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Maths 5 Marks Questions

Arithmetic Progressions · CBSE STD 10 (CBSE - English Medium). 88 questions.

Questions · Page 2 of 2

5 Marks Questions

Question 515 Marks
How many terms of the AP 9, 17, 25, ..... must be taken so that their sum is 636?
Answer
Here 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.
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Question 525 Marks
If the sum of first p term of an AP is $(ap^2+ bp)$, find its common difference.
Answer
Let $S_p$​​​​​​​ denotes the sum of first p terms of the AP.
$\therefore$ $S_p = ap^2 + bp$
$\Rightarrow S_{p - 1} = a(p - 1)^2 + b(p - 1)$
$= a(p^2 - 2p + 1) + b(p - 1)$
$= ap^2 - (2a - b)p + (a - b)$
Now,
$p^{th}$ term of the $AP, a_p = S_p - S_{p - 1}$
$= (ap^2 + bp) - [ap^2 - (2a - b)p + (a - b)]$
$= ap^2 + bp - ap^2+ (2a - b)p - (a - b)$
$= 2ap - (a - b)$
Let d be the common difference of the AP.
$\therefore$ $d = a_p- a_{p - 1}$
$= [2ap - (a - b)] - [2a(p - 1) - (a - b)]$
$= 2ap - (a - b) - 2a(p - 1) + (a - b)$
$= 2a$
Hence, the common difference of the AP is 2a.
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Question 535 Marks
In a potato race, a bucket is placed at the starting point, which is 5m from the first potato, and the other potatoes are placed 3m 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?
Answer
Distance he ran to pic the first potato
$=5 m+5 m=10 m$
Distance he ran to pic the second potato
$=5 m+3 m+5 m+3 m=16 m$
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 370m.
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Question 545 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.
Answer
Let 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,
$T_{12} = 29$
$\Rightarrow a + 11d = 29$
$\Rightarrow -4 + 11d = 29$
$\Rightarrow 11d = 33$
$\Rightarrow d = 3$
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Question 555 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.
Answer
Let 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, $\text{S}'_{50}-\text{S}_{50}=\frac{50}{2}\big[2(8)+49\text{d}\big]-\frac{50}{2}\big[2(3)+49\text{d}\big]$
$=25\big[16+49\text{d}\big]-25\big[6+49\text{d}\big]$
$25\big[16+49\text{d}-6-49\text{d}\big]$
$=25\times10$
$=250$
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Question 565 Marks
If the sum of first n term is $(3n^2 + 5n)$, find its common difference.
Answer
Let $S_n$ denotes the sum of first n terms of the AP.
$\therefore$ $S_n = 3n^2 + 5n$
$\Rightarrow S_{n - 1} = 3(n - 1)^2 + 5(n - 1)$
$= 3(n^2 - 2n + 1) + 5(n - 1)$
$= 3n^2 - n - 2$
Now,
$n^{th}$​​​​​​​ term of the $AP, a_n = S_n - S_{n - 1}$
$= (3n^2 + 5n) - (3n^2 - n - 2)$
$= 6n + 2$
Let d be the common difference of the AP.
$\therefore$ $d = a_n - a_{n -1}$
$= (6n + 2) - [6(n - 1) + 2]$
$= 6n + 2 - 6(n - 1) - 2$
$= 6$
Hence, the common difference of the AP is 6.
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Question 575 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.
Answer
The 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) from (ii), we get
$4d = 20$
$\Rightarrow d = 5$
Substituting (i), we get a = -13
So, the sum of the first 10 terms
$=\frac{10}{2}\big[2(-13)+9(5)\big]$
$=2\big[-26 + 45\big]$
$=95$
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Question 585 Marks
An AP 8, 10, 12, ....has 60 terms. Find its last term. Hence, find the sum of its last 10 terms.
Answer
The 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 8, 10, 12, .....
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
$=\frac{60}{2}\big[2(8)+59(2)\big]-\frac{50}{2}\big[2(8)+49(2)\big]$
$=30\big[16+118\big]-25\big[16+98\big]$
$=4020-2850$
$=1170$
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Question 595 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.
Answer
Let 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,
$T_{16}= 45$
$\Rightarrow a + 15d = 45$
$\Rightarrow 5 + 15d = 45$
$\Rightarrow 15d = 40$
$\Rightarrow\text{d}=\frac{40}{15}=\frac{8}{3}$
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Question 605 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.
Answer
Let 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) from (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, ......
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Question 615 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?
