Question 13 Marks
The $7^{th}$ term of an $AP$ is $-4$ and its $13^{th}$ term is $-16$. Find the $AP.$
AnswerIn the given $AP,$ let the first term $= a$ common difference $= d$
Then, $T_n= a + (n - 1)d$
$⇒ T_7= a + (7 - 1)d$, and $T_{13}= a + (13 - 1)d$
$⇒ T_7= -4 ⇒ a + 6d = -4 .....(1)$
$⇒ T_{3}= -16 ⇒ a + 12d = -16 .....(2)$
Subtracting $(1)$ from $(2),$ we get
$6d = -12 ⇒ d = -2$
Putting $d = -2$ in $(1),$ we get
$a + 6 × (-2) = -4 ⇒ a - 12 = -4 ⇒ a = 8$
Thus, $a = 8, d = -2$
So the required $AP$ is $8, 6, 4, 2, 0 .....$
View full question & answer→Question 23 Marks
Find the sum of first $n$ even natural numbers.
AnswerThe first n even natural numbers are $2, 4, 6, 8, 10, ..., n.$
Here, $a = 2$ and $d = (4 - 2) = 2$
Sum of $n$ terms of an $AP$ is given by
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$=\big(\frac{\text{n}}{2}\big)\times\big[2\times2+(\text{n}-1)\times2\big]$
$=\big(\frac{\text{n}}{2}\big)\times\big[4+2\text{n}-2\big]=\big(\frac{\text{n}}{2}\big)\times(2\text{n}+2)=\text{n}(\text{n}+1)$
Hence, the required is $n(n + 1).$
View full question & answer→Question 33 Marks
How many term are there in the $AP 6, 10, 14, 18,....174?$
AnswerIn the given $AP,$ we have $a = 6$ and $d = (10 - 6) = 4$
Suppose there are n term in the given $AP,$ then
$T_n= 174 ⇒ a + (n - 1)d = 174$
$⇒ 6 + (n - 1)4 = 174$
$⇒ 6 + 4n - 4 = 174$
$⇒ 2 + 4n = 174 ⇒ \text{n}=\frac{172}{4}=43$
Hence there are $43$ terms in the given $AP$
View full question & answer→Question 43 Marks
Find the common difference of an $AP$ whose first term is $5$ and the sum of its first four terms is half the sum of the next four termes.
AnswerThe genaral term of an $AP$ is given by $a_n= a + (n - 1)d$
and $S_n= \frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big].$
Given that $a = 5$
Also, $\text{S}_4=\frac{1}{2}(\text{S}_8-\text{S}_4)$
$\Rightarrow\frac{4}{2}\big[2(5)+3\text{d}\big]$
$=\frac{1}{2}\Big[\frac{8}{2}[2(5)+7\text{d}]-\frac{4}{2}[2(5)+3\text{d}]\Big]$
$\Rightarrow4\big[10+3\text{d}\big]=\big[4(10+7\text{d})-2(10+3\text{d})\big]$
$\Rightarrow40+12\text{d}=40+28\text{d}-20-6\text{d}$
$\Rightarrow-10\text{d}=-20$
$\Rightarrow\text{d}=2$
Thus, the common difference is $2.$
View full question & answer→Question 53 Marks
How many term are there in the $AP 41, 38, 35,....8?$
AnswerIn the given $AP,$ we have $a = 41$ and $d = 38 - 41 = -3$
Suppose there are n term in the given $AP,$ then
$T_n= 8 ⇒ a + (n - 1)d = 8$
$⇒ 41 + (n - 1)(-3) = 8$
$⇒ 41 - 3n + 3 = 8$
$⇒ -3n = -36 ⇒ n = 12$
Hence there are $12$ terms in the given $AP$
View full question & answer→Question 63 Marks
If the $\mathrm{n}^{\text {th }}$ term of a progression is $(4 n-10)$ show that it is an $AP.$ Find its:
- First term.
- Common difference.
- $16^{\text {th }}$ term.
Answer$T_n=(4 n-10) $
$ \Rightarrow T_1=(4 \times 1-10)=-6 $
$ \text { and } $
$ T_2=(4 \times 2-10)=-2$
Thus, we have
- First term $= -6$
- Common difference $=\left(T_2-T_1\right)=(-2+6)=4$
- $16^{\text {th }}$ term $=a+(16-1) d$, where $a=-6$ and $d=4$
$= (-6 + 15 × 4) = 54$ View full question & answer→Question 73 Marks
Find the sum of the following APs:
$0.6, 1.7, 2.8, ..... $ to $100$ terms.
