Question 13 Marks
The sum of $n$ term of an $A.P$ is $3 n^2+5 n$. Find the $A.P$ and its $15^{\text {th }}$ term.
Answer$S_n=3 n^2+5 n$
$S_1=3(1)^2+5(1)=8$
As we know, $S_1=a_1=a=8$
$S_2=3(2)^2+5(2)=12+10=22$
So, $a_2=S_2-S_1=22-8=14$
$S_3=3(3)^2+5(3)=27+15=42$
So, $a_3=S_3-S_2=42-22=20$
So, the $A.P$ is $8,14,20$, with common difference $d=a_2-a_1=14-8=6$
Now, we will find the $15^{\text {th }}$ term of this $A.P.$
$a_n=a+(n-1) d$
$\Rightarrow a_{15}=8+(15-1) \times 6$
$\Rightarrow a_{15}=8+14 \times 6$
$\Rightarrow a_{15}=92$
So, $15^{\text {th }}$ term of the $A.P$ is $92 .$
View full question & answer→Question 23 Marks
Find the sum of terms of the series $\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+\ldots \ldots$
AnswerBreaking the given series in two series, we'll get
$\Rightarrow(4+4+4 \ldots .)-\left(\frac{1}{n}+\frac{2}{n}+\frac{3}{n} \ldots .\right)$
$\Rightarrow(4+4+4 \ldots .)-\frac{1}{n}(1+2+3+\ldots .)$
We Know that sum of an $A.P$ is given by the formula
$S_n=\frac{n}{2}(2 a+(n-1) d)$
For the sequence $(4+4+4 \ldots)$.
$a=4, d=0$
$\therefore S_n=\frac{n}{2}(2 \times 4+(n-1) 0)$
$\Rightarrow S_n=\frac{n}{2}(8)=4 n$
For the sequence $\frac{1}{n}(1+2+3 \ldots \ldots)$
$a=1, d=1$
$\Rightarrow S_n=\frac{1}{n} \times \frac{n}{2}(2 \times 1+(n-1) 1)$
$\Rightarrow S_n=\frac{1}{2}(2+n-1)$
$\Rightarrow S_n=\frac{1}{2}(n+1)$
Now, combining sum of both sequences with respect to $(4+4+4 \ldots)-.$
$\left(\frac{1}{n}+\frac{2}{n}+\frac{3}{n} \ldots\right)$, we get;
$\Rightarrow 4 n-\frac{1}{2}(n+1)$
$\Rightarrow \frac{8 n-n-1}{2} $
$\Rightarrow \frac{7 n-1}{2}$
Hence, the sum of the $n^{t h}$ term of the series is $\frac{7 n-1}{2}$.
View full question & answer→Question 33 Marks
If $m^{t h}$ term of an $A.P.$ is $\frac{1}{n}$ and $n^{t h}$ term is $\frac{1}{m}$, then find the sum of its first $m n$ terms.
AnswerWe know that $n^{\text {th }}$ term of an $A.P$ is given by
$a_n=a+(n-1) d$
Where, $a$ is first term and $d$ is the common difference of the given $A.P.$
So, $n^{\text {th }}$ term $=a+(n-1) d=\frac{1}{m}$
Or, $a=\frac{1}{m}-(n-1) d \ldots \ldots(1)$
And, $m^{t h}$ term $=a+(m-1) d=\frac{1}{n}$
Or, $a=\frac{1}{n}-(m-1) d \ldots \ldots(2)$
Equating $(i)$ and $(ii)$, we'll get
$\Rightarrow \frac{1}{m}-(n-1) d=\frac{1}{n}-(m-1) d$
$\Rightarrow \frac{1}{m}-\frac{1}{n}=(n-1) d-(m-1) d$
$\Rightarrow \frac{n-m}{m n}=n d-d-m d+d$
$\Rightarrow \frac{n-m}{m n}=(n-m) d$
$\Rightarrow d=\frac{1}{m n} \ldots \ldots(3)$
Putting the value of $d$ in equation $(i)$, we'll get
$a=\frac{1}{m}-(n-1)\left(\frac{1}{m n}\right)$
$\Rightarrow a=\frac{1}{m}-\frac{n}{m n}+\frac{1}{m n}$
$\Rightarrow a=\frac{1}{m}-\frac{1}{m}+\frac{1}{m n}$
$\Rightarrow a=\frac{1}{m n} \ldots \ldots(4)$
Now, the $m n^{\text {th }}$ term will be given by
$m n^{\text {th }} \text { term }=a+(m n-1) d$
Putting the value of a andd in the above equation.
