Question 514 Marks
How many terms of the $AP 9, 17, 25, .....$ must be taken so that their sum is $636?$
AnswerHere $a = 9, d = (17 - 9) = 8$
Let the required number of terms be n.
Then, $\text{S}_\text{n}=636$
$\Rightarrow\frac{\text{}\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]=636$
$\Rightarrow\frac{\text{}\text{n}}{2}\big[2(9)+(\text{n}-1)\text{8}\big]=636$
$\Rightarrow\text{n}\big[18+8\text{n}-8\big]=1272$
$\Rightarrow\text{n}\big[8\text{n}+10\big]=1272$
$\Rightarrow8\text{n}^2+10\text{n}-1272=0$
$\Rightarrow4\text{n}^2+5\text{n}-636=0$
$\Rightarrow4\text{n}^2+53\text{n}-48\text{n}-636=0$
$\Rightarrow\text{n}(4\text{n}+53)-12(4\text{n}+53)=0$
$\Rightarrow(4\text{n}+53)(\text{n}-12)=0$
$\Rightarrow4\text{n}+53=0$ or $\text{n}=12$
$\Rightarrow\text{n}=\frac{-53}{4}$ or $\text{n}=12$
Since number of terms cannot neither be negative nor fraction, $n = 12$
Hence, the required number of terms is $12.$
View full question & answer→Question 524 Marks
If the sum of first $p$ term of an AP is $(ap^2+ bp),$ find its common difference.
AnswerLet $S_p$ denotes the sum of first $p$ terms of the $A P$.
$\therefore S_p=a p^2+b p $
$ \Rightarrow S_{p-1}=a(p-1)^2+b(p-1) $
$ =a\left(p^2-2 p+1\right)+b(p-1) $
$ =a p^2-(2 a-b) p+(a-b)$
Now,
$p^{\text {th }} \text { term of the } A P, a_p=S_p-S_p-1 $
$ =\left(a p^2+b p\right)-\left[a p^2-(2 a-b) p+(a-b)\right] $
$ =a p^2+b p-a p^2+(2 a-b) p-(a-b) $
$ =2 a p-(a-b)$
Let $d$ be the common difference of the $AP.$
$\therefore d=a_p-a_{p-1} $
$ =[2 a p-(a-b)]-[2 a(p-1)-(a-b)] $
$ =2 a p-(a-b)-2 a(p-1)+(a-b) $
$ =2 a$
Hence, the common difference of the $AP$ is $2a.$
View full question & answer→Question 534 Marks
In a potato race, a bucket is placed at the starting point, which is $5\ m$ from the first potato, and the other potatoes are placed $3\ m$ apart in a straight line. There are $10$ potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato to the bucket to drop it in, and he continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
AnswerDistance he ran to pic the first potato
$= 5m + 5m = 10m$
Distance he ran to pic the second potato
$= 5m + 3m + 5m + 3m = 16m$
Distance he ran to pic the third potato
$= 8m + 3m + 8m + 3m = 22m$
So, $a = 10, d = 6$
Total distance covered by the gardener is given by $S_n$, where $n = 25.$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\Rightarrow\text{S}_\text{10}=\frac{\text{10}}{2}\big[2(10)+(10-1)6\big]$
$\Rightarrow\text{S}_\text{10}=5\big[20+54\big]$
$\Rightarrow\text{S}_\text{10}=370$
Thus, the total distance is $370\ m.$
View full question & answer→Question 544 Marks
In an $AP$, the first term is $-4$, the last term is $29$ and the sum of all its terms is $150$. Find the common difference.
AnswerLet the given $A,P$. contains n terms.
First terms, $a = -4$
Last term, $l = 29$
$S_n= 150$
$\Rightarrow\frac{\text{n}}{2}\big[\text{a}+\text{l}\big]=150$
$\Rightarrow\frac{\text{n}}{2}\big[-4+29\big]=150$
$\Rightarrow\text{n}\times25=300$
$\Rightarrow\text{n}=12$
Thus. the given $A.P.$ contain $12$ terms.
Let d be the common difference of the given $A.P.$
Then,
$T_{12}= 29$
$\Rightarrow a + 11d = 29$
$\Rightarrow -4 + 11d = 29$
$\Rightarrow 11d = 33$
$\Rightarrow d = 3$
View full question & answer→Question 554 Marks
Two $APs$ have the same common difference. If the first term of these $APs$ be $3$ and $8$ respectively, find the difference between the sums of their first $50$ terms.
AnswerLet a and a' be the first term of two $APs$ respectively.
Then, $a = 3$ and $a' = 8$
Let d be the common difference of two $APs.$
Let $S_{50}$ and $S'_{50}$ denote the sum of their first 50 temms.