AnswerHere, $a = 0.6, d = 1.7 - 0.6 = 1.1$ and $n = 100$
Now, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\therefore\text{S}_\text{100}=\frac{\text{100}}{2}\big[2\times0.6+(100-1)(1.1)\big]$
$=50\big[1.2+108.9\big]$
$=50\times110.1$
$=5505$
View full question & answer→Question 83 Marks
Which term of the AP $121, 117, 113, .... $ is its first negative term?
AnswerThe given AP is $121, 117, 113, ...,$
Common difference $= 117 - 121 = -4$
The general term of an AP is given by
$a_n= a + (n - 1)d$
$⇒ a + (n - 1)d < 0$
$⇒ 121 + (n - 1)(-4) <0$
$⇒ 121 - 4n + 4 <0$
$⇒ -4n + 125 <0$
$⇒ -4n < - 125$
$\Rightarrow\text{n}>\frac{125}{4}=31.25$
So, $n > 31.25$
Hence, the first negative term would be the $32^{nd}$ term.
View full question & answer→Question 93 Marks
If the $10^{th}$ term of an $AP$ is $52$ and $17^{th}$ term is $20$ more than its $13^{th}$ term, find the $AP.$
AnswerIn the given $AP$ let first term $= a$
And common difference $= d$
Then, $T_n=a+(n-1) d$
$T_{10}=a+(10-1) d, T_{17}=a+(17-1) d$ and $T_{13}=a+(13-1) d$
$T_{10}=a+9 d, T_{17}=a+16 d$ and $T_{13}=a+12 d$
Now, $T_{10}=52 \Rightarrow a+9 d=57$.....(1)
and $T_{17}=T_{13}+20 \Rightarrow a+16 d=a+12 d+20$
$⇒ 4d = 20 ⇒ d = 5$
Putting $d = 5$ in $(1),$ we get
$a + 9 × 5 = 52 ⇒ a = 52 - 45 ⇒ a = 7$
Thus, $a = 7$ and $d = 5$
So the required $AP$ is $7, 12, 17, 22....$
View full question & answer→Question 103 Marks
Find the sum of all odd numbers between $0$ and $50.$
AnswerOdd numbers between $0$ to $50$ are
$1, 3, 5, 7, ....., 49$
Here
First term $= a = 1$
last term $= l = 49$
There are $25$ such terms
So, $n = 25$
We need to find sum
So, we can use formula
$\text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}+\text{l})$
Putting value in the formula
$\text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}+\text{l})$
$=\frac{25}{2}(1+49)$
$=\frac{25}{2}\times50$
$=625$
Therefore, the sum of odd number between $0$ & $50$ is $625$.
View full question & answer→Question 113 Marks
For what value of $n$, the $n^{th}$ terms of the arithmetic progressions $63, 65, 67, ....$ and $3, 10, 17, ....$are equal$?$
AnswerFirst $AP$ is $63, 65, 67...$
First term $= 63,$ common difference $= 65 - 63 = 2$
$\therefore$ $n^{th}$ term $= 63 + (n - 1)2 = 63 + 2n - 2 = 2n + 61$
Second $AP$ is $3, 10, 17...$
First term $= 3,$ common difference $= 10 - 3 = 7$
$n^{th}$ term $= 3 + (n - 1)7 = 3 + 7n - 7 = 7n - 4$
The two $n^{th}$ term are equal
$\therefore 2n + 61 = 7n - 4$ or $5n = 61 + 4 = 65$
$\Rightarrow\text{n}=\frac{65}{4}=13$
View full question & answer→Question 123 Marks
Which term of the $AP \frac{5}{6},1,1\frac{1}{6},1\frac{1}{3},\ ...$ is $3?$
Answer$\text{a}=\frac{5}{6};\text{d}=\Big(1-\frac{5}{6}\Big)=\frac{1}{6}$
In the given $AP,$ we have
Suppose there are $n$ terms in given $AP,$ we have
Then,
$\text{T}_\text{n}=3$
$\Rightarrow\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}=3$
$\Rightarrow\frac{5}{6}+(\text{n}-1)\frac{1}{6}=3$
$\Rightarrow\frac{5}{6}+\frac{1}{6}\text{n}-\frac{1}{6}=3$
$\Rightarrow4+\text{n}=18\Rightarrow\text{n}=14$
$\therefore\text{n}=14$
Thus, $14^{th}$ term in the given $AP$ is $3$
View full question & answer→Question 133 Marks
Determine the $n^{th}$ term of the $AP$ whose $7^{th}$ term is $-1$ and $16^{th}$ term is $17.$
AnswerThe general term of an $AP$ is given by
$a_n= a + (n - 1)d$
Given that $a_7= -1$
