Question 12 Marks
How many terms of $ A.P. 27,24,21, \ldots$. should be taken so that their sum is zero $( 0 )$?
AnswerHere $a=27$ and $d=24-27=-3$,
According to the Question, the sum should be zero i.e. $S_n=0$
Applying the formula for sum of $n $ terms of an $A.P,$
$S_n=\frac{n}{2}(2 a+(n-1) d)$
$\Rightarrow 0=\frac{n}{2}(2 \times 27+(n-1)(-3))$
$\Rightarrow 0=\frac{n}{2}\{54+(-3 n+3)\}$
$\frac{n}{2}=0 $ or $ 54+(-3 n+3)=0$
$n=0 $ or $ 57-3 n=0$
Ignoring $n=0,$ we get
$57-3 n=0 $
$\Rightarrow 3 n=57 $
$\Rightarrow n=19$
Therefore $,19$ terms of this $A.P$.
should be taken so that their sum is zero $( 0 ).$
View full question & answer→Question 22 Marks
Which term of the progression $20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots \ldots$ is the first negative term?
AnswerGiven,
First term $a=20$
Common difference
$d=19 \frac{1}{4}-20$
$=\frac{77-80}{4}=-\frac{3}{4}$
Let $n^{\text {th }}$ term of $AP$ be the first negative term
$\Rightarrow a_n<0$
$\Rightarrow a_n<0$
$\Rightarrow 20+(n-1)\left(-\frac{3}{4}\right)<0 $
$\Rightarrow 20+\left(\frac{3}{4}-\frac{3 n}{4}\right)<0$
$\Rightarrow 20+\frac{3}{4}-\frac{3 n}{4}<0 $
$\Rightarrow \frac{83}{4}-\frac{3 n}{4}<0$
$\Rightarrow 83-3 n<0 $
$\Rightarrow 3 n>83 $
$\Rightarrow n>27 \frac{2}{3} $
$\Rightarrow n \geq 28$
Therefore, $28^{\text {th }}$ term is the first negative term of the $A.P$.
View full question & answer→Question 32 Marks
How many terms of the $A.P. 18, 16, 14 \ldots$ be taken so that their sum is zero?
AnswerThe first term of the $AP$ is $18$
Common difference $=16-18=-2$
Let the sum of the first $x$ terms of the $AP$ be $0.$
So, the sum of the first $x$ terms is given by$-$
$\frac{x}{2}[2 \times 18+(x-1)(-2)]=0$
$\Rightarrow \frac{x}{2}[36+(-2 x+2)]=0$
$\Rightarrow \frac{x}{2}[36-2 x+2]=0$
$\Rightarrow \frac{x}{2}[38-2 x]=0$
$\Rightarrow x[19-x]=0$
So, we get,
$x=0 \text { or } 19-x=0$
Ignoring $x=0$ we get,
$x=19$
Hence, the sum of first $19$ terms of the $AP$ is $0.$
View full question & answer→Question 42 Marks
The $4^{\text {th }}$ term of an $A.P$. is zero. Prove that the $25^{\text {th }}$ term of an $A.P$. is three times its $11^{\text {th }}$ term.
AnswerGiven, $4^{\text {th }}$ term is zero.
So, $a+3 d=0$
$\Rightarrow a=-3 d$
$n^{\text {th }}$
Term of an $A.P$. is given by,
$a_n=a+(n-1) d$
$a_{11}=a+10 d$
Substitute the value of $a$,
$a_{11}=-3 d+10 d=7 d$
Similarly, $a_{25}=a+24 d$
$a_{25}=-3 d+24 d=21 d$
Or $, a_{25}=3 \times 7 d$
$a_{25}=3 \times a_{11}$
Hence proved.
View full question & answer→Question 52 Marks
In an $AP,$ if $S_5+S_7=167$ and $S_{10}=235, $ then find the $AP,$ where $S_n$ denotes the sum of first terms.
