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 $a_8= 0$
$\Rightarrow a + 7d = 0$
$\Rightarrow a = -7d$
Now,
$a_{38} = a + (n - 1)d$
$\Rightarrow a_{38} = -7d + 37d$
$\Rightarrow a_{38} = 30d ...(i)$
Now,
$3a_{18} = 3(a + 17d)$
$\Rightarrow 3a_{18} = 3(-7d + 17d)$
$\Rightarrow 3a_{18} = 3(10d)$
$\Rightarrow 3a_{18} = 30d ....(ii)$
From (i) and (ii), we get
$a_{38} = 3a_{18}$
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 $n ^{\text {th }}$ term.
AnswerThe genaral term of an AP is given by $a_n=a+(n-1) d$
Given that $a _{17}=2 a _8+5$
$\Rightarrow a+16 d=2(a+7 d)+5 \\
\Rightarrow a+16 d=2 a+14 d+5 \\
\Rightarrow a-2 d=-5 \ldots . .(i)$
Next, $a_{11}=43$
$\Rightarrow a+10 d=43 \ldots \text { (ii) }$
Subtracting (i) from (ii), we get
$12 d=48 \Rightarrow d=4$
Substituting in (i), we get $a=3$.
So, the $n^{\text {th }}$ term is $3+4(n-1)=4 n-1$.
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 $a _{24}=2 a _{10}$
$\Rightarrow a+23 d=2(a+9 d)$
$\Rightarrow a+23 d=2 a+18 d$
$\Rightarrow a=5 d$
Next, $a_{72}=a+71 d$
$\Rightarrow a_{72}=5 d+71 d=76 d \ldots . . .(i)$
$4 a_{15}=4(a+14 d)$
$=4 a+56 d$
$=4(5 d)+56 d$
$=76 d \ldots \text { (ii) }$
from (i) and (ii), we have
$a_{72}=4 a_{15}$
Hence proved.
View full question & answer→Question 62 Marks
Find:
The $20^{\text {th }}$ term of the AP $9,13,17,21, \ldots$.
AnswerThe given AP is $9,13,17,21, \ldots$.
$a =9$ and $d =13-9=4$
$a=a+(n-1) d$
$\Rightarrow a_{20}=9+19(4)$
$\Rightarrow a _{20}=85$
So, the $20^{\text {th }}$ 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 $5,11,17,23, \ldots . . a=5$ and $d=11-5=6$ The $n^{\text {th }}$ term is given by $a_n=a+(n-1) d$
$\Rightarrow a_n=5+(n-1)(6)$
$\Rightarrow a_n=5+6 n-6$
$\Rightarrow a_n=6 n-1$
So, the $n ^{\text {th }}$ term is $6 n -1$.
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 $\sqrt{2},\sqrt{18},\sqrt{50},\sqrt{98},\ ...$
⇒ 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 $4^{\text {th }}$ term of an $A P$ is 11 . The sum of the $5^{\text {th }}$ and $7^{\text {th }}$ term of this $A P$ is 34 . Find its common difference.
AnswerThe general term of an AP is given by
$a_n=a+(n-1) d$
Given that $a_4=11$
$\Rightarrow a+3 d=11 \ldots \text { (i) }$
Now,
$a_5=a_7=34$
$\Rightarrow a+4 d+a+6 d=34$
$\Rightarrow 2 a+10 d=34 \ldots . \text { (ii) }$
Multiply (i) by $2$ and subtract from (ii).
$2 a+6 d=22 \text { and } 2 a+10 d=34$
$\Rightarrow 4 d=12$
$\Rightarrow d=3$
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 nth 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?
AnswerAP: $8, 14, 20, 26, ....$
$a = 8$
$d = 14 - 8 = 6$
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 $n^{th}$ term is given by
$a_n = a + (n - 1)d$
$\Rightarrow a_n = 16 + (n - 1)(-7)$
$\Rightarrow a_n = 16 - 7n + 7$
$\Rightarrow a_n = 9 - 7n$
So, the $n^{th}$ term is $9 - 7n.$
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
⇒ 6p - 2p = 10 - 2
⇒ 4p = 8
$\Rightarrow\text{p}=\frac{8}{4}$
⇒ 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 $= [l - (n - 1)d]$
$= [184 - (8 - 1)3]$
$= [184 - 7 \times 3]$
$= [184 - 21] = 163$
Hence. the $8^{th}$ term from the end is $163$.
View full question & answer→