Answer
Student 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.
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Question 625 Marks
Find the sum of n terms of the following series:
$\Big(4-\frac{1}{\text{n}}\Big)+\Big(4-\frac{2}{\text{n}}\Big)+\Big(4-\frac{3}{\text{n}}\Big)+\ ...$
Answer
Sum 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)
$=4\text{n}-\frac{1}{\text{n}}\times\frac{\text{n}(\text{n}+1)}{2}$
$=4\text{n}-\frac{(\text{n}+1)}{2}$
$=\frac{[8\text{n}-(\text{n}+1)]}{2}\dots.\text{taking L.C.M}$
$=\frac{(7\text{n}-1)}{2}$
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Question 635 Marks
Find the sum of all multiples of 9 lying between 300 and 700.
Answer
All 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, $S_n = 21978$
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Question 645 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.
Answer
The 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
$\Rightarrow\text{S}_{12}=\frac{12}{2}\big[2(-11+11(3)\big]$
$\Rightarrow S_{12} = 6[-22 + 33]$
$\Rightarrow S_{12} = 6(11)$
$\Rightarrow S_{12} = 66$
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Question 655 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?
Answer
The 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.
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Question 665 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.
Answer
The 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) from (ii), we get
10d = 200
$\Rightarrow d = 20$
Substituting in (i), we get
$2a + 29(20) = 1600$
$\Rightarrow 2a = 1020$
$\Rightarrow a = 510$
Hence, the first installment he paid was Rs. 510.
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Question 675 Marks
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Answer
The 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, $\text{S}_{51}=\frac{51}{2}\big[2(10)+(50)4\big]$
$\Rightarrow\text{S}_{51}=\frac{51}{2}\big[20+200\big]$
$\Rightarrow\text{S}_{51}=\frac{51}{2}\big[220\big]$
$\Rightarrow\text{S}_{51}=51\times110$
$\Rightarrow\text{S}_{51}=5610$
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Question 685 Marks
Find the sum of the first 15 multiples of 8.
Answer
Multiples 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
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Question 695 Marks
The $n^{th}$ term of an AP is (7 - 4n). Find its common difference.
Answer
We have:
$T_n= (7 - 4n)$
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.
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Question 705 Marks
In an AP , the first terms is $22, n ^{\text {th }}$ term is -11 and sum of first n terms is 66 . Find n and hence find the common difference.
Answer
First term of an AP, a = 22
Last term $= n ^{\text {th }}$ 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
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Question 715 Marks
How many three-digit natural numbers are divisible by 9?
Answer
The 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,
$a_n = 999$
$\Rightarrow 108 + (n - 1) \times 9 = 999 [a_n = a + (n - 1)d]$
$\Rightarrow 9n + 99 = 999$
$\Rightarrow 9n = 999 - 99 = 900$
$\Rightarrow n = 100$
Hence, there are 100 three-digit numbers divisible by 9.
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Question 725 Marks
How many three-digit natural numbers are divisible by 7?
Answer
The 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,
$a_n = 994$
$\Rightarrow 105 + (n - 1) \times 7 = 994$
$\Rightarrow 7n + 98 = 994 [a_n = a + (n - 1)d]$
$\Rightarrow 7n = 994 - 98 = 896$
$\Rightarrow n = 128$
Hence, there are 128 three-digit numbers divisible by 7.
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Question 735 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.
Answer
There are 2 AP's with different first term and common difference.
For the first $A P$
Let first term be a common difference be $d$
Sum of n term $= S _{ n }=\frac{ n }{2}(2 a +( n -1) d ) \& n ^{\text {th }}$ term $= a _{ n }= a +( n -1) d$
Similarly for second AP
Let first term $= A$, common difference
$S_n=\frac{n}{2}(2 A+(n-1) D) \& n^{\text {th }} \text { term }=A_n=A+(n-1) D$
We need to find ratio of $12^{\text {th }}$ 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.
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Question 745 Marks
If $a_n$ denotes the $n^{\text {th }}$ term of the AP $2,7,12,17, \ldots$, find the value of $\left(a_{30}-a_{20}\right)$.
Answer
The given AP is 2, 7, 12, 17, ....
Here, a = 2 and d = 7 - 2 = 5
$\therefore$ $a_{30} - a_{20}$
$= [2 + (30 - 1) \times 5] - [2 + (20 - 1) \times 5] [a_n = a + (n - 1)d]$
$= 147 - 97$
$= 50$
Hence, the required value is 50.