Answer$S_5+S_7=167 \quad ($Given$)$
$S_{10}=235 \quad ($Given$)$
Sum of $n$ terms is,
$S_n=\frac{n}{2}\{2 a+(n-1) d\}$
Where,
$n=$ Number of terms
$a=$ First term
$b=$ Common difference
$S_5=\frac{5}{2}\{2 a+(5-1) d\} $ and $ S_7=\frac{7}{2}\{2 a+(7-1) d\}$
$\frac{5}{2}\{2 a+(5-1) d\}+\frac{7}{2}\{2 a+(7-1) d\}=167$
On simplifying
$5 a+10 d+7 a+21 d=167$
$12 a+31 d=167 \ldots \ldots(1)$
For the sum of first ten terms,
$\frac{10}{2}\{2 a+(10-1) d\}=235$
$10 a+45 d=235$
$2 a+9 d=47 \ldots \ldots(2)$
Multiply equation $(ii)$ by $6$ and subtract equation $(i)$ from it
$6(2 a+9 d)-(12 a+31 d)=6(47)-167$
$12 a+54 d-12 a-31 d=282-167$
$23 d=115$
$d=\frac{115}{23}=5$
Substitute the value of $d$ in equation $(ii),$ we get
$2 a+9(5)=47$
$2 a+45=47$
$2 a=47-45=2$
$a=\frac{2}{2}=1$
So the values of $a$ and $b$ are $d=5 a$
Hence, the $AP$ is $1,6,11,16,21,26, \ldots$.
View full question & answer→Question 62 Marks
Find the middle term of the $A.P. 6,13,20, \ldots$, $216.$
Answer$6,13,20, \ldots, 216$
In the given sequence $a=6, a_n=216$ and $d=7$
We know that $a_n=\alpha+(n-1) d$
Substitute $a=6, a_n=216$ and $d=7$ in the formula to find the number of terms we get,
$216=6+(n-1) 7$
$\Rightarrow 216=6+7 n-7$
$\Rightarrow 216=7 n-1$
$\Rightarrow 217=7 n$
$\Rightarrow n=\frac{217}{7}=31$
Therefore the number of terms in the given sequence is $31$ which is odd.
Therefore the middle term will be $\left(\frac{n+1}{2}\right)$ th term of the $A.P$.
$\Rightarrow \frac{31+1}{2}=16 ^{th}$ term
We know that nth term in an $A.P$. is calculated by the formula $a_n=a+(n-1) d$
$\Rightarrow 16^{ th}$ term $=6+(16-1) 7=6+105$
Now here $n=16$
Hence, the middle term is $111$.
View full question & answer→Question 72 Marks
The first and the last terms of an $AP$ are $5$ and $45$ respectively. If the sum of all its terms is $400 ,$ find its common difference.
AnswerLet $a$ be the first term and $d$ be the common difference.
Given
$a=5$
$a_n=45$
$S_n=400$
We know $a_n=a+(n-1) d$
$45=5+(n-1) d]$
$40=(n-1) d \ldots \ldots .$
And $S_n=\frac{n}{2}\left(a+a_n\right)$
$400=\frac{n}{2}(5+45)$
$\frac{n}{2}=\frac{400}{50}$
$n=2 \times 8=16$
On substituting $n=16$ in $(1)$
$40 =(16-1) d$
$40 =(15) d$
$d =\frac{40}{15}=\frac{8}{3}$
Thus, the common difference is $\frac{8}{3}$.
View full question & answer→Question 82 Marks
Find the number of natural numbers between $101$ and $999$ which are divisible by both $2$ and $5.$
AnswerNumbers that are divisible by both $2$ and $5$ will be are multiples of $10 .$
The natural numbers between $101$ and $999$
which are divisible by both $2$ and $5$ will be : $110, 120, 130, \ldots ., 990$.
This forms an $A.P.$ with first term as $110 ,$ common difference $10$ and last term as $990 .$
$n^{\text {th }}$ term of an $A.P. a_n=a+(n-1) d$
Substitute the value of $a, d$ and $a_n$
$\Rightarrow 990=110+(n-1) 10$
$\Rightarrow 990-110=10 n-10$
$\Rightarrow 880=10 n-10$
$\Rightarrow 10 n=890$
$\Rightarrow n=89$
Therefore, $89$ natural numbers lie between $101$ and $999$ which are divisible by $2$ and $5.$
View full question & answer→Question 92 Marks
How many three digit natural numbers are divisible by $7 ?$
AnswerFollowing are the three digit natural numbers divisible by $7 :$
$105,112,119,126.$ The given series is $A.P.$
First term $(a)=105$
Common diff $(d)=7$
$n^{\text {th }} \text { term }\left(a_n\right)=994$
It is known that the term of an $A.P.$ is given by,
$a_n=a+(n-1) d$
Substituting $a=105, d=7$ and $a_n=994$
$994=105+(n-1) 7$
$\Rightarrow 994=105+7 n-7$
$\Rightarrow 994=98+7 n$
$\Rightarrow 7 n=994-98$
$\Rightarrow 7 n=896$
$\Rightarrow n=128$
Thus $128$ three digit natural numbers are divisible by $7 .$
View full question & answer→Question 102 Marks
How many two-digit numbers are divisible by $3 ?$
AnswerNumbers divisible by $3$ are multiples of $3 .$
$3,6,9,12,15 \ldots$
The smallest two-digit number divisible by $3$ is $12$. And the largest $2-$digit number divisible by $3$ is $99 .$
So, the series of the $2-$digit multiples of $3$ starts with $12$ and ends with $99$ . The difference between the numbers is $3$ .