$⇒ a + 6d = -1 ...(i)$
Now,
$a_{16}= 17$
$⇒ a + 15d = 17...(ii)$
Subtract from $(i)$ from $(ii).$
$9d = 18$
$⇒ d = 2$
Substituting in $(i),$ we get $a = -13.$
$\text { Thus, the } \mathrm{n}^{\text {th }} \text { term will be, }$
$a_n=a+(n-1) d$
$\Rightarrow a_n=-13+(n-1)(2)$
$\Rightarrow a_n=-13+2 n-2$
$\Rightarrow a_n=2 n-15 .$
View full question & answer→Question 143 Marks
The sum of the $2^{nd}$ and the $7^{th}$ terms of an $AP$ is $30.$ If its $15^{th}$ term is $1$ less than twice its $8^{th}$ term, find the $AP.$
AnswerThe genaral term of an $AP$ is given by $a_n= a + (n - 1)d$
Given that $a_2+ a_7= 30$
$⇒ a + d +a + 6d = 30$
$⇒ 2a + 7d = 30 .....(i)$
Next, $a_{15}= 2a_8- 1$
$⇒ a + 14d = 2(a + 7d) - 1$
$⇒ a + 14d = 2a + 14d - 1$
$⇒ a = 1$
Substituting in $(i),$ we get $d = 4.$
So, the AP is $a, a + d, a + 2d, a + 3d.$
that is, $1, 5, 9, 13,.....$
View full question & answer→Question 153 Marks
Find the sum of the following arithmetic series:
$(-5) + (-8) + (-11) + .... + (-230)$
Answerwe have
$a = -5,$
$d = -8 - (-5) = -8 + 5 = -3$
Let the total number of terms be n.
Then $ T_n= -230$
$⇒ \text{a} + (\text{n} - 1)\text{d} = -230$
$\Rightarrow-5+(\text{n}-1)(-3)=-230$
$\Rightarrow(\text{n}-1)(-3)=-225$
$\Rightarrow\text{n}-1=\frac{-225}{-3}$
$\Rightarrow\text{n}-1=75$
$\Rightarrow\text{n}=76$
Now, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[\text{a}+\text{l}\big]$
$\therefore\text{S}_\text{76}=\frac{\text{76}}{2}\big[-\text{5}+(-230)\big]$
$=38\times(-235)=-8930$
View full question & answer→Question 163 Marks
The $16^{th}$ term of an $AP$ is $5$ times its $3^{rd}$ term. If its $10^{th}$ term is $41,$ find the sum of its first $15$ terms.
AnswerThe general term of an $AP$ is given by
$a_n=a+(n-1) d$
Given that $a_{16}=5 a_3$
$⇒ a + 15d = 5(a + 2d)$
$⇒ a + 15d = 5a + 10d$
$⇒ 4a = 5d ....(i)$
Now,
$a_{10}= 41$
$⇒ a + 9d = 41$
$⇒ 4a + 36d = 164$
$⇒ 5d + 36d = 164 ....($from $(i))$
$⇒ 41d = 164$
$⇒ d = 4$
substituting in $(i)$, we gwt $a = 5.$
We know that, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
Sum of the first 15 term
$=\text{S}_{15}$
$=\frac{15}{2}\big[2(5)+14(4)\big]$
$=5[10+56]$
$=495$
View full question & answer→Question 173 Marks
The $4^{th}$ term of an AP is zero. Prove that its $25^{\text {th }}$ term is triple its $11^{\text {th }}$ term.
AnswerIn the first term of given $A P=a$ and common difference $=d$
Then, $T_n=a+(n-1) d$
$ \Rightarrow T_4=a+(4-1) d, T_{25}=a+(25-1) d, \text { and } T_{11}=a+(11-1) d $
$ \Rightarrow T_4=a+3 d, T_{25}=a+24 d, \text { and } T_{11}=a+10 d$
Now, $T_4=0 \Rightarrow a+3 d=0 \Rightarrow a=-3 d$
$\therefore T_{25}=a+24 d=(-3 d+24 d)=21 d $
$ \text { And } T_{11}=a+10 d=-3 d+10 d=7 d $
$ \therefore T_{25}=21 d=3 \times(7 d)=3 \times T_{11}$
Then, $T_n=a+(n-1) d$
$ \Rightarrow T_4=a+(4-1) d, T_{25}=a+(25-1) d, \text { and } T_{11}=a+(11-1) d $
$ \Rightarrow T_4=a+3 d, T_{25}=a+24 d, \text { and } T_{11}=a+10 d$
Now, $T_4=0 \Rightarrow a+3 d=0 \Rightarrow a=-3 d$
$ \therefore T_{25}=a+24 d=(-3 d+24 d)=21 d $
$ \text { And } T_{11}=a+10 d=-3 d+10 d=7 d $
$ \therefore T_{25}=21 d=3 \times(7 d)=3 \times T_{11}$
Hence $25^{\text {th }}$ term is triple its $11^{\text {th }}$ term
View full question & answer→Question 183 Marks
The $13^{th}$ term of an $AP$ is $4$ times its $3^{rd}$ term. If its $5^{th}$ term is $16,$ find the sum of its first $10$ terms.