$\Rightarrow m n^{t h} \text { term }=\frac{1}{m n}+(m n-1)\left(\frac{1}{m n}\right)$
$\Rightarrow m n^{t h} \text { term }=\frac{1}{m n}+\frac{m n}{m n}-\frac{1}{m n}$
$\Rightarrow m n^{t h}=a_{m n}=\text { term }=1 \ldots \ldots(5)$
We know that the sum of $n$ terms of $A.P$ is given by;
$S_n=\frac{n}{2}\left(a+a_n\right)$
Therefore, sum of first mn terms is given by;
$S_{m n}=\frac{m n}{2}\left(a+a_{m n}\right)$
Put values of $a$ and $a_{m n}$ from equation $(iv)$ and $(v)$ we get;
$S_{m n}=\frac{m n}{2}\left(\frac{1}{m n}+1\right)$
$\Rightarrow S_{m n}=\frac{1}{2}+\frac{m n}{2}$
$\Rightarrow S_{m n}=\frac{1}{2}(1+m n)$
Hence, the sum of first $m n$ terms is
$S_{m n}=\frac{1}{2}(1+m n)$
View full question & answer→Question 43 Marks
Which term of $A.P. 3,15,27,39, \ldots \ldots$ will be $132$ more than its $54^{\text {th }}$ term.
AnswerHere, $a=3$ and $d=15-3=12$.
Now, we first find the $54^{\text {th }}$ term of the given $A.P$.
As we know that $n^{\text {th }}$ term is given by $a_n=a+(n-1) d$ where $n=54$
$\Rightarrow a_{54}=3+(54-1) 12 $
$\Rightarrow a_{54}=3+53 \times 12$
$\Rightarrow a_{54}=3+636 $
$\Rightarrow a_{54}=639$
Let's say $a$ is the term which is $132$ more than $54^{\text {th }}$ term.
So, according to the Question,
$a_n=a_{54}+132$
$\Rightarrow a+(n-1) d=639+132$
$\Rightarrow 3+(n-1) 12=771 $
$\Rightarrow 3+12 n-12=771$
$\Rightarrow 12 n-9=771 $
$\Rightarrow 12 n=771+9$
$\Rightarrow 12 n=780 $
$\Rightarrow n=\frac{780}{12}$
$\Rightarrow n=65$
Therefore, $65^{\text {th }}$ term will be $132$ more than $54^{\text {th }}$ term.
View full question & answer→Question 53 Marks
If the sum of first $7$ terms of an $A.P$. is $49$ and that of its first $17$ terms is $289 ,$ find the sum of first $n$ terms of the $A.P$.
AnswerLet us suppose that the first term and the common difference of the given $AP$ is $'a\ '$ and $'d\ '$ respectively.
We are given the sum of first $7$ terms, $S_7=49$
We know that
$S=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \frac{7}{2}(2 a+6 d)=49$
$\Rightarrow \frac{7}{2} \times 2(a+3 d)=49$
$\Rightarrow a+3 d=7 \ldots \ldots(1)$
Sum of the first $17 $ terms, $S_{17}=289$
$\frac{17}{2}(2 a+16 d)=289$
$\Rightarrow \frac{17}{2} \times 2(a+8 d)=289$
$\Rightarrow a+8 d=\frac{289}{17}=17$
$\Rightarrow a+8 d=17 \ldots \ldots(2)$
Subtracting $(2)$ from $(1),$
$5 d=10$
$\Rightarrow d=2$
Substituting the value of $d$ in $(1),$ we get,
$\Rightarrow a+3(2)=7$
$\Rightarrow a=7-6=1$
Now sum of first $n$ terms of the $A.P$. will be
$S_n=\frac{n}{2}[2 a+(n-1) d]$
$=\frac{n}{2}[2 \times 1+2(n-1)]$
$=n(1+n-1)=n^2$
Hence, the sum of the first $n$ terms of the $AP$ is $n^2$.
