Question 12 Marks
Find the sum of the following APs:
$9, 7, 5, 3, ....$ to $14$ terms.
AnswerHere, $a = 9, d = 7 - 9 = -2$ and $n = 14$
Now, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\therefore\text{S}_\text{14}=\frac{\text{14}}{2}\big[2\times9+(\text{14}-1)(-\text{2)}\big]$
$=7\big[18-26\big]$
$=7\times(-8)$
$=-56$
View full question & answer→Question 22 Marks
If the sixth term of an $AP$ is zero then show that its $33^{rd}$ term is three times its $15^{th}$ term.
AnswerThe general term of an $AP$ is given by
$a_n=a+(n-1) d$
Given that $\mathrm{a}_8=0$
$\Rightarrow a+7 d=0$
$ \Rightarrow a=-7 d$
Now，
$ a_{38}=a+(n-1) d $
$ \Rightarrow a_{38}=-7 d+37 d $
$ \Rightarrow a_{38}=30 d \ldots(i)$
Now,
$ 3 a_{18}=3(a+17 d) $
$ \Rightarrow 3 a_{18}=3(-7 d+17 d) $
$ \Rightarrow 3 a_{18}=3(10 d)$
$ \Rightarrow 3 a_{18}=30 d \ldots .(i i)$
From $(i)$ and $(ii)$, we get
$a_{38}=3 a_{18}$
View full question & answer→Question 32 Marks
The $17^{\text {th }}$ term of $A P$ is 5 more than twich its $8^{\text {th }}$ term. If the $11^{\text {th }}$ term of the $A P$ is $43$ , find its $\mathrm{n}^{\text {th }}$ term.
AnswerThe genaral term of an $AP$ is given by $a_n= a + (n - 1)d$
Given that $a_{17}= 2a_8+ 5$
$\Rightarrow a + 16d = 2(a + 7d) + 5$
$\Rightarrow a + 16d = 2a + 14d + 5$
$\Rightarrow a - 2d = -5 ....(i)$
Next, $a_{11}= 43$
$\Rightarrow a + 10d = 43 ....(ii)$
Subtracting $(i)$ from $(ii)$, we get
$12d = 48 \Rightarrow d = 4$
Substituting in $(i)$, we get $a = 3.$
So, the $n^{th}$ term is $3 + 4(n - 1) = 4n - 1.$
View full question & answer→Question 42 Marks
Find the sum of the following $APs:$
$-37, -33, -29, ....$ to $12$ terms.
AnswerHere, $a = -37, d = -33 - (-37) = -33 + 37 = 4$ and $n = 12$
Now, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\therefore\text{S}_\text{12}=\frac{\text{12}}{2}\big[2\times(-37)+(\text{12}-1)\text{4}\big]$
$=6\big[-74+44\big]$
$=6\times(-30)$
$=-180$
View full question & answer→Question 52 Marks
The $24^{\text {th }}$ term of an $AP$ is twice its $10^{\text {th }}$ term. Show that its $72^{\text {nd }}$ term is 4 times its $15^{\text {th }}$ term.
AnswerThe genaral term of an AP is given by $a_n= a + (n - 1)d$
Given that $a_{24}= 2a_{10}$
$\Rightarrow a + 23d = 2(a + 9d)$
$\Rightarrow a + 23d = 2a + 18d$
$\Rightarrow a = 5d$
Next, $a_{72}= a + 71d$
$\Rightarrow a_{72}= 5d + 71d = 76d ....(i)$
$4a_{15}= 4(a + 14d)$
$= 4a + 56d$
$= 4(5d) + 56d$
$= 76d ...(ii)$
from (i) and (ii), we have
$a_{72}= 4a_{15}$
Hence proved.
View full question & answer→Question 62 Marks
Find:
The $20^{th}$ term of the $AP\ 9, 13, 17, 21, ....$
AnswerThe given $AP$ is $9, 13, 17, 21, ....$
$a = 9$ and $d = 13 - 9 = 4$
$a = a + (n - 1)d$
$\Rightarrow a_{20}= 9 + 19(4)$
$\Rightarrow a_{20}= 85$
So, the $20^{th}$ term is $85$.
View full question & answer→Question 72 Marks
Show that the progressions given below is an $AP$. Find the first term, common difference and next term.
$\sqrt{20},\sqrt{45},\sqrt{80},\sqrt{125},\ ....$
AnswerA sequence in which each term differs from its preceding term by a constant is called an $AP$.
$\sqrt{45}-\sqrt{20}=3\sqrt{5}-2\sqrt{5}=\sqrt{5}$
$\sqrt{80}-\sqrt{45}=4\sqrt{5}-3\sqrt{5}=\sqrt{5}$
$\sqrt{125}-\sqrt{80}=5\sqrt{5}-4\sqrt{5}=\sqrt{5}$
clearly, the progression an $AP$.