AnswerThe general term of an $AP$ is given by
$a_n=a+(n-1) d$
Given that $a_{13}=4 a_3$
$⇒ a + 12d = 4(a + 2d)$
$⇒ a + 12d = 4a + 8d$
$⇒ 3a = 4d ....(i)$
Now,
$a_5=16$
$⇒ a + 4d = 16$
$⇒ 3a + 12d = 48$
$⇒ 4d + 12d = 48 ....($from $(i))$
$⇒ 16d = 48$
$⇒ d = 3$
substituting in $(i),$ we gwt $a = 4.$
We know that, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
Sum of the first 10 term
$=\text{S}_{10}$
$=\frac{10}{2}\big[2(4)+9(3)\big]$
$=5[8+27]$
$=175$
View full question & answer→Question 193 Marks
Find the sum of first $n$ terms of an $AP$ whose nth term is $(5 - 6n).$ Hence, find the sum of its first $20$ terms.
AnswerThe $n^{\text {th }}$ term of the given $A.P.$ is given by, $T_n=5-6 n$
Let a be the first term and d be the common difference of this $A.P..$
Then,
$ a=T_1=5-6(1)=5-6=-1 $
$T_2=5-6(2)=5-12=-7$
$ \Rightarrow d=T_2-T_1=-7=-7-(-1)+1=-6$
$\therefore$ Sum of first $n$ terms $=\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$=\frac{\text{n}}{2}\big[2(-1)+(\text{n}-1)(-6)\big]$
$=\frac{\text{n}}{2}\big[-2-6\text{n}+6)\big]$
$=\frac{\text{n}}{2}\big[4-6\text{n}\big]$
$=\text{n}\big[2-3\text{n}\big]$
$ \therefore \text { Sum of first } 20 \text { terms }=S_{20}=20[2-3 \times 20]=20 \times(-58)=-1160$
View full question & answer→Question 203 Marks
The $19^{th}$term of an $AP$ is equal to $3$ times its $6^{th}$ term. If its $9^{th}$ term is $19,$ find the $AP.$
AnswerThe genaral term of an $AP$ is given by $a_n=a+(n-1) d$
Given that $a_{19}=3 a_6$
$ \Rightarrow a+18 d=3(a+5 d) $
$ \Rightarrow a+18 d=3 a+15 d $
$ \Rightarrow 2 a=3 d \ldots . .(i)$
Next, $a_9=19$
$ \Rightarrow a+8 d=19 $
$ \Rightarrow 2 a+16 d=38 $
$ \Rightarrow 3 d+16 d=38 $
$ \Rightarrow 19 d=38 $
$ \Rightarrow d=2$
Substituting in $(i)$, we get $a=3$
Thus, the $AP$ is $3,5,7,9, \ldots$.
View full question & answer→Question 213 Marks
Which term of the $AP\ 72, 68, 64, 60, .....$ is $0?$
AnswerIn the given $AP,$ we have $a =72$ and $d = 68 - 72 = -4$
Suppose there are $n$ terms in given $AP,$ we have
$T_n= 0 ⇒ a + (n - 1)d = 0 $
$⇒ 72 + (n - 1)(-4) = 0$
$⇒ 72 - 4n + 4 = 0$
$⇒ 4n = 76 ⇒ n = 19$
Hence, the $19^{th}$ term in the given $AP$ is $0.$
View full question & answer→Question 223 Marks
How many terms of the$\text{AP}\ 20, 19\frac{1}{3},18\frac{2}{3},\ ...$must be taken so that their sum is $300?$ Explain the double answer.