View full question & answer→Question 63 Marks
If the ratio of the sum of first $n$ terms of two $A.P\ '$ is $(7 n+1):(4 n+27)$ find the ratio of their $m ^{\text {th }}$ terms.
AnswerSum of first $n$ terms of an $A.P$ is,
$S_n=\frac{n}{2}(2 a+(n-1) d)$
According to the Question,
$\frac{\frac{n}{2}(2 a+(n-1) d)}{\frac{n}{2}\left(2 a^{\prime}+(n-1) d\right)}=\frac{(7 n+1)}{(4 n+27)}$
$\Rightarrow \frac{(2 a+(n-1) d)}{\left(2 a^{\prime}+(n-1) d^{\prime}\right)}=\frac{(7 n+1)}{(4 n+27)} \ldots \ldots(1)$
term of an $A.P$. is $a_n=a+(n-1) d$
So the ratio of the $m^{\text {th }}$ terms is
$a_m: a_m^{\prime}=a+(m-1) d: a^{\prime}+(m-1) d^{\prime}$
Multiply both sides by $2$
$a_m: a_m^{\prime}=2 a+2[(m-1) d]: 2 a^{\prime}+2\left[(m-1) d^{\prime}\right]$
$=2 a+(2 m-2) d: 2 a^{\prime}+(2 m-2) d^{\prime}$
Replace $n$ by $(2 m-1)$ in equation $(1)$ we get;
$\frac{(2 a+(2 m-2) d)}{\left(2 a^{\prime}+(2 m-2) d^{\prime}\right)}=\frac{(7(2 m-1)+1)}{(4(2 m-1)+27)}$
$\frac{2(a+(m-1) d)}{2\left(a^{\prime}+(m-1) d^{\prime}\right)}=\frac{(14 m-6)}{(8 m+23)}$
$\frac{a_m}{a_m}=\frac{(14 m-6)}{(8 m+23)}$
Hence, ratio of the $m^{\text {th }}$ terms is $\frac{(14 m-6)}{(8 m+23)}$
View full question & answer→Question 73 Marks
The $14^{\text {th }}$ term of an $AP$ is twice its $8^{\text {th }}$ term. If its $6^{\text {th }}$ term is $-8$ , then find the sum of its first $20$ terms.
Answer$14 t h$ term of the $AP =2 \times 8$ th term of the $AP$
$($Given$)$
Also, $a_n=a+(n-1) d$
Given that $ a_{14}=2 \times a_8$
$\Rightarrow a+(14-1) d=2\{a+(8-1) d\}$
$\Rightarrow a+13 d=2 a+14 d$
$ a+d=0 \ldots \ldots(a)$
Also,
$a_6=-8 \quad\quad ($give$)$
$a+(6-1) d=-8$
$a+5 d=-8 \ldots \ldots(b)$
Subtract equation $(a)$ from $(b)$
$a+5 d-(a+d)=-8-0$
$\Rightarrow a+5 d-a-d=-8-0$
$\Rightarrow 4 d=-8$
$\Rightarrow d=\frac{-8}{4}=-2$
Substitute the value of $d$ in equation $(a)$
$a-2=0$
$\Rightarrow a=2$
So the values of $a$ and $d$ are $2$ and $-2$ respectively.
Sum of first $n$ term is given by;
$S_n=\frac{n}{2}\{2 a+(n-1) d\}$
$S_{20}=\frac{20}{2}\{2 \times 2+(20-1)(-2)\}$
$S_{20}=10(4-38)$
$S_{20}=-340$
View full question & answer→Question 83 Marks
If $S_n$ denotes the sum of first n terms of an $A.P$., prove that $S_{12}=3\left(S_8-S_4\right)$.