Now, $\text{S}'_{50}-\text{S}_{50}=\frac{50}{2}\big[2(8)+49\text{d}\big]-\frac{50}{2}\big[2(3)+49\text{d}\big]$
$=25\big[16+49\text{d}\big]-25\big[6+49\text{d}\big]$
$25\big[16+49\text{d}-6-49\text{d}\big]$
$=25\times10$
$=250$
View full question & answer→Question 564 Marks
If the sum of first n term is $(3n^2+ 5n),$ find its common difference.
AnswerLet $S_n$ denotes the sum of first n terms of the $AP.$
$ \therefore S_n=3 n^2+5 n $
$ \Rightarrow S_{n-1}=3(n-1)^2+5(n-1) $
$ =3\left(n^2-2 n+1\right)+5(n-1) $
$ =3 n^2-n-2$
Now,
$\mathrm{n}^{\text {th }} \text { term of the } A P, a_n=S_n-S_{n-1}$
$ =\left(3 n^2+5 n\right)-\left(3 n^2-n-2\right) $
$ =6 n+2$
Let d be the common difference of the $AP.$
$ \therefore d=a_n-a_{n-1} $
$ =(6 n+2)-[6(n-1)+2] $
$ =6 n+2-6(n-1)-2 $
$ =6$
Hence, the common difference of the $AP$ is $6.$
View full question & answer→Question 574 Marks
The sum of the $4^{\text {th }}$ and the $8^{\text {th }}$ term of an $AP$ is $24$ and the sum of its $6^{\text {th }}$ and $10^{\text {th }}$ terms is $44 .$ Find the sum of its first $10$ terms.
AnswerThe general term of an AP is given by $a_n= a + (n - 1)d$
and $\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big].$
Given that $a_4 + a_8 = 24$
$\Rightarrow a + 3d + a + 7d = 24$
$\Rightarrow 2a + 10d = 24 ....(i)$
Also, $a_6+ a_{10}= 44$
$\Rightarrow a + 5d + a + 9d = 44$
$\Rightarrow 2a + 14d = 44 ....(ii)$
Subtracting $(i)$ and $(ii)$,, we get
$4d = 20$
$\Rightarrow d = 5$
Substituting $(i)$ we get $a = -13$
So, the sum of the first 10 terms
$=\frac{10}{2}\big[2(-13)+9(5)\big]$
$=2\big[-26 + 45\big]$
$=95$
View full question & answer→Question 584 Marks
An $AP 8, 10, 12, ...$.has $60$ terms. Find its last term. Hence, find the sum of its last $10$ terms.
AnswerThe general term of an $AP$ is given by
$a_n=a+(n-1) d $
$\text { and } S_n=\frac{n}{2}[2 a+(n-1) d] .$
The AP is $8,10,12, \ldots \ldots$
So, $a=8$ and $d=2$
Given that $a_{60}=a+(n-1) d$
$\Rightarrow a_{60}=8+59(2)$
$ \Rightarrow a_{60}=126$
So, its last term is $126$.
Sum of its last $10$ terms
= sum of $60$ terms - sum of $50$ terms
$=\frac{60}{2}\big[2(8)+59(2)\big]-\frac{50}{2}\big[2(8)+49(2)\big]$
$=30\big[16+118\big]-25\big[16+98\big]$
$=4020-2850$
$=1170$
View full question & answer→Question 594 Marks
The first and the last terms of an $AP$ are $5$ and $45$ respectively. If the sum of all its term is $400$, find the common difference and the number of terms.
AnswerLet the given $A.P$. contains n terms.
First term, $a = 5$
Last term, $l = 45$
$S_n= 400$
$\Rightarrow\frac{\text{n}}{2}\big[\text{a}+\text{l}\big]=400$
$\Rightarrow\frac{\text{n}}{2}\big[\text{5}+\text{45}\big]=400$
$\Rightarrow\text{n}\times50=800$
$\Rightarrow\text{n}=16$
Thus, the given $A.P.$ contains $16$ terms.
