =m^2+n^2$

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Question 52 Marks

How many multiples of $4$ lie between $10$ and $250?$

Answer

Numbers between $10$ and $250$ which are multiple of $4$ are as follows:
$12, 16, 20, 24,.........248$
Clearly this forms an A.P. with first term $a = 12,$
common difference $d = 4$ and last term $I = 248$
$I = a + (n - 1)d$
$\Rightarrow 248=12+(n-1) \times 4$
$ \Rightarrow 236=(n-1) \times 4 $
$ \Rightarrow n-1=59$
$\Rightarrow n=60$
Thus, $60$ multiples of $4$ lie between $10$ and $250.$

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Question 62 Marks

Find the sum of first 12 natural numbers each of which is a multiple of 7.

Answer

First 12 natural numbers which are multiples of 7 are as follows
7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84
Clearly thus forms an A.P with first term a = 7, common difference d = 7 and last term l = 84
Sum of first n terms $=S=\frac{n}{2}[a+l]$
$\Rightarrow$ Sum of first 12 natural numbers whihc are multiple of 7 = $\frac{12}{2}[7+84]$
= 6 x 91
= 546

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Question 72 Marks

Find the sum of the first $22$ terms of the A.P. $: 8, 3, -2, ………$

Answer

The given A.P is $8, 3, -2, .....$
Here $a = 8, d= 3 - 8 = -5$ and $n = 22$
$\therefore S=\frac{n}{2}[2 a+(n-1) d] $
$ =\frac{22}{2}[2 \times 8+(22-1) \times(-5)] $
$=11[16+21 \times(-5)] $
$=11[16-105] $
$=11 \times(-89) $
$=-979$

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Question 82 Marks

Find the value of p, if x, 2x + p and 3x + 6 are in A.P

Answer

Since x, 2x + p and 3x + 6 are in A.P, we have
(2x + p) - x = (3x + 6) - (2x + p)
$\Rightarrow$ 2x + p - x = 3x + 6 - 2x - p
$\Rightarrow p+p=x-x+6$
$\Rightarrow 2 p=6$
$\Rightarrow p=3$

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Question 92 Marks

If the pth term of an A.P. is $(2p + 3),$ find the A.P.

Answer

pth term of an AP. $= 2p + 3$
$\Rightarrow t_p=2 p+3$
Putting $t=1,2,3$, ....., we get
$t_1=2 \times 1+3=2+3=5$
$ t_2=2 \times 2+3=4+3=7$
$t_3=2 \times 3+3=6+3=9$ and so on
Thus the A.P is $5, 7, 9, ....$

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Question 102 Marks

The nth term of the sequence is (2n – 3), find its fifteenth term.

Answer

$n^{\text {th }}$ term of A.P $=(2 n-3)$
$\Rightarrow t_n=2 n-3$
$\therefore 15^{\text {th }}$ term $=t_{15}=2 \times 15-3=30-3=27$

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Question 112 Marks

Which of the following are in arithmetic progression
$\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}$

Answer

$\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5} $
$ d_1=\frac{1}{3}-\frac{1}{2}=\frac{2-3}{6}=-\frac{1}{6}$
$d_2=\frac{1}{4}-\frac{1}{3}=\frac{3-4}{12}=-\frac{1}{12}$
Since $d_1 \neq d_2$, the given sequence is not in arithematic progression

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Question 122 Marks

Which of the following are in arithmetic progression

5, 9, 12, 18

Answer

$5,9,12,18$
$d_1=9-5=4$
$d_2=12-9=3$
Since $d_1 \neq d_2$ the given sequence is not in arithematic progression

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Question 132 Marks

Which of the following are in arithmetic progression $15, 12, 9, 6$

Answer

$15, 12, 9, 6$
$d_1=12-15=-3$
$d_2=9-12=-3$
$d_3=6-9=-3$
Since $d_1=d_2=d_3$, the given sequence is in arithemetic progression

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