AnswerTo prove: $S_{12}=3\left(S_8-S_4\right)$
Proof: The sum of $n$ terms of an $A P$ is given by the formula,
$S_n=\frac{n}{2}[2 a+(n-1) d]$
Where $'n\ '$ denotes the number of terms in an $AP\, ' a\ '$ denotes the first term and $'d\ '$ denotes the common difference of $AP$.
$S_8$ denotes the sum of first $ 8$ terms of an $AP$.
$S_8=\frac{8}{2}[2 a+(8-1) d]$
$\Rightarrow S_8=4[2 a+7 d]=8 a+28 d \ldots \ldots(1)$
$S_4$ denotes the sum of first $4$ terms of an $AP$.
$S_4=\frac{4}{2}[2 a+(4-1) d]$
$\Rightarrow S_4=2[2 a+3 d]=4 a+6 d \ldots \ldots(2)$
Now, from equations $(i)$ and $(ii)$;
$3\left(S_8-S_4\right)=3\{8 a+28 d-(4 a+6 d)\}$
$3\left(S_8-S_4\right)=3\{8 a+28 d-(4 a+6 d)\}$
$=3(4 a+22 d)$
$=6(2 a+11 d)$
$=\frac{12}{2}[2 a+(12-1) d]$
$=S_{12}$
Hence proved.
View full question & answer→Question 93 Marks
The sum of the $5^{\text {th }}$ and the $9^{\text {th }}$ terms of an $AP$ is $30$ . If its $8^{\text {th }}$ term is three times its $25^{\text {th }}$ term, find the $AP$.
Answer$5^{\text {th }}$ term of an $A.P. =a+4 d$
$9^{\text {th }}$ term of an $A.P. =a+8 d$
Sum of $5^{\text {th }}$ and $9^{\text {th }}$ term of an $A.P. =30$
$\Rightarrow a+4 d+a+8 d=30$
Add the like terms
$\Rightarrow 2 a+12 d=30$
$\Rightarrow 2(a+6 d)=30$
$\Rightarrow a+6 d=\frac{30}{2}=15$
$\Rightarrow a+6 d=15 \ldots .(1)$
$25^{t h} $ term of an ${A.P. }=a+24 d$
$8^{t h} $ term of an ${A.P. }=a+7 d$
Given that the $25^{\text {th }}$ term is three times its $8^{\text {th }}$ term
$\Rightarrow a+24 d=3(a+7 d)$
$\Rightarrow a+24 d=3 a+21 d$
$\Rightarrow 3 a-a=24 d-21 d$
$\Rightarrow 2 a=3 d \ldots \ldots(2)$
Equation $(1)$ can bewritten as
$a+2(3 d)=15 \ldots \ldots(3)$
Substitute the value of $3 d$ from equation $(2)$ in the above equation $(3)$.
Therefore,
$a+2(2 a)=15$
$a+4 a=15$
$5 a=15$
$a=3$
Substitute the value of a in equation $(2)$
$2 \times 3=3 d$
$d=2$
View full question & answer→Question 103 Marks
If the seventh term of an $A P$ is $\frac{1}{9}$ and its ninth term is $\frac{1}{9}$, find its $63^{-d}$ term.
AnswerLet $a$ be the first term and $d$ be the common difference of the given $A.P.$
$a_7=\frac{1}{9}$
$a_9=\frac{1}{7}$
$a_7=a+(7-1) d=\frac{1}{9}$
$a+6 d=\frac{1}{9} \ldots . .(1)$
$a_9=a+(9-1) d=\frac{1}{7}$
$a+8 d=\frac{1}{7} \ldots . .(2)$
Subtracting $(1)$ from $(2)$
$2 d=\frac{2}{63} $
$\Rightarrow d=\frac{1}{63}$
Put $d=\frac{1}{63}$ in the equation $(1)$
$a+\left(6 \times \frac{1}{63}\right)=\frac{1}{9}$
$\Rightarrow a=\frac{1}{63}$
$a_{63}=a+(63-1) d=\frac{1}{63}+62\left(\frac{1}{63}\right)=\frac{63}{63}=1$
Hence, $a_{63}=1$
View full question & answer→Question 113 Marks
Find the number of terms of the $A.P. 18,15 \frac{1}{2}, 13, \ldots \ldots \ldots . .,-49 \frac{1}{2}$ and find the sum of all its terms.