The first term $=\sqrt{20}=2\sqrt{5}$
common difference $=\sqrt{5}$
The next term $\sqrt{125}+\sqrt{5}=5\sqrt{5}+\sqrt{5}=6\sqrt{5}=\sqrt{180}$
View full question & answer→Question 82 Marks
Find the $n^{th}$ term of the following $APs:$
$5, 11, 17, 23,....$
AnswerThe given $AP$ is $5, 11, 17, 23,....$
$a = 5$ and $d = 11 - 5 = 6$
The $n^{th}$ term is given by
$a_n= a + (n - 1)d$
$\Rightarrow a_n= 5 + (n - 1)(6)$
$\Rightarrow a_n= 5 + 6n - 6$
$\Rightarrow a_n= 6n - 1$
So, the $n^{th}$ term is $6n - 1.$
View full question & answer→Question 92 Marks
Find:
The $9^{th}$ term of the AP $\frac{3}{4},\frac{5}{4},\frac{7}{4},\frac{9}{4},\ ...$
AnswerThe given AP is $\frac{3}{4},\frac{5}{4},\frac{7}{4},\frac{9}{4},\ ...$
$\text{a}=\frac{3}{4}$ and $\text{d}=\frac{5}{4}-\frac{3}{4}=\frac{2}{4}=\frac{1}{4}$
$\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow\text{a}_{9}=\frac{3}{4}+8\Big(\frac{1}{2}\Big)$
$\Rightarrow\text{a}_{9}=\frac{3}{4}+4$
$\Rightarrow\text{a}_{9}=\frac{19}{4}$
So, the $9^{th}$ term is $\frac{19}{4}.$
View full question & answer→Question 102 Marks
Find:
The $18^{th}$ term of the $AP$ $\sqrt{2},\sqrt{18},\sqrt{50},\sqrt{98},\ ....$
AnswerThe given $AP$ is $\sqrt{2},\sqrt{18},\sqrt{50},\sqrt{98},\ ...$
$\Rightarrow$ The $AP$ can be rewritten as $\sqrt{2},3\sqrt{2},5\sqrt{2},7\sqrt{2},\ ...$
$\text{a}=\sqrt{2}$ and $\text{d}=3\sqrt{2}-\sqrt{2}=2\sqrt{2}$
$\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow\text{a}_{18}=\sqrt{2}+17(2\sqrt{2})$
$\Rightarrow\text{a}_{18}=\sqrt{2}+34\sqrt{2}$
$\Rightarrow\text{a}_{18}=\sqrt{2450}$
So, the $18^{th}$ term is $\sqrt{2450}.$
View full question & answer→Question 112 Marks
Find the sum of the following APs:
$\frac{1}{15},\frac{1}{12},\frac{1}{10},\ ....$ to $11$ terms.
Answer$\frac{1}{15},\frac{1}{12},\frac{1}{10},\ ....$to $11$ terms.
Here, $\text{a}=\frac{1}{15},\text{d}=\frac{1}{12}-\frac{1}{15}=\frac{15-12}{180}=\frac{3}{180}=\frac{1}{60}$ and $n = 11$
Now, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\therefore\text{S}_\text{11}=\frac{\text{11}}{2}\big[2\times\frac{1}{15}+(\text{11}-1)\frac{1}{60}\big]$
$=\frac{11}{2}\Big[\frac{2}{15}+\frac{1}{6}\Big]$
$=\frac{11}{2}\Big[\frac{12+15}{90}\Big]$
$=\frac{11}{2}\times\frac{27}{90}$
$=\frac{11}{2}\times\frac{3}{10}$
$=\frac{33}{20}$
View full question & answer→Question 122 Marks
Show that the progressions given below is an $AP$. Find the first term, common difference and next term.
$9, 15, 21, 27, ....$
AnswerA sequence in which each term differs from its preceding term by a constant is called an $AP$.
$15 - 9 = 6$
$12 - 15 = 6$
$27 - 21 = 6$
clearly, the progression an $AP$.
The first term $= 9$
common difference $= 27 + 6 = 33$
View full question & answer→Question 132 Marks
Show that the progressions given below is an $AP.$ Find the first term, common difference and next term.
$-1,\frac{-5}{6},\frac{-2}{3},\frac{-1}{2},....$
AnswerA sequence in which each term differs from its preceding term by a constant is called an AP.
$\frac{-5}{6}-(-1)=\frac{-5}{6}+1=\frac{1}{6}$
$\frac{-2}{3}=\Big(\frac{-5}{6}\Big)=\frac{1}{6}$
$\frac{-1}{2}-\Big(\frac{-2}{3}\Big)=\frac{1}{6}$
clearly, the progression an $AP$.
