Question 14 Marks
The angles of a quadrilateral are in AP whose common difference is 10º. Find the angles.Hint: Let these angles be xº, (x + 10)º, (x + 20)º and (x + 30)º.
Their sum is 360º.
AnswerLet the required angles be (a - 15)º, (a - 5)º, (a + 5)º, and (a + 15)º as the common difference is 10 (given).
Then (a - 15)º + (a - 5)º + (a + 5)º + (a + 15)º = 360º
⇒ 4a = 360
⇒ a = 90
Hence, the required angles of a quadrilateral are
(90 - 15)º, (90 - 5)º, (90 + 5)º and (90 + 15)º; or 75º, 85º, 95º and 105º.
View full question & answer→Question 24 Marks
If (2p - 1), 7, 3p are in AP, find the value of p.
AnswerLet (2p - 1), 7 and 3p be three consecutive terms of an AP.
Then 7 - (2p - 1) = 3p - 7
⇒ 5p = 15
⇒ p = 3
$\therefore$ When p = 3, (2p - 1), 7 and 3p form three consecutive terms of an AP.
View full question & answer→Question 34 Marks
Find the value of x for which (x + 2), 2x, (2x + 3) are three consecutive terms of an AP.
AnswerSince (x + 2), 2x and (2x + 3) are in AP, we have:
2x - (x + 2) = (2x + 3) - 2x
⇒ x - 2 = 3
⇒ x = 5
$\therefore$ x = 5
View full question & answer→Question 44 Marks
The sum of first $q$ term of an AP is $(63q - 3q^2).$ If its $p^{th}$ term is $-60$, find the value of $p$. Also, find the $11^{th}$ term of its AP.
AnswerLet $S_q$ be the sum of the first $m$ terms of the $A P$.
$S_q=63 q-3 q^2$
$\Rightarrow S_{q-1}=63(q-1)-3(q-1)^2$
$\Rightarrow S_{q-1}=63(q-1)-3(q-1)^2$
$\Rightarrow S_{q-1}=63 q-63-3\left(q^2-2 q+1\right)$
$\Rightarrow S_{q-1}=63 q-63-3 q^2+6 q-3$
$\Rightarrow S_{q-1}=-3 q^2+69 q-66$
Let $a_q$ be the $q^{\text {th }}$ term of the AP.
$\therefore a_q=S_q=S_{q-1}$
$\therefore a_q=\left(63 q-3 q^2\right)-\left(-3 m^2+69 q-66\right)$
$\therefore a_q=63 q-3 q^2+3 q^2-69 q+66$
$\therefore a_q=-6 q+66 \ldots . .(i)$
Given that $a_q=-60$.
$\Rightarrow-6 p+66=-60$
$\Rightarrow-6 p=-126$
$\Rightarrow p=21$
The $11^{\text {th }}$ term $=a_{11}=-6(11)+66=0$.
View full question & answer→Question 54 Marks
Find the value of x for which the numbers (5x + 2), (4x - 1) and (x + 2) are in AP.
AnswerIt is given that (5x + 2), (4x - 1) and (x + 2) are in AP.
$\therefore$ (4x - 1) - (5x + 2) = (x + 2) = (x + 2) - (4x - 1)
⇒ 4x - 1 - 5x - 2 = x + 2 - 4x + 1
⇒ - x - 3 = -3x + 3
⇒ 3x - x = 3 + 3
⇒ 2x = 6
⇒ x = 3
Hence, the value of x is 3.
View full question & answer→Question 64 Marks
Find the sum of all natural numbers between $200$ and $400$ which are divisible by $7$.
AnswerNatural numbers between $200$ and $400$ divisible by $7$ are as follows:
$203, 210, 217, ....399$
Here, $T_n = 399$
$\Rightarrow a + (n - 1)d = 399$
$\Rightarrow 203 + (n - 1)(7) = 399$
$\Rightarrow (n - 1)7 = 196$
$\Rightarrow n - 1 = 28$
$\Rightarrow n = 29$
$\therefore\text{S}_{29}=\frac{29}{2}\big[2\times203+28\times7\big]$
$=\frac{29}{2}\big[406+196]$
$=\frac{29}{2}\times602$
$=8729$
View full question & answer→Question 74 Marks
What is the $5^{th}$ term from the end of the AP $2, 7, 12, ...47?$
AnswerThe given AP is $2, 7, 12, ..., 47.$
Let us re-write the given AP in reverse order i.e. $47, 42, .... 12, 7, 2.$
Now, the $5^{th}$ term from the end of the given AP is equal to the $5^{th}$ term from beginning of the AP $47, 42, ...., 12, 7, 2.$
Consider the AP $47, 42, ....., 12, 7, 2.$
Here, $a = 47$ and $42 - 47 = -5$
$5^{th}$ term of this AP
$= 47 + (5 - 1) × (-5)$
$= 47 - 20$
$= 27$
Hence, the $5^{th}$ term from the end of the given AP is $27$.
View full question & answer→Question 84 Marks
The first and last terms of an AP are a and l respectively. Show that the sum of the $n^{th}$ term from the beginning and the $n^{th}$ term from the end is $(a + l)$.
AnswerLet a be the first term and d be the common difference
$\therefore$ $n^{th}$ term ftom the beginning $= a + (n - 1)d ....(1)$
$n^{th}$ term from end $= l - (n - 1)d ....(2)$
Addind (1) and (2),
Sum of the $n^{th}$ term from the beginning and $n^{th}$ term from the end $= [a + (n - 1)d] + [l - (n - 1)d] = a + l$
View full question & answer→Question 94 Marks
There are $25$ trees at equal distances of $5m$ in a line with a water tank, the distance of the water tank from the nearest tree being $10m$ A gardener waters all the trees separately, starting from the water tank and returning back to the water tank after watering each tree to get water for the next.