AnswerHere$\text{a}=20,\ \text{d}=19\frac{1}{3}-20=\frac{58}{3}-20$
$=\frac{58-60}{3}=\frac{-2}{3}$
Let the required number of terms be $ n.$
Then, $\text{S}_\text{n}=300$
$\Rightarrow\frac{\text{}\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]=300$
$\Rightarrow\frac{\text{n}}{2}\Big[2(20)+(\text{n}-1)\Big(\frac{-2}{3}\Big)\Big]=300$
$\Rightarrow\text{n}\big[40-\frac{2}{3}\text{n}+\frac{2}{3}\big]=600$
$\Rightarrow\text{n}\big[\frac{122}{3}-\frac{2}{3}\text{n}\big]=600$
$\Rightarrow\frac{122}{3}\text{n}-\frac{2}{3}\text{n}^2=600$
$\Rightarrow122\text{n}-2\text{n}^2=1800$
$\Rightarrow61\text{n}-\text{n}^2=900$
$\Rightarrow\text{n}^2-61\text{n}+900=0$
$\Rightarrow\text{n}^2-25\text{n}-36\text{n}+900=0$
$\Rightarrow\text{n}(\text{n}-25)-36(\text{n}-25)=0$
$\Rightarrow(\text{n}-25)(\text{n}-36)=0$
$\Rightarrow\text{n}-25=0$ or $\text{n}-36=0$
$\Rightarrow\text{n}=25$ or $\text{n}=36$
$\therefore$ Sum of first $25$ terms $=$ Sum of first $36$ terms $= 300$
This means that the sum of all terms from $26^{nd}$ to $36^{th}$ term is zero.
View full question & answer→Question 233 Marks
Find the $6^{th}$ term from the end of the AP $17, 14, 11, ...., (-40).$
AnswerHere $a = 17, d = (14 - 17) = -3, l = -40$
And $n = 6$
Now, $n^{th}$ term from the end $= [l - (n - 1)d]$
$= [-40 - (6 - 1)(-3)]$
$= [-40 + 5 × 3]$
$= [-40 + 15] = -25$
Hence. the $6^{th}$ term from the end is $-25.$
View full question & answer→Question 243 Marks
Find the middel term of the AP $10, 7, 4, ....., (-62).$
AnswerThe given $AP$ is $10, 7, 4, .... ,(-62)$
$a = 10$ and $d = 7 - 10 = -3$
$a_n= a + (n - 1)d$
$⇒ -62 = 10 + (n - 1)(-3)$
$⇒ -72 = (n - 1)(-3)$
$⇒ -72 = -3n + 3$
$⇒ 3n = 75$
$⇒ n = 25$
So, there are $25$ term in the $AP.$
The middel term $=\Big(\frac{\text{n}+1}{2}\Big)=\Big(\frac{26}{2}\Big)=13$
So, the $13^{th}$ term is given by
$a_{13}= a + 12d = 10 + 12(-3) = -26.$
View full question & answer→Question 253 Marks
How many two-digit numbers are duvisible by $3?$
AnswerThe two-digit numbers divisible by $3$ start from
$12, 15, 18, 21, ..., 99$
Here,
$a = 12$
$d = 3$
$a_n= a + (n - 1)d$
$⇒ 99 = 12 + (n - 1)(3)$
$⇒ 99 = 12 + 3n - 3$
$⇒ 90 = 3n$
$⇒ n = 30$
This, $30$ two-digit number are divisible by $3.$
View full question & answer→Question 263 Marks
Find:
The $15^{th}$ term of the AP $-40, -15, 10, 35, ....$
AnswerThe given $AP$ is $-40, -15, 10, 35, ....$
$a = -40$ and $d = -15 - (-40)$
$= -15 + 40 = 25$
$a_n= a + (n - 1)d$
$⇒ a_{15}= -40 + 14(25)$
$⇒ a_{15}= 310$
So, the $15^{th}$ term is $310.$
View full question & answer→Question 273 Marks
Find the sum of the following arithmetic series:
$5 + (-41) + 9 + (-39) + 13 + (-37) + 17 + ..... + (-5) + 81 + (-3).$
Answerlet there be two series in the given series
first $5 + 9 + 13 .... 81$
and the second be $-41 + (-39) + (-37) .... -3$
number of terms in first series
$a_n= a + (n - 1)d$
$81 = 5 + (n - 1)d$
$76 = (n - 1)4$
$19 = n - 1$
$n = 20$
similarly in second series there are $20$ terms
now, sum of the two series
sum of given series $=$ sum of first series $+$ sum of second series
$=\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}+\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}$
$=\frac{\text{n}}{2}\big[\{2\times5+19\times4\}+\{\text{2}\times-41+19\times2)\}\big]$
$=10\big[\{10+76\}+\{-82+32\}\big]$
$=10\times\big[86-44\big]$
$=10\times42$
$=420$
View full question & answer→Question 283 Marks
Which term of the $AP$ $5, 15, 25, ....$ will be $130$ more than its $31^{st}$ term$?$
AnswerThe given $AP$ is $5, 15, 25 ....$
$\therefore a = 5, d = 15 - 5 = 10$
We have, $T_n= 130 + T_{31}$
$⇒ a + (n - 1)d = 130 + 5 + (31 - 1) × 10$
$⇒ 5 + (n - 1) × 10 = 130 + 5 + (31 - 1) × 10$
$⇒ 5 + 10n - 10 = 135 + 300$
$⇒ 10n - 5 = 435$ or $10n = 453 + 5$
$\therefore\text{n}=\frac{440}{10}=44$
Thus, the required term is $44^{th}$
View full question & answer→Question 293 Marks
If the $p^{th}$ term of an $AP$ is $q$ and its $q^{th}$ term is $p$ then show that its $(p + q)^{th}$ term is zero.