The first term $= -1$
common difference $=\frac{1}{6}$
The next term $=\frac{-1}{2}+\Big(\frac{1}{6}\Big)=\frac{-1}{3}$
View full question & answer→Question 142 Marks
The $4^{th}$ term of an $AP$ is $11$. The sum of the $5^{th}$ and $7^{th}$ term of this $AP$ is $34$. Find its common difference.
AnswerThe general term of an $AP$ is given by
$a_n=a+(n-1) d$
Given that $\mathrm{a}_4=11$
$\Rightarrow a+3 d=11 \ldots(i)$
Now,
$a_5=a_7=34$
$\Rightarrow a + 4d + a + 6d = 34$
$\Rightarrow 2a + 10d = 34 ....(ii)$
Multiply $(i)$ by $2$ and subtract from $(ii)$.
$2a + 6d = 22$ and $2a + 10d = 34$
$\Rightarrow 4d = 12$
$\Rightarrow d = 3$
So, the common dofference is $3$.
View full question & answer→Question 152 Marks
Show that the progressions given below is an $AP$. Find the first term, common difference and next term.
$\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},\ ....$
AnswerA sequence in which each term differs from its preceding term by a constant is called an $AP$.
$\sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}$
$\sqrt{18}-(\sqrt{8})=3\sqrt{2}-(2\sqrt{2})=\sqrt{2}$
$\sqrt{32}-(\sqrt{18})=4\sqrt{2}-3\sqrt{2}=\sqrt{2}$
Clearly, the progression an $AP$.
The first term $=\sqrt{2}$
Common difference $=\sqrt{2}$
The next term $\sqrt{32}+\sqrt{2}=4\sqrt{2}+\sqrt{2}=5\sqrt{2}=\sqrt{50}$
View full question & answer→Question 162 Marks
Find the $37^{th}$term of the $AP$ $6,7\frac{3}{4},9\frac{1}{2},11\frac{1}{4},\ ...$
AnswerThe given $AP$ is $6,7\frac{3}{4},9\frac{1}{2},11\frac{1}{4},\ ...$
First term $= 6$, common difference $=\Big(7\frac{3}{4}-6\Big)$
$=\Big(\frac{31}{4}-6\Big)=\frac{7}{4}$
$\therefore\text{a}=6,\text{d}=\frac{7}{4}$
The $n^{th}$ term is given by
$\text{T}_\text{n }=\text{ a} + (\text{n} - 1)\text{d}$
$\text{T}_{37} = 6 + (37 - 1)\frac{7}{4}$
$=6+63=69$
Hence, $37^{th}$ term is $69$.
View full question & answer→Question 172 Marks
Which term of the $AP$ $8, 14, 20, 26, ....$ will be $72$ more than its $41^{st}$ term?
Answer$AP:$ $ 8, 14, 20, 26, ....$
$a = 8$
$d = 14 - 8 = 6$
Let $T_n=T_{41}+72$
$a + (n - 1) = a + 40d + 72$
$\Rightarrow a + (n -1)d - a - 40d = 72$
$\Rightarrow d(n - 1 - 40) = 72$
$\Rightarrow 6 (n - 41) = 72$
$n - 41 = 12$
$n = 12 + 41$
$n = 53$
Required term $= 53$
$53^{th}$ term of AP will be $72$ more than its $41^{th}$ term.
View full question & answer→Question 182 Marks
Find the sum of the following $APs:$
$2, 7, 12, 17, ....$ to $19$ terms.
AnswerHere, $a = 2, d = 7 - 2 = 5$ and $n = 19$
Now, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\therefore\text{S}_\text{19}=\frac{\text{19}}{2}\big[2\times2+(\text{19}-1)\text{5}\big]$
$=\frac{19}{2}\big[4+90\big]$
$=\frac{19}{2}\times94$
$=893$
View full question & answer→Question 192 Marks
Find the $25^{th}$ term of the AP $5,4\frac{1}{2},4,3\frac{1}{2},3,...$
AnswerThe given $AP$ is $5,4\frac{1}{2},4,3\frac{1}{2},3,...$
First term $= 5$
common difference $=4\frac{1}{2}-5\Rightarrow\frac{9}{2}-5$
$\Rightarrow\frac{9-10}{2}=-\frac{1}{2}$
$\therefore\text{a}=5,\text{d}=-\frac{1}{2}$
Now, $\text{T}_\text{n }=\text{ a} + (\text{n} - 1)\text{d}$
$\text{T}_{25} = \text{a} + (25 - 1)\text{d}$
$=\text{a}+24\text{d}$
$=5+24\times\Big(-\frac{1}{2}\Big)=5-12=-7$
$\therefore25^\text{th}\text{ term}=-7$
View full question & answer→Question 202 Marks
Find:
The $35^{th}$ term of the $AP$ $20, 17, 14, 11, ....$
AnswerThe given $AP$ is $20, 17, 14, 11, ....$
$a = 20$ and $d = 17 - 20 = -3$
$an = a + (n - 1)d$
$ \Rightarrow a_{35}=20+34(-3) $
$ \Rightarrow a_{35}=-82 $
So, the $20^{th}$ term is $-82$.