Let d be the common difference of the given $A.P.$
Then,
$T_{16}= 45$
$\Rightarrow a + 15d = 45$
$\Rightarrow 5 + 15d = 45$
$\Rightarrow 15d = 40$
$\Rightarrow\text{d}=\frac{40}{15}=\frac{8}{3}$
View full question & answer→Question 604 Marks
The sum of first $10$ term of an $AP$ is $-150$ and the sum of its next $10$ terms is $-550$. Find the $AP.$
AnswerLet a be the first term and d be the common difference of the given $A.P.$
Then, we have
$\text{S}_{10}=-150$
$\Rightarrow\frac{10}{2}\big[2\text{a}+9\text{d}\big]=-50$
$\Rightarrow5\big[2\text{a}+9\text{d}\big]=-50$
$\Rightarrow2\text{a}+9\text{d}=-30\dots(\text{i})$
Clearly, the sum of first 20 terms $= -150 + (-550) = -700$
$\therefore\text{S}_{20}=-700$
$\Rightarrow\frac{20}{2}\big[2\text{a}+19\text{d}\big]=-700$
$\Rightarrow10\big[2\text{a}+19\text{d}\big]=-700$
$\Rightarrow2\text{a}+19\text{d}=-70\dots(\text{ii})$
Subtracting $(i)$ and $(ii)$,, we get
$10\text{d}=-40$
$\Rightarrow\text{d}=-4$
$\Rightarrow2\text{a}=-30-9(-4)=-30+36=6$
$\Rightarrow\text{a}=3$
Thus, we have First term $= a + d = 3 + (-4) = -1$
Third term $= a + 2d = 3 + 2(-4) = 3 - 8 = -5$
Fourth term $= a + 3d = 3 + 3(-4) = 3 - 12 = -9$
Thus, the given $AP$ is $3, -1, -5, -9, ......$
View full question & answer→Question 614 Marks
In a school, students decided to plant trees in and around the school to reduce air pollution. it was decided that number of trees that each section of each class will plant will be double of the class in which they are studying. If there are $1$ to $12$ classes in the school and each class has two sections, find how many trees were planted by students. Which value is shown in the question?
AnswerStudent of first section of class $1$ will plant $2$ trees.
Student of second section of class $1$ will plant $2$ trees.
Thus, students of class $1$ will plant $4$ trees.
Student of first section of class $2$ will plant $4$ trees.
Student of second section of class $2$ will plant $4$ trees.
Thus, students of class $2$ will plant $8$ trees.
Student of first section of class $3$ will plant $6$ trees.
Student of second section of class $3$ will plant $6$ trees.
Thus, students of class $3$ will plant $12$ trees.
Thus, the number of trees planted by the students,
From an $AP : 4, 8, 12, .....$
Thus, $a = 4$ and $d = 4$
Let us find the number of trees planted in total.
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\Rightarrow\text{S}_\text{12}=\frac{\text{12}}{2}\big[2\times4+(\text{12}-1)\text{4}\big]$
$\Rightarrow\text{S}_\text{12}=6\big[8+44\big]$
$\Rightarrow\text{S}_\text{12}=312$
Thus, the total number of trees is $312.$
We shouid conserve the nature around us and bring about awareness to save trees.
View full question & answer→Question 624 Marks
Find the sum of n terms of the following series:
$\Big(4-\frac{1}{\text{n}}\Big)+\Big(4-\frac{2}{\text{n}}\Big)+\Big(4-\frac{3}{\text{n}}\Big)+\ ...$
AnswerSum of $4-\frac{1}{\text{n}},4-\frac{2}{\text{n}},4-\frac{3}{\text{n}}$ up to the nth term
$= (4 + 4 + 4 + 4 + 4 + .........$ up to n terms) + $\frac{1}{\text{n}}$$(1 + 2 + 3 +4 .........$ upto n terms)
$= 4 ( 1 + 1 + 1 + 1..........$ upto n terms) - $\frac{1}{\text{n}}$$(1 + 2 + 3 +4 .........$ upto n terms)
$=4\text{n}-\frac{1}{\text{n}}\times\frac{\text{n}(\text{n}+1)}{2}$
$=4\text{n}-\frac{(\text{n}+1)}{2}$
$=\frac{[8\text{n}-(\text{n}+1)]}{2}\dots.\text{taking L.C.M}$
$=\frac{(7\text{n}-1)}{2}$
View full question & answer→Question 634 Marks
Find the sum of all multiples of $9$ lying between $300$ and $700.$
AnswerAll numbers between $300$ and $700$ that are multiples of $9$ are $306, 315, 324, 333, ..., 693$
This is an $AP$ in which $a = 306, d = (315 - 306) = 9, l = 693$
Let the number of these terms be n, then
$T_n= 693$
$\Rightarrow a + (n - 1)d = 693$
$\Rightarrow 306 + (n - 1) \times 9 = 693$
$\Rightarrow 9(n - 1) = 387$
$\Rightarrow (n - 1) = 43$
$\Rightarrow n = 44$
Required $\text{sum}=\frac{\text{n}}{2}(\text{a}+\text{l})$
$=\frac{44}{2}(306+693)$
$\Rightarrow 22 \times 999$
$\Rightarrow 22 \times (1000 - 1)$
$\Rightarrow 22 \times 1000 - 22$
$\Rightarrow 22000 - 22 = 21978$
Hence, $S_n= 21978$
View full question & answer→Question 644 Marks
Find the number of term of the $AP -12, -9, -6, ....,21$. If i is added to each term of this $AP$ then find the sum of all terms of the $AP$ thus obtained.
