1 Marks Question

Question 11 Mark

What is the common difference of an $A.P.$ in which $a_{21}-a_7=84$ ?

Answer

Let be the first term $d$ be the common difference of Arithmetic progression.
As,
$a_n=a+(n-1) d$
$\therefore a_{21}=a+(21-1) d=a+20 d$
$a_7=a+(7-1) d=a+6 d$
Given $a_{21}-a_7=84$
$\Rightarrow(a+20 d)-(a+6 d)=84$
$\Rightarrow a+20 d-a-6 d=84 $
$\Rightarrow 20 d-6 d=84$
$\Rightarrow 14 d=84$
Dividing both sides by $14$
$\Rightarrow \frac{14 d}{14}=\frac{84}{14} $
$\Rightarrow d=6$
Therefore, the common difference of $A.P.$ is $6$

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Question 21 Mark

Find the $9^{t h}$ term from the end $($towards the first term$)$ of the $A.P.\ 5, 9, 13,..........185.$

Answer

The last term of the $AP$ from the end,
$l=185$ Common difference, $d=9-5=4$
To find the $9^{\text {th }}$ term from the end,
rearrange the $A.P.\ 185,181, \ldots \ldots ., 13,9,5$
The first term of this $A.P.$ is $185$ and the common difference is $-4.$
The $n^{\text {th }}$ term of the $AP$ is given by $a_n=a+(n-1) d$.
So, the $9^{\text {th }}$ term of the $A.P.$ is
$=185+(9-1)(-4)$
$=185-8 \times 4=185-32=153$
Hence, the $9^{\text {th }}$ term from the end is $153.$

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Question 31 Mark

For what value of $k$ will $k+9,2 k-1,2 k+7$ are the consecutive terms of an $A.P.$?

Answer

Arithmetic mean, $2 b=a+c$
$a=k+9, b=2 k-1, c=2 k+7$
By putting the values of $a, b$ and $c$ in equation
$2(2 k-1)=k+9+2 k+7$
$4 k-2=3 k+16$
$4 k=3 k+18$
$k=18$
Hence, for $k=18 ; k+9,2 k-1,2 k+7$ will be the consecutive terms of an $A.P.$

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Question 41 Mark

In an $AP,$ if the common difference $(d)=-4,$ and the seventh term $\left(a_7\right)$ is $4 ,$ then find the first term.

Answer

The $n^{th}$ term $=a+(n-1) d$
Where a $=$ first term, $n=7, d=-4$
Putting these values,
$a+(7-1)(-4)=4$
$a+6(-4)=4$
$\Rightarrow a=4+24$
$\Rightarrow a=28$
So the first term is $28 .$

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Question 51 Mark

Find the $11^{\text {th }}$ term of the A.P. $-27,-22,-17$, -12........

Answer

$\begin{aligned} \text { AP : }-27,-22,-17, & -12 \ldots \ldots \ldots \ldots \\ a & =-27 \\ d & =-22+27 \\ d & =+5 \\ n & =11 \\ T_{11} & = a +10 d \\ & =-27+10(5)=-27+50 \\ T_{11} & =22\end{aligned}$

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Question 61 Mark

Find the value of $x$ so that $-6, x, 8$ are in $A.P.$

Answer

$a_1=-6$
$a_2=x$
$a_3=8$
$a_2-a_1=a_3-a_2$
$x-(-6)=8-x$
$x-(-6)=8-x$
$ x+6 =8-x$
$2 x =2$
$x =1$

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Question 71 Mark

Find the number of terms in the $ A.P: \ 18, 15 \frac{1}{2}, 13, \ldots,-47$.

Answer

Given, first term $(a)=18$
last term $(l) =-47$
Common difference $(d)$
$=15 \frac{1}{2}-18=\frac{31}{2}-18=\frac{31-36}{2}=\frac{-5}{2}$
Let the number of terms in $A.P$ be $n$
$\because l=a+(n-1) d$
$-47=18+(n-1) \times\left(\frac{-5}{2}\right)$
$\Rightarrow-47=18-\frac{5}{2} n+\frac{5}{2}$
$\Rightarrow-47=-\frac{5}{2} n+\frac{41}{2}$
$\Rightarrow-\frac{5}{2} n=-47-\frac{41}{2}$
$\Rightarrow-\frac{5}{2} n=\frac{-94-41}{2}$
$\Rightarrow-5 n=-135 $
$\Rightarrow n=\frac{135}{5}=27$
Hence, the number of terms in $A \cdot P =27$

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