AnswerLet $a$ be the term and d be the common difference
$p^{th}$ term $= a + (p - 1)d = q($given$) .....(1)$
$q^{th}$ term $= a + (q - 1)d = p($given$) .....(2)$
subtracting $(2)$ from $(1)$
$(p - q)d = q - p$
$(p - q)d = -(p - q)$
$\therefore d = -1$
Putting $d = -1$ in $(1)$
$a - (p - 1) = q$
$\therefore a = p + q - 1$
$\therefore (p + q)^{th}$ term $= a + (p + q - 1)d$
$= (p + q - 1) = (p + q - 1) = 0$
View full question & answer→Question 303 Marks
A sum of $₹\ 2800$ is to be used to award four prizes. If each prize after the first is $₹\ 200$ less than the preceding prize, find the value of each of the prizes.
AnswerLet the first price be $₹\ a.$
Since each prize after the first is $₹\ 200$
Less than the preceding prize,
So, the prizes from the first to the last are
$₹\ a, ₹(a - 200), ₹(a - 400), ₹(a - 600)$
The common difference $= (a - 200) - a = -200$
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow2800=\frac{4}{2}\big[2\text{a}+3(-200)\big]$
$\Rightarrow1400=2\text{a}-600$
$\Rightarrow2\text{a}=2000$
$\Rightarrow\text{a}=1000$
So, the prizes are $₹\ 1000, ₹\ 800, ₹\ 600$ and $₹\ 400.$
View full question & answer→Question 313 Marks
Which term of the AP $21, 18, 15, ....$ is $-81?$
AnswerThe given AP is $21, 18, 15, ...$
$ a=21 \text { and } d=18-21=-3$
$a_n=a+(n-1) d$
$\Rightarrow-18=21+(n-1)(-3)$
$\Rightarrow-18=21+(n-1)(-3)$
$\Rightarrow-18=21-3 n+3$
$\Rightarrow 3 n=105$
$\Rightarrow n=35$
$\text { So, }-18 \text { is the } 35^{\text {th }} \text { term. }$
View full question & answer→Question 323 Marks
How many terms of the $AP\ 63, 60, 57, 54, .....$ must be taken so that their sum is $693?$ Explain the double answer.
AnswerHere $a = 63, d = (60 - 63) = -3$
Let the required number of terms be $n.$
Then, $\text{S}_\text{n}=693$
$\Rightarrow\frac{\text{}\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]=693$
$\Rightarrow\frac{\text{n}}{2}\big[2(63)+(\text{n}-1)(-3)\big]=693$
$\Rightarrow\text{n}\big[126-3\text{n}+3\big]=1386$
$\Rightarrow\text{n}\big[129-3\text{n}\big]=1386$
$\Rightarrow129\text{n}-3\text{n}^2=1386$
$\Rightarrow3\text{n}^2-129\text{n}+1386=0$
$\Rightarrow\text{n}^2-43\text{n}+462=0$
$\Rightarrow\text{n}^2-21\text{n}-22\text{n}+462=0$
$\Rightarrow\text{n}(\text{n}-21)-22(\text{n}-21)=0$
$\Rightarrow(\text{n}-21)(\text{n}-22)=0$
$\Rightarrow\text{n}=21=0$ or $\text{n}-22=0$
$\Rightarrow\text{n}=21$ or $\text{n}=22$
$\therefore$ Sum of first $21$ terms $=$ Sum of first $22$ terms $= 693$
This means that the $22^{nd}$ term is zero.
View full question & answer→Question 333 Marks
The sum of the first n terms of an $AP$ is given by $Sn = (3n^2- n).$ Find its:
- $n^{th}$ term
- First term
- Common difference.