AnswerThe general term of an $AP$ is given by $a_n= a + (n - 1)d$
and $\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$d = -9 - (-12) = 3$
So, $21 = -12 + (n - 1)(3)$
$\Rightarrow 33 = 3n - 3$
$\Rightarrow 36 = 3n$
$\Rightarrow n = 12$
If is added to each term of this AP,
Then the AP becomes $-11, -8, -5, ...., 20.$
$d = -8 - (-11) = 3$
$\Rightarrow\text{S}_{12}=\frac{12}{2}\big[2(-11+11(3)\big]$
$\Rightarrow S_{12}=6[-22+33] $
$ \Rightarrow S_{12}=6(11) $
$\Rightarrow S_{12}=66$
View full question & answer→Question 654 Marks
A contract on constrution job specifies a penalty for delay of completion beyond a certain date as follows: $₹ 200$ for the first day, $₹ 250$ for the second day, $₹ 300$ for the third day, etc, the penalty for each succeeding day being $₹ 50$ more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by $30$ days?
AnswerThe penalty is given to be:
$Rs. 200$ for the first day,
$Rs. 250$ for the second day,
$Rs.300$ for the third day, etc
Since the penalty for each succeeding day is $Rs. 50$ more than for the preceding day,
The common difference = $Rs. 50$
Consider the work to be delayed for $30$ days.
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow\text{S}_\text{30}=\frac{\text{30}}{2}\big[2\text{(200)}+29\text{(50)}\big]$
$\Rightarrow\text{S}_\text{30}=15\big[400+1450\big]$
$\Rightarrow\text{S}_\text{30}=27750$
Hence, the contractor has to pay $Rs. 27750$ as penalty.
View full question & answer→Question 664 Marks
A man arranges to pay off a debt of $₹ 36000$ by $40$ monthly instalments which form an arithmetic series. When $30$ of the instalments are paid, he dies leaving one-third of the debt unpaid. Find the value of the first instalment.
AnswerThe man arranges to pay off a debt of $Rs. 36000$ by $40$ monthly installments.
So, $n = 40$ and $S_{40}= 36000$
Let the first installment be Rs. a, and let d be the common difference.
One-third debt is unpaid, that means two-third is paid.
$\frac{2}{3}(36000)=\text{Rs. }24000$
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow\text{S}_\text{30}=\frac{\text{30}}{2}\big[2\text{a}+29\text{d}\big]$
$\Rightarrow24000=15[2\text{a}+29\text{d}]$
$\Rightarrow1600=2\text{a}+29\text{d}$
$\Rightarrow2\text{a}+29\text{d}=1600\dots(\text{i})$
$\Rightarrow\text{S}_\text{40}=\frac{\text{40}}{2}\big[2\text{a}+39\text{d}\big]$
$\Rightarrow36000=20[2\text{a}+39\text{d}]$
$\Rightarrow1800=2\text{a}+39\text{d}$
$\Rightarrow2\text{a}+39\text{d}=1800\dots(\text{ii})$
Subtracting $(i)$ and $(ii)$,, we get
$10d = 200$
$⇒ d = 20$
Substituting in $(i)$, we get
$2a + 29(20) = 1600$
$⇒ 2a = 1020$
$⇒ a = 510$
Hence, the first installment he paid was $Rs. 510.$
View full question & answer→Question 674 Marks
Find the sum of first $51$ terms of an $AP$ whose second and third terms are $14$ and $18$ respectively.
AnswerThe general term of an $AP$ is given by $a_n= a + (n - 1)d$
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
Given that $a_2=14$ and $a_3=18$
So, $d=a_3-a_2=18-14=4$.
Now, $a_2= 14 \Rightarrow a + 4 = 14 \Rightarrow a = 10$
Also, $\text{S}_{51}=\frac{51}{2}\big[2(10)+(50)4\big]$
$\Rightarrow\text{S}_{51}=\frac{51}{2}\big[20+200\big]$
$\Rightarrow\text{S}_{51}=\frac{51}{2}\big[220\big]$
$\Rightarrow\text{S}_{51}=51\times110$
$\Rightarrow\text{S}_{51}=5610$
View full question & answer→Question 684 Marks
Find the sum of the first $15$ multiples of $8.$
AnswerMultiples of $8$ are
$8, 16, 24, ....$
Since fifference is same, it is an $AP$
We need to find sum of first $15$ multiples
We use formula
$\text{S}_\text{n}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})$
Here, $n = 15, a = 8 \& d = 16 - 8 = 8$
Putting values in formula
$\text{S}_\text{n}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})$
$=\frac{15}{2}(2\times8+(15-1)\times8)$
$=\frac{15}{2}(16+14\times8)$
$=\frac{15}{2}(16+122)$
$=\frac{15}{2}\times128$
$=960$
Therefore, the sum of first $15$ multiples of $8$ is $960$
View full question & answer→Question 694 Marks
The $\mathrm{n}^{\text {th }}$ term of an $AP$ is $(7 - 4n).$ Find its common difference.