AnswerThe given $A.P$. is:
$18,15 \frac{1}{2}, 13, \ldots \ldots \ldots \ldots,-49 \frac{1}{2}$
And,
First term $\left(a_1\right)=18$
Second term $\left(a_2\right)=15 \frac{1}{2}$
Common difference $(d)=\alpha_2-\alpha_1$
$=15 \frac{1}{2}-18=\frac{31-36}{2}$
$\Rightarrow d=-\frac{5}{2}$
Let the $A.P$. has $n$ terms.
$a_n=-49 \frac{1}{2}$
Also, $a_n=a+(n-1) d$
Thus, $-49 \frac{1}{2}=\alpha+(n-1) d$
$\frac{-99}{2}=18+(n-1) \frac{-5}{2}$
$\frac{-99}{2}=18+\left(\frac{-5 n}{2}+\frac{5}{2}\right)$
$-99=41-5 n$
$5 n=140$
$n=28$
$\therefore$ The number of terms in the given $A.P.$ is $28$
Sum of $n$ terms $\left(S_n\right):$ Activate Windows
$S_n=\frac{n}{2}[2 a+(n-1) d]$
$S_{28}=\frac{28}{2}\left[2 \times 18+(28-1) \frac{-5}{2}\right]$
$S_{28}=14\left[\frac{72-135}{2}\right]$
$S_{28}=7 \times-63$
$S_{28}=-441$
Thus the sum of all the terms of the $A.P$. is $-44$1 .
View full question & answer→Question 123 Marks
Find the sum of all multiples of $7$ lying between $500$ and $900$ .
AnswerBetween $500$ and $900 ,$ the first multiple of $7 $ is $a=504$ and the last multiple of $7$ is $l=896$ and the common difference is $d=7$
Hence, the $A.P$ will be
$504,511,518, \ldots ., 896$
The number of terms in this $A.P$ is given by;
$a_n=a+(n-1) d$
$\Rightarrow 896=504+(n-1) 7 $
$\Rightarrow(n-1) 7=392$
$\Rightarrow n-1=56$
$\Rightarrow n=57$
Therefore, the sum of all the multiples of $7$ lying between $500$ and $900$ is given by
$S_n=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow S_{57}=\frac{57}{2}[2 \times 504+(57-1) 7]$
$\Rightarrow S_{57}=\frac{57}{2}[1008+56 \times 7]$
$\Rightarrow S_{57}=\frac{57}{2}[1008+392] $
$\Rightarrow S_{57}=\frac{57}{2} \times 1400$
$\Rightarrow S_{57}=57 \times 700 $
$\Rightarrow S_{57}=39900$
Therefore, the sum of all multiples of $7$ lying between $500$ and $900$ is $39900$ .
View full question & answer→Question 133 Marks
Determine the $AP$ whose fourth term is $18$ and the difference of the ninth term from the fifteenth term is $30$ .
AnswerGiven that, fourth term $a_4=18$ and the difference of the ninth term from the fifteenth term is $30 ,$
$a_{15}-a_9=30$
Clearly,
$a_4=18$
$\Rightarrow a+3 d=18$
$a_{15}-a_9=30$
$a+14 d-(a+8 d)=30$
$a+14 d-a-8 d=30$
$\Rightarrow 6 d=30$
$\Rightarrow d=\frac{30}{6}=5$
Substitute the value of $d$ in equation $(i)$
$a+3(5)=18$
$a=18-15=3$
First term $a_1=3$
Second term $a_2=a_1+d=3+5=8$
Third term $a_3=a_1+2 d=3+2(5)=13$
Therefore, the $AP$ is $3,8,13, \ldots$
View full question & answer→Question 143 Marks
Find the value of the middle term of the following $AP: -6,-2,2, \ldots ., 58$.
AnswerClearly, $-6,-2,2, \ldots ., 58$ is an $AP$ with the first term and common difference $d=-2-(-6)=-2+$
$6=4$.