Therefore, the $A.P.$ is $12,15,18,21,24 \ldots 90,93,96,99$.
In the sequence, the first term, $a=12$.
The last term, $l=99$.
The common difference, $d=3$. The $n^{\text {th }}$ term is $a_n=99$.
Now, $a_n=a_1+(n-1) d$
$\Rightarrow 99=12+(n-1) 3 \Rightarrow 99=12+3 n-3$
$\Rightarrow 99=3 n+9 \Rightarrow 3 n=90$
$\Rightarrow n=30$
Therefore, there are $30$ two$-$digit numbers divisible by $3 .$
View full question & answer→Question 112 Marks
Find whether $-150$ is a term of the $AP \ 17,12,7$, $2, \ldots$ ?
AnswerGiven, the $AP \ 17,12,7,2, \ldots$
We have, $a=17, d=12-17=-5$
Let, $a_n=-150$
But, $a_n=a+(n-1) d$
Substitute the value of $a_n$ in the above equation.
$-150=17+(n-1)(-5)$
$-150=17-5 n+5$
$-150=22-5 n$
$-150-22=-5 n$
$-172=-5 n$
$\Rightarrow n=\frac{172}{5}$
But $"n "$ cannot be a fractions. $" n"$ is always a whole number.
Therefore, $-150$ is not a term of the given $AP .$
View full question & answer→Question 122 Marks
Find the sum of first 20 terms of an A.P. whose $n ^{\text {th }}$ term is given as $a_n=5-2 n$.
AnswerGiven, $a_n=5-2 n$
for $n=1, a_1=5-2(1)=3$
$
n=2, a_2=5-2(2)=1
$
$\therefore$ Common difference $=1-(3)=-2$
Sum of first n terms:
$
S_n=\frac{n}{2}[2 z+(n-1) d]
$ $\therefore$ Sum of first 20 terms is:
$
\begin{aligned}
S_n & =\frac{20}{2}[2(3)+(20-1)(-2)] \\
& =10(6-28) \\
& =10 \times(-32)=-320
\end{aligned}
$
Hence, sum of first 20 terms is -320 .
View full question & answer→Question 132 Marks
Determine the $A.P.$ whose third term is $5$ and seventh term is $9$ .
AnswerGiven, $T_3=5$ and $T_7=9$
We know that, $n^{\text {th }}$ term of an $A.P.$ is
$T_n=a+(n-1) d$
$5=a+4 d$
$\therefore 9=a+6 d$ and $9=a+6 d$
$Eq(i)-Eq \text { (ii), }$
$-4=-2 d $
$\Rightarrow d=2$
From eq $(i),$
$5=a+4(2)$
$\Rightarrow a=5-8=-3$
So, required $A.P.$ is $-3,-1,1,3,5,7,9, \ldots$.
View full question & answer→Question 142 Marks
In an $A.P.$ if the sum of third and seventh term is zero, find its $5^{\text {th }}$ term.
AnswerGiven, sum of third and seventh term of $A .P$. is zero.
We know that, $n ^{\text {th }}$ term of an $A.P.$ is
$T_n=a+(n-1) d$
$\therefore T_3+T_7=0$
$\Rightarrow a+2 d+a+6 d=0$
$\Rightarrow 2 a+8 d=0$
$\Rightarrow a+4 d=0$
Now, $T_5=a+(5-1) d=a+4 d=0$
Hence, $5^{\text {th }}$ term of $A.P.$ is zero.