AnswerWe have:
$T_n=(7-4 n)$
Common difference $=T_2-T_1$
$T_1=7-4 \times 1=3$
$ T_2=7-4 \times 2=-1$
$ d=-1-3=-4$
Hence, the common difference is$ -4.$
View full question & answer→Question 704 Marks
In an $AP$, the first terms is $22, n^{th}$ term is $-11$ and sum of first n terms is $66$. Find n and hence find the common difference.
AnswerFirst term of an $AP, a = 22$
Last term $= n^{th}$ term $= -11$
Sum of n terms $=\text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}+\text{l})=66$
$\Rightarrow \frac{\text{n}}{2}(\text{22}+\text{11})=66$ or $\frac{\text{n}}{2}\times11=66$
$\therefore\text{n}=\frac{66\times2}{11}=12$
$n^{th}$ term $= l = a + (n - 1)d$
$\therefore-11=22+(12-1)\times\text{d}$ or $-11=22+11\text{d}$
$\Rightarrow11\text{d}=-22-11$
$\Rightarrow11\text{d}=-33$
$\therefore\text{d}=\frac{-33}{11}=-3$
Thus, $n = 12, d = -3$
View full question & answer→Question 714 Marks
How many three-digit natural numbers are divisible by $9?$
AnswerThe three-digit natural numbers divisible by $9$ are $108, 117, 126, ..., 999.$
Clearly, three number are in AP.
Here, $a = 108$ and $d = 117 - 108 = 9$
Let this AP contains n terms. Then,
$ a_n=999 $
$ \Rightarrow 108+(n-1) \times 9=999\left[a_n=a+(n-1) d\right] $
$ \Rightarrow 9 n+99=999 $
$ \Rightarrow 9 n=999-99=900 $
$ \Rightarrow n=100$
Hence, there are $100$ three-digit numbers divisible by $9.$
View full question & answer→Question 724 Marks
How many three-digit natural numbers are divisible by $7?$
AnswerThe three-digit natural numbers divisible by $7$ are $105, 112, 119, ..., 994.$
Clearly, three number are in $AP.$
Here, $a = 105$ and $d = 112 - 105 = 7$
Let this AP contains n terms. Then,
$ \mathrm{a}_{\mathrm{n}}=994 $
$ \Rightarrow 105+(\mathrm{n}-1) \times 7=994 $
$ \Rightarrow 7 \mathrm{n}+98=994\left[\mathrm{a}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}\right] $
$ \Rightarrow 7 \mathrm{n}=994-98=896 $
$ \Rightarrow \mathrm{n}=128$
Hence, there are $128$ three-digit numbers divisible by $7.$
View full question & answer→Question 734 Marks
The sum of first $n$ terms of two $APs$ are in the ratio $(3 n+8):(7 n+15)$. Find the ratio of their $12^{\text {th }}$ terms.
AnswerThere are $2 AP$ 's with different first term and common difference.
For the first $AP$
Let first term be a common difference be d
Sum of n term $=\text{S}_\text{n}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})$ & $n^{th}$ term $=\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
Similarly for second $AP$
Let first term = A, common difference
$\text{S}_\text{n}=\frac{\text{n}}{2}(2\text{A}+(\text{n}-1)\text{D})$ & $\mathrm{n}^{\text {th }}$ term $=\mathrm{A}_{\mathrm{n}}=\mathrm{A}+(\mathrm{n}-1) \mathrm{D}$
We need to find ratio of $12^{th}$ term
$\text{i.e.}\frac{\text{a}_{12}\text{ of first AP}}{\text{A}_{12}\text{ of second AP}}$
$=\frac{\text{a}+(12-1)\text{d}}{\text{A}+(12-1)\text{D}}$
$=\frac{\text{a}+11\text{d}}{\text{A}+11\text{D}}$
It is given that
$\frac{\text{Sum of n terms of 1st AP}}{\text{Sum of n terms of 2nd AP}}=\frac{3\text{n}+8}{7\text{n}+15}$
$\frac{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]}{\frac{\text{n}}{2}[2\text{A}+(\text{n}-1)\text{D}]}=\frac{3\text{n}+8}{7\text{n}+15}$
$\frac{[2\text{a}+(\text{n}-1)\text{d}]}{[2\text{A}+(\text{n}-1)\text{D}]}=\frac{3\text{n}+8}{7\text{n}+15}$
$\frac{2\big(\text{a}+\big(\frac{\text{n}-1}{2}\big)\text{d}\big)}{2\big(\text{A}+\big(\frac{\text{n}-1}{2}\big)\text{D}\big)}=\frac{3\text{n}+8}{7\text{n}+15}$
$\frac{\big(\text{a}+\big(\frac{\text{n}-1}{2}\big)\text{d}\big)}{\big(\text{A}+\big(\frac{\text{n}-1}{2}\big)\text{D}\big)}=\frac{3\text{n}+8}{7\text{n}+15}\dots(1)$
We need to find $\frac{\text{a}+11\text{d}}{\text{A}+11\text{D}}$
Hence $\frac{\text{n}-1}{2}=11$
$\text{n}-1=22$
$\text{n}=23$
Putting $n = 23$ in $(1)$
$\frac{\text{a}+\big(\frac{\text{23}-1}{2}\text{d}\big)}{\text{A}+\big(\frac{\text{23}-1}{2}\big)\text{D}}=\frac{3\times23+8}{7\times23+15}$
$\frac{\text{a}+\big(\frac{\text{22}}{2}\text{d}\big)}{\text{A}+\big(\frac{\text{22}}{2}\big)\text{D}}=\frac{69+8}{161+15}$
$\frac{\text{a}+11\text{d}}{\text{A}+11\text{D}}=\frac{77}{176}$
$\frac{\text{a}+11\text{d}}{\text{A}+11\text{D}}=\frac{7}{16}$
Hence ratio of their $12^{th}$ term is $\frac{7}{16}$ i.e. $7 : 16.$
View full question & answer→Question 744 Marks
If $a_n$ denotes the $n^{th}$ term of the AP $2, 7, 12, 17, ....,$ find the value of $(a_{30}- a_{20})$.