Answer$\quad \mathrm{S}_{\mathrm{n}}=\left(3 \mathrm{n}^2-\mathrm{n}\right)$ (given)
$\therefore S_{n-1}=\left[3(n-1)^2-(n-1)\right] $
$ =\left[3 n^2+3-6 n-n+1\right] $
$ =\left[3 n^2-7 n+4\right]$
- i. The $\mathrm{n}^{\text {th }}$ term is given by
$ T_n=\left(S_n-S_{n-1}\right) $
$ =\left[\left(3 n^2-n\right)-\left(3 n^2-7 n+4\right)\right] $
$ =6 n-4 $
$ \therefore n^{\text {th }} \text { term }=6 n-4$
- ii. Putting $\mathrm{n}=1$ in $(1),$ we get
$T_1=(6 \times 1-4)=2$
$\therefore$ First term $=2$
- iii. Putting $n=1$ in $(1),$ we get $T_2=(6 \times 2-4)=8$
$\therefore d=\left(T_1-T_2\right)=8-2=6$ View full question & answer→Question 343 Marks
Is $184$ a term of the $AP\ 3, 7, 11, 15, ....?$
AnswerThe given $AP$ is $3, 7, 11, 15$
Common difference $= 7 - 3 = 4$
So, every term of the $AP$ will be an odd number,
Since even $+$ odd $=$ odd
$184$ is an even number and hence cannot be a term of the given $AP.$
View full question & answer→Question 353 Marks
The sum of the first $n$ terms of an $AP$ is $\Big(\frac{5\text{n}^2}{2}+\frac{3\text{n}}{2}\Big).$ Find the $n^{th}$ term and the $20^{th}$ term of this $AP.$
Answer$\text{S}_\text{n}=\Big(\frac{5\text{n}^2}{2}+\frac{3\text{n}}{2}\Big)\dots(1)$
It is given that
Now, $20^{th}$term
$= ($sum of first $20$ term$) - ($sum of first $19$ terms$)$
Putting $- 20$ in $(1)$ we get
$\text{S}_\text{20}=\Big[\frac{5\times(20)^2}{2}+\frac{3\times20}{2}\Big]=[1000+30]$
$=1030$
Putting $n = 19$ in $(1),$ we get
$\text{S}_\text{19}=\Big[\frac{5\times(19)^2}{2}+\frac{3\times19}{2}\Big]=\Big[\frac{5\times19\times19}{2}+\frac{57}{2}\Big]$
$\Big[\frac{1805}{2}+\frac{57}{2}\Big]=\Big[\frac{1862}{2}\Big]=931$
$\therefore T_{20}=\left(\mathrm{S}_{20}-\mathrm{S}_{19}\right)=(1030-931)=99$
$\text { Hence, the } 20^{\text {th }} \text { term is } 99$
View full question & answer→Question 363 Marks
The sum of the first $7$ term of an $AP$ is $49$ and the sum of its first $17$ terms is $289.$ Find the sum of its first $n$ terms.
AnswerLet $a$ be the first term and $d$ be the common difference of the given $A.P.$
Then, we have
$\text{S}_7=49$
$\Rightarrow\frac{7}{2}[2\text{a}+6\text{d}]=49$
$\Rightarrow\frac{7\times2}{2}[\text{a}+3\text{d}]=49$
$\Rightarrow\text{a}+3\text{d}=7\dots\text{(i)}$
Also, $\text{S}_{17}=289$
$\Rightarrow\frac{17}{2}[2\text{a}+16\text{d}]=289$
$\Rightarrow\frac{17\times2}{2}[\text{a}+8\text{d}]=289$
$\Rightarrow\text{a}+8\text{d}=17\dots\text{(ii)}$
Subtracting $(i)$ from $(ii),$ we get
$5\text{d}=10$
$\Rightarrow\text{d}=2$
$\Rightarrow\text{a}=7-3(2)=7-6=1$
$\therefore\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$=\frac{\text{n}}{2}\big[2\text{(1)}+(\text{n}-1)\text{2}\big]$
$=\frac{\text{n}}{2}\big[2+2\text{n}-2\big]$
$=\frac{\text{n}}{2}\times2\text{n}$
$=\text{n}^2$
Thus, the first term is $1$ and the common difference is $2.$
View full question & answer→Question 373 Marks
Find the sum of two middel most term of the $\text{AP}-\frac{4}{3},-1,\frac{-2}{3},\ ....4\frac{1}{3}.$
AnswerThe given $AP$ is $\frac{4}{3},-1,\frac{-2}{3},\ ....4\frac{1}{3}.$
$\text{a}=-\frac{4}{3}$ and $\text{d}=-1-\Big(-\frac{4}{3}\Big)=-1+\frac{4}{3}=\frac{1}{3}$
$\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow4\frac{1}{3}=-\frac{4}{3}+(\text{n} - 1)\Big(\frac{1}{3}\Big)$
$\Rightarrow\frac{13}{3}=-\frac{4}{3}+\frac{\text{n}}{3}-\frac{1}{3}$
$\Rightarrow\frac{13}{3}+\frac{4}{3}+\frac{1}{3}=\frac{\text{n}}{3}$
$\Rightarrow\text{n}=18$
So, there are $18$ term in the $AP.$
The two middel most term will be $\Big(\frac{\text{n}}{2}\Big)^\text{th}$ and $\Big(\frac{\text{n}}{2}+1\Big)^\text{th}$ term
So, the terms are the $9^{th}$ and $10^{th}$ term.