Let there be $n$ terms in the given $AP.$ Then,
$a_n=58$
$a_n=a+(n-1) d$
$a+(n-1) d=58$
$(-6)+(n-1) 4=58$
$-6+4 n-4=58$
$4 n-10=58$
$4 n=58+10$
$\Rightarrow 4 n=68 $
$\Rightarrow n=\frac{68}{4}=17$
Here, $n$ is odd, so, the middle te $rm$ is $\left(\frac{n+1}{2}\right)^{t h}$ term
i.e. $\left(\frac{17+1}{2}\right)^{t h}=\left(\frac{18}{2}\right)^{t h} $
$\Rightarrow 9^{t h}$ term is the middle term and is given by,
$a_9=a+(9-1) d$
$=-6+8(4)=-6+32=26$
The value of the middle term of the given $AP$ is $26$.
View full question & answer→Question 153 Marks
Find the middle term of the $ A.P. 7,13,19, \ldots, 247.$
Answer$AP: 7,13,19, \ldots .247$
$a=7$
$d=13-7=6$
$T_n=247$
$T_n=a+(n-1) d$
$\Rightarrow 247=7+(n-1) 6$
$\Rightarrow 240=6 n-6$
$\Rightarrow 246=6 n$
$\Rightarrow n=41$
Since, $n$ is odd, so medium term is $\frac{n+1}{2}$
$\Rightarrow \frac{41+1}{2}$
$\Rightarrow 21^{\text {st }} $ term
$T_{21}=a+20 d$
$=7+20 \times 6$
$=7+120$
$\Rightarrow T_{21}=127$
View full question & answer→Question 163 Marks
Which term of the $A.P. 20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4} ,.. \quad$ is the first negative term.
Answer$AP: 20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4} \ldots \ldots .$
$ a=20$
$ d=19 \frac{1}{4}-20$
$\Rightarrow \frac{77}{4}-20$
$\Rightarrow \frac{77-80}{4}$
$\Rightarrow \frac{-3}{4}$
Let the $n ^{\text {th }}$ term of the $AP$ be the first negative term.
Then $T_{n}<0$
$\Rightarrow[a+(n-1) d]<0$
$\Rightarrow\left[20+(n-1)-\frac{3}{4}\right]<0$
$\Rightarrow\left[20-\frac{3 n}{4}+\frac{3}{4}\right]<0$
$\Rightarrow \frac{80-3 n+3}{4}<0$
$\Rightarrow \frac{83-3 n}{4}<0$
$\Rightarrow \frac{83}{4}-\frac{3 n}{4}<0$
$\Rightarrow \frac{83}{4}<\frac{3 n}{4}$
$\Rightarrow n>27.3$
$\Rightarrow n=28$
Hence the $28^{\text {th }}$ term is the first negative term of the $A.P$.
View full question & answer→Question 173 Marks
Find the sum of first 16 terms of the A.P. whose $n ^{\text {th }}$ term is given by $a _{ n }=5 n -3$.
AnswerGiven, $n ^{\text {th }}$ term of A.P. is
$
\begin{array}{ll}
& a_n=5 n-3 \\
\Rightarrow & a_1=5(1)-3=2 \\
\text { and } & a_2=5(2)-3=7
\end{array}
$
$\therefore$ Common difference,
$
\therefore d=7-2=5
$
$\therefore$ Sum of n terms of A.P.,
$
\begin{aligned}
S_{n} & =\frac{n}{2}[2 a+(n-1) d] \\
\therefore \quad S_{16} & =\frac{16}{2}[2(2)+(16-105] \\
& =8(4+75)=8 \times 79=632
\end{aligned}
$
View full question & answer→Question 183 Marks
If the sum of the first $14$ terms of an $A.P.$ is $1050$ and its first term is $10 ,$ then find the $21^{\text {st }}$ term of the $A.P.$
Answer$S _{14}=1050, a =10$
$\Rightarrow \frac{14}{2}[2 a+13 d]=1050$
$\Rightarrow 2 \times 10+13 d=150$
$\Rightarrow 13 d$
$\Rightarrow=130$
$\Rightarrow=10$
$T_{21}= a +20 d$
$=10+20 \times 10$
$T_{21}=210$
View full question & answer→