AnswerThe given AP is $2, 7, 12, 17, ....$
Here, $a = 2$ and $d = 7 - 2 = 5$
$ \therefore a_{30}-a_{20} $
$ =[2+(30-1) \times 5]-[2+(20-1) \times 5]\left[a_n=a+(n-1) d\right] $
$ =147-97 $
$ =50$
Hence, the required value is $50.$
View full question & answer→Question 754 Marks
If $\frac{4}{5},\text{a},2$ are in $AP$, find the value of a.
AnswerIf $\frac{4}{5},\text{a}$ and $2$ are three consecutive terms of an $AP$, then we have:
$\text{a}-\frac{4}{5}=2-\text{a}$
$\Rightarrow2\text{a}=2+\frac{4}{5}$
$\Rightarrow2\text{a}=\frac{14}{5}$
$\Rightarrow\text{a}=\frac{7}{5}$
View full question & answer→Question 764 Marks
If the numbers $a, 9, b, 25$ form an $AP$, find $a$ and $b.$
AnswerIt is given that the numbers $a, 9, 25$, form an $AP.$
$\therefore 9 - a = b - 9 = 25 - b$
So,
$b - 9 = 25 - b$
$\Rightarrow 2b = 34$
$\Rightarrow b = 17$
Also,
$9 - a = b - 9$
$\Rightarrow a = 18 - b$
$\Rightarrow a = 18 - 17 (b = 17)$
$\Rightarrow a = 1$
Hence, the required value of a and b are $1$ and $17$, respectively.
View full question & answer→Question 774 Marks
If $k, (2k - 1)$ and $(2k + 1)$ are the three successive term of an $AP$, find the value of $k.$
AnswerIt is given that $k,(2 k-1)$ and $(2 k+1)$ are the three successive terms of an $AP.$
$\therefore (2k - 1) - k = (2k + 1) - (2k - 1)$
$\Rightarrow k - 1 = 2$
$\Rightarrow k = 3$
Hence, the value of $k$ is $3.$
View full question & answer→Question 784 Marks
A ma saved $₹ 33000$ in $10$ month. In each month after the first, he saved ₹ $100$ more than he did in the preceding month. How much did he save in the first month?
AnswerA man saved $Rs.\ 33000$ in $10$ month
That is, $S_{10}= Rs.\ 33000$
Common difference $= d = Rs.\ 100$
Let the amount he saved in the first month be Rs. a.
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}$
$\text{S}_\text{10}=\frac{\text{10}}{2}\big[2\text{a}+9(100)\big]$
$\Rightarrow33000=5[2\text{a}+900]$
$\Rightarrow33000=10\text{a}+4500$
$\Rightarrow10\text{a}=28500$
$\Rightarrow\text{a}=\text{R.s }2850$
Hence, he saved $Rs. 2850$ in the first month.
View full question & answer→Question 794 Marks
The $n^{th}$ term of an $AP$ is $(3n + 5)$. Find its common difference.