View full question & answer→Question 152 Marks
Find the sum of first $8$ multiple of $3 .$
AnswerThe first $8$ multiples of $3$ are $3,6,9,12,15,18, 21, 24 .$
It form an $AP$ with the first term $(a)=3$,
the difference $(d)=3$ and the last term $(l)=24$.
The sum of as an $AP:$
$S_n=\frac{n}{2}(a+l)$
$=\frac{8}{2}(3+24)=4 \times 27$
$\Rightarrow S_n=108$
So the sum of the first $8$ multiples of $3$ is equal to $108.$
View full question & answer→Question 162 Marks
Write the next two terms of the A.P. :
$\sqrt{27}, \sqrt{48}, \sqrt{75}, \ldots \ldots$
AnswerGiven A.P. is $\sqrt{27}, \sqrt{48}, \sqrt{75} \ldots \ldots$
Here, $a _1=\sqrt{27}=3 \sqrt{3}$
$
a_2=\sqrt{48}=4 \sqrt{3}
$ $\therefore$ Common difference $=4 \sqrt{3}-3 \sqrt{3}$
$
=\sqrt{3}(4-3)=\sqrt{3}
$
Now, Given $a _3=\sqrt{75}=5 \sqrt{3}$
$
\therefore \quad a_4=6 \sqrt{3}=\sqrt{108}
$
and $a_5=7 \sqrt{3}=\sqrt{147}$
Hence, next two terms are $\sqrt{108}$ and $\sqrt{147}$
View full question & answer→Question 172 Marks
Determine the $36^{\text {th }}$ term of the A.P. whose first two terms are -3 and 4 respectively.
AnswerGiven, first two terms of AP are -3 and 4 .
$\therefore a_1=-3$ and $a_2=4$
$\therefore$ Common difference, $d =4-(-3)=4+3=7$
We know that, $n ^{\text {th }}$ term of A.P. is
$
\therefore \quad \begin{aligned}
T_{n} & =a+(n-1) d \\
T_{n} & =a+(n-1) d \\
& =-3+(36-1) d \\
& =-3+35(7) \\
& =-3+245 \\
& =242
\end{aligned}
$
View full question & answer→Question 182 Marks
Determine the $A.P.$ Whose third term is 16 and $7^{\text {th }}$ term exceeds the $5^{\text {th }}$ term by $12.$
Answer$\Rightarrow$ we know that $n ^{\text {th }}$ term of $AP$
$a_x=a+(n-1) d$
where $a =$ first term $d =$ common difference
So,
$a_3 =a+(3-1)$
$a_3 =a+2 d$
$16 =a+2 d\quad\left(\text { Given } 3^{\text {rd }} \text { term is } 16\right)$
$a+2 d =16 \ldots \ldots(1)$
$16=a+2 d$
Also, $a _7= a +(7-1) d$
$a_7=a+6 d \ldots \ldots(2)$
$\text { Similarly } a_0=a+(5-1) d$
$a_0=a+4 d \ldots \ldots(3)$
Given that
$7^{\text {th }}$ term exceed the $5^{\text {th }}$ term by $12$
$7^{\text {th }}$ term -5 th term $=12$
$a _7- a _{ j }=12$
$\Rightarrow a +6 d- a -4 d=12$
$[$from equation $(ii) \ (iii)]$
$\Rightarrow 2 d=12 \Rightarrow d=\frac{12}{2}=6$
putting the value of $d^{\prime}$ ' in equation $(i),$ we get
$a+2 \times 6=16$
$\Rightarrow a+16-12=4$
Hence, first term of $A.P. =4$
Second term of $A.P.$ $=$ First term $+$ Common difference $=4+6=10$
Third term of A.P. $=$ Second term $+$ Common difference $=10+6=16$ And So On,
So, the $A.P.$ is $4,10,16, \ldots-$
View full question & answer→Question 192 Marks
How many multiples of $4$ lie between $10$ and $205$ ?
AnswerLet the number of multiples of $4$ lie between $10$ and $205$ be $n$.
$\therefore$ first multiples $(a) =12$
last multiples $( l )=204$
Common difference $( d )=4$
$\because l=a+(n-1) d$
$204=12+(n-1) 4$
$\Rightarrow 4(n-1)=204-12$
$\Rightarrow 4(n-1)=192 $
$\Rightarrow(n-1)=\frac{192}{4}=48$
$\Rightarrow n=48+1=49$
Hence, required number of multiples $=49$.
View full question & answer→