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Question 755 Marks
If $\frac{4}{5},\text{a},2$ are in AP, find the value of a.
Answer
If $\frac{4}{5},\text{a}$ and 2 are three consecutive terms of an AP, then we have:
$\text{a}-\frac{4}{5}=2-\text{a}$
$\Rightarrow2\text{a}=2+\frac{4}{5}$
$\Rightarrow2\text{a}=\frac{14}{5}$
$\Rightarrow\text{a}=\frac{7}{5}$
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Question 765 Marks
If the numbers a, 9, b, 25 form an AP, find a and b.
Answer
It is given that the numbers a, 9, 25, form an AP.
$\therefore$ 9 - a = b - 9 = 25 - b
So,
b - 9 = 25 - b
⇒ 2b = 34
⇒ b = 17
Also,
9 - a = b - 9
⇒ a = 18 - b
⇒ a = 18 - 17 (b = 17)
⇒ a = 1
Hence, the required value of a and b are 1 and 17, respectively.
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Question 775 Marks
If k, (2k - 1) and (2k + 1) are the three successive term of an AP, find the value of k.
Answer
It is given that k, (2k - 1) and (2k + 1) are the three successive terms of an AP.
$\therefore$ (2k - 1) - k = (2k + 1) - (2k - 1)
⇒ k - 1 = 2
⇒ k = 3
Hence, the value of k is 3.
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Question 785 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?
Answer
A 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.
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Question 795 Marks
The$ n^{th}$ term of an AP is (3n + 5). Find its common difference.
Answer
We have:
$T_n= (3n + 5)$
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.
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Question 805 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
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Question 815 Marks
If (2p + 1), 13, (5p - 3) are in AP, find the value of p.
Answer
Let (2p + 1), 13, (5p - 3) be three consecutive terms of an AP.
Then 13 - (2p + 1) = (5p - 3) - 13
⇒ 7p = 28
⇒ p = 4
$\therefore$ When p = 4, (2p + 1), 13 and (5p - 3) fprm three consecutive terms of an AP.
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Question 825 Marks
In an AP, it is given that $S_5+S_7=167$ and $S_{10}=235$, then find the $A P$, 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, ....
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Question 835 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.
Answer
Child will put 5 Rs on $1^{\text {st }}$ day, 10 RS ( $2 \times 5$ Rs) on $2^{\text {nd }}$ day, 15 Rs $\left(3 \times 5\right.$ Rs) on $3^{\text {rd }}$ day etc.
Total savings $=190$ coins $=190 \times 5=950$ Rs
The above problem can be written as Arithmetic progression series
$5,10,15,20, \ldots$
$\text { With } a=5, d=5, S_n=950$
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.
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Question 845 Marks
Find the sum of first 100 even natural numbers which are divisible by 5.
Answer
The first few even natural numbers which are divisible by 5 are 10, 20, 30, 40, ...
This is an AP in which a = 10, d = (20 - 10) = 10 and n = 100
The 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{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.
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Question 855 Marks
Write the next term of the $\text{AP}\sqrt{2},\sqrt{8},\sqrt{18},....$
Answer
The 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, $\text{T}_4=\text{a}+3\text{d}=\sqrt{2}+3\sqrt{2}=4\sqrt{2}=\sqrt{32}$
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Question 865 Marks
An AP 5, 12, 19, ....has 50 terms. Find its last term. Hence, find the sum of its last 15 terms.
Answer
The 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
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
$=\frac{50}{2}\big[2(5)+49(7)\big]-\frac{35}{2}\big[2(5)+34(7)\big]$
$=25\big[10+343\big]-\frac{35}{2}\big[10+68\big]$
$=8825-4340$
$=4485$
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Question 875 Marks
Find the sum of first forty positive intergers divisible by 6.
Answer
First forty positive intergers divisible by 6 are as follows:
6, 12, 18, 24, ....240
$\therefore\text{S}_{40}=\frac{40}{2}\big[6+240\big]$
$=20\times246$
$=4920$
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Question 885 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.
Answer
Given 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 $\text{S}_\text{mn}=\frac{\text{mn}}{2}\big[2\text{a}+(\text{mn}-1)\text{d}\big]$
$=\frac{\text{mn}}{2}\Big[\frac{2}{\text{mn}}+\frac{(\text{mn}-1)}{\text{mn}}\Big]$
$=\frac{\text{mn}}{2\text{mn}}\big[2+\text{mn}-1\big]$
$=\frac{1}{2}(\text{mn}+1)$
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