AnswerWe have:
$ T_n=(3 n+5) $
$ \text { Common difference }=T_2-T_1 $
$ T_1=3 \times 1+5=8 $
$ T_2=3 \times 2+5=11 $
$ d=11-8=3$
Hence, the common difference is $3.$
View full question & answer→Question 804 Marks
If $1 + 4 + 7 + 10 + .....+ x = 287$, find the value of $x.$
Answer$\text{S}_\text{n}=\frac{\text{n}}{2}(2\text{a}+(\text{n} - 1)\text{a})$
$287=\frac{\text{n}}{2}(2+3\text{n}-3)$
$574=2\text{n}+3\text{n}^2-3\text{n}$
$3\text{n}^2-\text{n}-574=0$
$ {-\text{b} \pm \sqrt{(\text{b}^2-4\text{ac})} \over 2\text{a}}$ $=\frac{-1\pm\sqrt{1^2-4(3)(-574)}}{2(3)}$
$=\frac{-1\pm\sqrt{1-12-574}}{6}$
$\Rightarrow-\frac{1\pm\sqrt{-12-574}}{6}$
$\Rightarrow\frac{-1+3\sqrt{65}}{6}$
$=\frac{-41}{3}\not=\text{n}$
$\Rightarrow\text{n}=14$
$\text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}+\text{a}_\text{n})$
$=\frac{14}{2}(1+\text{x})$
$587=\frac{14}{2}(1+\text{x})$
$\frac{587}{7}=1+\text{x}$
$41=1+\text{x}$
$\text{x}=40$
is the value
View full question & answer→Question 814 Marks
If $(2p + 1), 13, (5p - 3)$ are in $AP$, find the value of $p.$
AnswerLet $(2p + 1), 13, (5p - 3)$ be three consecutive terms of an $AP.$
Then $13 - (2p + 1) = (5p - 3) - 13$
$\Rightarrow 7p = 28$
$\Rightarrow p = 4$
$\therefore$ When $p = 4, (2p + 1), 13$ and $(5p - 3)$ fprm three consecutive terms of an $AP.$
View full question & answer→Question 824 Marks
In an $AP$, it is given that $S_5+S_7=167$ and $S_{10}=235$, then find the $AP$, where $S_n$ denotes the sum of its first $n$ terms.
Answer$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
Now, $\text{S}_5+\text{S}_7=167$
$\Rightarrow\frac{5}{2}\big[2\text{a}+4\text{d}\big]+\frac{7}{2}\big[2\text{a}+6\text{d}\big]=167$
$\Rightarrow\frac{5\times2}{2}\big[\text{a}+2\text{d}\big]+\frac{7\times2}{2}\big[\text{a}+3\text{d}\big]=167$
$\Rightarrow5\text{a}+10\text{d}+7\text{a}+21\text{d}=167$
$\Rightarrow12\text{a}+31\text{d}=167\dots(\text{i})$
also, $\text{S}_{10}=235$
$\Rightarrow\frac{10}{2}\big[2\text{a}+9\text{d}\big]=235$
$\Rightarrow5\big[2\text{a}+9\text{d}\big]=235$
$\Rightarrow10\text{a}+45\text{d}=235$
$\Rightarrow2\text{a}+9\text{d}=47\dots(\text{ii})$
Multiplying equation $(ii)$ by $6$, we get
$12\text{a}+54\text{d}=282\dots(\text{iii})$
Subtracting $(i)$ from $(iii)$, we get
$23\text{d}=115$
$\Rightarrow\text{d}=5$
$\Rightarrow2\text{a}+9(5)=47\dots[\text{From}(\text{ii})]$
$\Rightarrow2\text{a}=2$
$\Rightarrow\text{a}=1$
$?$ First term $= a = 1$
Second term $= a + d = 1 + 5 = 6$
Third term $= a + 2d = 1 + 2(5) = 11$
Thus, the $A.P,$ is $1, 6, 11, ....$
View full question & answer→Question 834 Marks
A child puts one five-rupee coin of her saving in the piggy bank on the first day. She increases her saving by one five-rupee coin daily. If the piggy bank can hold $190$ coins of five rupees in all, find the number of days she can contribute to put the five-rupee coins onto it and find the tatal money the saved.
AnswerChild will put 5 Rs on $1^{st}$ day, 10 RS (2 × 5 Rs) on $2^{nd}$ day,15 Rs (3 × 5 Rs) on $3^{rd}$ day etc.
Total savings $= 190$ coins $= 190 × 5 = 950$ Rs
The above problem can be written as Arithmetic progression series
$5, 10, 15, 20, ...$
With $a = 5, d = 5, S_n= 950$
Let n be the last day when piggy bank become full.
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$950=\frac{\text{n}}{2}\big[2\times5+(\text{n}-1)\text{5}\big]$
$1900=\text{n}\big[10+5\text{n}-5\big]$
$1900=\text{n}\big[5\text{n}+5\big]$
$1900=5\text{n}^2+5\text{n}$
Divide the equation by 5.
$380=\text{n}^2+\text{n}$
$\text{n}^2+\text{n}-380=0$
$\text{n}^2+20\text{n}-19\text{n}-380=0$
$\text{n}(\text{n}+20)-19(\text{n}+20)=0$
$(\text{n}+20)(\text{n}-19)=0$
$\text{n}+20=0$ or $\text{n}-19=0$
$\text{n}=-20$ or $\text{n}=19$
n cannot be negative, hence $n = 19$
She can put money for $19$ days.