View full question & answer→Question 212 Marks
Find the $n^{th}$ term of the following $APs$:
$16, 9, 2, -5,....$
AnswerThe given $AP$ is $16, 9, 2, -5,....$
$a = 16$ and $d = 9 - 16 = -7$
The $\mathrm{n}^{\text {th }}$ term is given by
$ a_n=a+(n-1) d $
$ \Rightarrow a_n=16+(n-1)(-7) $
$ \Rightarrow a_n=16-7 n+7 $
$ \Rightarrow a_n=9-7 n$
So, the $\mathrm{n}^{\text {th }}$ term is $9-7 n$.
View full question & answer→Question 222 Marks
Find the value of $p$ for which the numbers $2p - 1, 3p + 1, 11$ are in $AP.$
Hence, find the numbers.
AnswerThe given three numbers $(2p - 1), (3p + 1)$ and $11$ are in $AP$.
Then,
$2(3p + 1) = (2p - 1) + 11$
$6p + 2 = 2p -1 + 11$
$\Rightarrow 6p - 2p = 10 - 2$
$\Rightarrow 4p = 8$
$\Rightarrow\text{p}=\frac{8}{4}$
$\Rightarrow p = 2$
So, the value of $p$ is $2$.
Then, the given number be
$(2 × 2 - 1), (3 × 2 + 1)$ and $11$
i.e., $3, 7$ and $11$
So, we get an Arithmetic progression
$3, 7, 11.$
View full question & answer→Question 232 Marks
Which term of the $\text{AP } 20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4},\ ....$is its first negative term?
AnswerThe given $AP$ is $20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4},\ ....$
Common difference $19\frac{1}{4}-20=\frac{77}{4}-20=\frac{-3}{4}$
The general term of an $AP$ is given by
$\text{a}_\text{n} = \text{a} + (\text{n} - 1)\text{d}$
$\Rightarrow \text{a} + (\text{n} - 1)\text{d} < 0$
$\Rightarrow20+(\text{n}-1)\Big(\frac{-3}{4}\Big)<0$
$\Rightarrow20-\frac{3\text{n}}{4}+\frac{3}{4}<0$
$\Rightarrow-\frac{3\text{n}}{4}+\frac{83}{4}<0$
$\Rightarrow-3\text{n}+83<0$
$\Rightarrow-3\text{n}<-83$
$\Rightarrow\text{n}>\frac{83}{3}=27.67$
So,$n = 28$
Hence, the first negative term would be the $28^{th}$ term.
View full question & answer→Question 242 Marks
How many term are there in the $AP$ $18,15\frac{1}{2},13,...,-47?$
AnswerThe $AP$ is $18,15\frac{1}{2},13,...,-47$
Here,
$\text{a} = 18$
$\text{d}=15\frac{1}{2}-18=\frac{31}{2}-18=\frac{-5}{2}$
$\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow-47 = 18 + (\text{n}-1 )\Big(\frac{-5}{2}\Big)$
$\Rightarrow-47 = 18 -\frac{5\text{n}}{2}+\frac{5}{2}$
$\Rightarrow\frac{5\text{n}}{2}=47+18+\frac{5}{2}$
$\Rightarrow\frac{5\text{n}}{2}=\frac{94+36+5}{2}$
$\Rightarrow5\text{n}=135$
$\Rightarrow\text{n}=27$
So, the number of terms is $27$.
View full question & answer→Question 252 Marks
Show that the progressions given below is an $AP$. Find the first term, common difference and next term.
$11, 6, 1, -4, ....$
AnswerA sequence in which each term differs from its preceding term by a constant is called an $AP$.
$6 - 11 = -5$
$1 - 6 = -5$
$-4 - 1 = -5$
clearly, the progression an $AP$.
The first term $= 11$
common difference $= -5$
The next term $= -4 + (-5) = -9$
View full question & answer→Question 262 Marks
Find the $8^{th}$ term from the end of the $AP\ 7, 10, 13, ...., 184.$
AnswerHere $a = 7, d = (10 - 7) = 3, l = 184$
And $n = 8$
Now, $n^{th}$ term from the end $= [l - (n - 1)d]$
$= [184 - (8 - 1)3]$
$= [184 - 7 × 3]$
$= [184 - 21] = 163$
Hence. the $8^{th}$ term from the end is $163$.
View full question & answer→