$\text{a}_9=\text{a}+8\text{d}=-\frac{4}{3}+8\Big(\frac{1}{3}\Big)=-\frac{4}{3}+\frac{8}{3}=\frac{4}{3}$
$\text{a}_{10}=\text{a}+9\text{d}=-\frac{4}{3}+9\Big(\frac{1}{3}\Big)=-\frac{4}{3}+\frac{9}{3}=\frac{5}{3}$
So, the sum $=\frac{4}{3}+\frac{5}{3}=3.$
View full question & answer→Question 383 Marks
If $4$ times the $4^{th}$ term of an $AP$ is equal to $18$ times its $18^{th}$ term then find its $22^{nd}$ term.
AnswerThe general term of an $AP$ is given by
$a_n=a+(n-1) d$
Given that $4 a_4=18 a_{18}$
$⇒ 4(a + 3d) = 18(a + 17d)$
$⇒ 2(a + 3d) = 9(a + 17d)$
$⇒ 2a + 6d = 9a + 153d$
$⇒ 7a = -147d$
$⇒ a = -21a ....(i)$
Now,
$a_{22}=a+21 d$
$ \Rightarrow a_{22}=-21 d+21 d $
$\Rightarrow a_{22}=0$
Thus, the $22^{\text {nd }}$ term is $0.$
View full question & answer→Question 393 Marks
Find the sum of the following arithmetic series:
$7+10\frac{1}{2}+14+...+84$
Answerwe have
$\text{a}=7,\text{d}=10\frac{1}{2}-7=\frac{21}{2}-7=\frac{21-14}{2}=\frac{7}{2}$
Let the total number of terms be n.
Then $ T_n= 84$
$⇒ \text{a} + (\text{n} - 1)\text{d} = 84$
$\Rightarrow7+(\text{n}-1)\frac{7}{2}=84$
$\Rightarrow(\text{n}-1)\frac{7}{2}=77$
$\Rightarrow(\text{n}-1)=\frac{77\times2}{7}$
$\Rightarrow\text{n}-1=22$
$\Rightarrow\text{n}=23$
Now, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[\text{a}+\text{l}\big]$
$\therefore\text{S}_\text{23}=\frac{\text{23}}{2}\big[\text{7}+84\big]$
$=\frac{23}{2}\times91=\frac{2093}{2}=1046\frac{1}{2}$
View full question & answer→Question 403 Marks
The $9^{th}$ term of an $AP$ is $-32$ and The sum of its $11^{th}$ and $13^{th}$ terms is $-94$. Find the common difference of the $AP.$
AnswerThe general term of an $AP$ is given by
$a_n=a+(n-1) d$
Given that $a_9= -32$
$⇒ a + 8d = -32 ...(i)$
Now,
$a_{11}= a_{13}= -94$
$⇒ a + 10d + a + 12d = -94$
$⇒ 2a + 22d = -94$
$⇒ a + 11d = -47 ...(ii)$
Subtract from $(i)$ from $(ii)$
$3d = -15$
$⇒ d = -5$
So, the common difference is $-5.$
View full question & answer→Question 413 Marks
Find the middel term of the $AP$ $6, 13, 20, ....., 216.$
AnswerThe given $AP$ is $6, 13, 20, .... ,216$
$a = 6$ and $d = 13 - 6 = 7$
$a_n=a+(n-1) d$
$⇒ 216 = 6 + (n - 1)(7)$
$⇒ 210 = (n - 1)(7)$
$⇒ 210 = 7n - 7$
$⇒ 7n = 217$
$⇒ n = 31$
So, there are $31$ term in the $AP.$
The middel term $=\Big(\frac{\text{n}+1}{2}\Big)=\Big(\frac{32}{2}\Big)=16$
So, the $16^{th}$ term is given by
$a_{16}=a+15 d=6+15(7)=111$
View full question & answer→Question 423 Marks
Which term of the $AP$ $3, 8, 13, 18, .....$is $88?$
AnswerIn the given $AP,$ we have $a = 3$ and $d = 8 - 3 = 5$
Suppose there are $n$ terms in given $AP,$ then
$T_n= a + (n - 1)d $
$⇒ 3 + (n - 1)5 = 88$
$⇒ 3 + 5n - 5 = 88$
$⇒ 5n = 90 ⇒ \text{n}=\frac{90}{5}=18$
Hence, the $18^{th}$ term of given $AP$ is $88.$
View full question & answer→