Total savings is $950 Rs.$
View full question & answer→Question 844 Marks
Find the sum of first $100$ even natural numbers which are divisible by $5.$
AnswerThe first few even natural numbers which are divisible by $5$ are $10,20,30,40, \ldots$
This is an $AP$ in which $a =10, d=(20-10)=10$ and $n =100$
The sum of $n$ terms of an AP is qiven by
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$=\Big(\frac{100}{2}\Big)\times\big[2\times10(100-1)\times10\big]$$\big[\therefore\text{a}=10,\text{d}=10, \text{and } \text{n}=100\big] $
$=50\times[20+990]=50\times1010=50500$
Hence, the sum of the first hundred even natural numbers which are divisible by $5$ is $50500.$
View full question & answer→Question 854 Marks
Write the next term of the $\text{AP}\sqrt{2},\sqrt{8},\sqrt{18},....$
AnswerThe given $AP$ is $\sqrt{2},\sqrt{8},\sqrt{18},....$
On simplifying the terms, we get :
$\sqrt{2},2\sqrt{2},3\sqrt{2},....$
Here, $\text{a}=\sqrt{2}$ and $\text{d}=(2\sqrt{2}-\sqrt{2})=\sqrt{2}$
$\therefore$ Next term, $\text{T}_4=\text{a}+3\text{d}=\sqrt{2}+3\sqrt{2}=4\sqrt{2}=\sqrt{32}$
View full question & answer→Question 864 Marks
An $AP 5, 12, 19, ....$ has $50$ terms. Find its last term. Hence, find the sum of its last $15$ terms.
AnswerThe general term of an $AP$ is given by
$a_n= a + (n - 1)d$
and $\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big].$
The $AP$ is $5, 12, 19, .....$
So, $a = 5$ and $d = 7$
$ \text { Given that } a_{60}=a+(n-1) d $
$ \Rightarrow a_{50}=5+49(7) $
$ \Rightarrow a_{50}=348$
So, its last term is $348.$
Sum of its last $15$ terms
= sum of $50$ terms - sum of $35$ terms
$=\frac{50}{2}\big[2(5)+49(7)\big]-\frac{35}{2}\big[2(5)+34(7)\big]$
$=25\big[10+343\big]-\frac{35}{2}\big[10+68\big]$
$=8825-4340$
$=4485$
View full question & answer→Question 874 Marks
Find the sum of first forty positive intergers divisible by $6.$
AnswerFirst forty positive intergers divisible by $6$ are as follows:
$6, 12, 18, 24, ....240$
$\therefore\text{S}_{40}=\frac{40}{2}\big[6+240\big]$
$=20\times246$
$=4920$
View full question & answer→Question 884 Marks
If mth term of an is $\frac{1}{\text{n}}$ and nth term is $\frac{1}{\text{m}}$ then find the sum of its first mn terms.
AnswerGiven that $\text{a}_\text{m}=\frac{1}{\text{n}}$
$\Rightarrow\text{a}+(\text{m}-1)\text{d}=\frac{1}{\text{n}}$
$\Rightarrow\text{an}+\text{mnd}-\text{nd}=1\dots(1)$
$\text{a}_\text{n}=\frac{1}{\text{m}}$
$\Rightarrow\text{a}+(\text{n}-1)\text{d}=\frac{1}{\text{m}}$
$\Rightarrow\text{am}+\text{mnd}-\text{md}=1\dots(2)$
From $(1)$ and $(2)$, we get
$\text{an}+\text{mnd}-\text{nd}=\text{am}+\text{mnd}-\text{md}$
$\Rightarrow\text{a}(\text{n}-\text{m})-(\text{n}-\text{m})\text{d}=0$
$\Rightarrow\text{a}(\text{n}-\text{m})=(\text{n}-\text{m})\text{d}$
$\therefore\text{a}=\text{d}$
Consider $(1)$, $\text{an}+\text{mnd}-\text{nd}=1$
$\text{dn}+\text{mnd}-\text{nd}=1$
$\therefore\text{d}=\frac{1}{\text{mn}}$
Hence $\text{a}=\frac{1}{\text{mn}}$
Sum of mn term of $AP$ is $\text{S}_\text{mn}=\frac{\text{mn}}{2}\big[2\text{a}+(\text{mn}-1)\text{d}\big]$
$=\frac{\text{mn}}{2}\Big[\frac{2}{\text{mn}}+\frac{(\text{mn}-1)}{\text{mn}}\Big]$
$=\frac{\text{mn}}{2\text{mn}}\big[2+\text{mn}-1\big]$
$=\frac{1}{2}(\text{mn}+1)$
View full question & answer→