Find the total distance covered by the gardener in order to water all the trees.
AnswerAssume that gardener is standing near the well initially and he did not return to the well after watering the last tree.
Distance covered by gardener to water $1^{st}$ tree and return to the initial position
$= 10m + 10m = 20m$
Distance covered by gardener to water $2^{nd}$ tree and return to the initial position
$= 15m + 15m = 30m$
Distance covered by gardener to water $3^{rd}$ tree and return to the initial position
$= 20m + 20m = 40m$
$\therefore$ Distances covered by the gardener to water the plants are in AP.
Here $a = 20, d = 10$
Total distance covered the gardener is given by Sn, where $n = 25.$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\Rightarrow\text{S}_\text{25}=\frac{\text{25}}{2}\big[2(20)+(25-1)10\big]=3500$
Thus, the total distance covered by the gardener is $3500m$.
View full question & answer→Question 104 Marks
Determine k so that (3k - 2), (4k - 6) and (k + 2) are three consecutive terms of an AP.
AnswerIt is given that (3k - 2), (4k - 6) and (k + 2) are three consecutive terms of an AP.
$\therefore$ (4k - 6) - (3k - 2) = (k + 2) - (4k - 6)
⇒ 4k - 6 - 3k + 2 = k + 2 - 4k + 6
⇒ k - 4 = -3k + 8
⇒ k + 3k = 8 + 4
⇒ 4k = 12
⇒ k = 3
Hence, the value of k is 3.
View full question & answer→Question 114 Marks
The sum of three consecutive terms of an AP is $21$ and the sum of the squares of these terms is $165$. Find these terms.
AnswerLet the required terms be $(a - d), a$ and $(a + d)$.
Then $(a - d) + a + (a + d) = 21$
$\Rightarrow 3a = 21$
$\Rightarrow a = 7$
Also, $(a - d)^2 + a^2 + (a + d)^2 = 165$
$\Rightarrow 3a^2 + 2d^2 = 165$
$\Rightarrow (3 \times 49 + 2d^2) = 165$
$\Rightarrow 2d^2 = 165 - 147 = 18$
$\Rightarrow d^2 = 9$
$\Rightarrow\text{d}=\pm3$
Thus, $a = 7$ and $\text{d}=\pm3$
Hence, the required numbers are $(4, 7, 10)$ or $(10, 7, 4).$
View full question & answer→Question 124 Marks
How many three-digit numbers are divisible by $9$?
AnswerThe two-digit numbers divisible by $9$ start from
$108, 117, 126, 135, ..., 999$
Here,
$a = 108$
$d = 9$
$a_n = a + (n - 1)d$
$\Rightarrow 999 = 108 + (n - 1)(9)$
$\Rightarrow 999 = 108 + 9n - 9$
$\Rightarrow 900 = 9n$
$\Rightarrow n = 100$
This, $100$ two-digit number are divisible by $9$.
View full question & answer→Question 134 Marks
Divide $24$ in three parts such that they are in AP and their product is $440$.
AnswerLet the required parts of $24$ be $(a - d), a$ and $(a + d)$ such that they are in AP.
Then $(a - d) + a + (a + d) = 24$
$\Rightarrow 3a = 24$
$\Rightarrow a = 8$
Also, $(a - d) \times a \times (a + d) = 440$
$\Rightarrow a(a^2 - d^2) = 440$
$\Rightarrow 8(64 - d^2) = 440$
$\Rightarrow d^2 = 64 - 55 = 9$
$\Rightarrow\text{d}=\pm3$
Thus, $a = 8$ and $\text{d}=\pm3$
Hence, the required parts of $24$ are $(5 , 8, 11)$ or $(11, 8, 5)$.
View full question & answer→Question 144 Marks
Find the sum of all three-digit natural numbers which are divisible by $13$.
AnswerAll three-digit numbers which are divisible by $13$ are $104, 117, 130, 143, .…… 938$.
This is an AP in which $a = 104, d = (117 – 104) = 13$ and $l = 938$
Let the number of terms be n
Then $T_n = 938$
$\Rightarrow a + (n - 1)d = 988$
$\Rightarrow 104 + (n - 1) \times 13 = 988$
$\Rightarrow 13n = 897$
$\Rightarrow n = 69$
$\therefore\text{Required sum}=\frac{\text{n}}{2}(\text{a}+\text{l})$
$=\frac{69}{2}\big[104+988\big]=69\times546=37674 $
Hence, the required sum is $37674$.
View full question & answer→Question 154 Marks
Find three numbers in AP whose sum is $15$ and product is $80$.
Hint: Let the numbers be $(a - d), a, (a + d).$
AnswerLet the required numbers be $(a - d)$, $a$ and $(a + d).$
Then $(a - d) + a + (a + d) = 15$
$\Rightarrow 3a = 15$
$\Rightarrow a = 5$
Also, $(a - d) \times a \times (a + d) = 80$
$\Rightarrow a(a^2 - d^2) = 80$
$\Rightarrow 5(25 - d^2) = 80$
$\Rightarrow d^2 = 25 - 16 = 9$
$\Rightarrow\text{d}=\pm3$
Thus, a = 5 and $\text{d}=\pm3$
Hence, the required numbers are $(2, 5$ and $8)$ or$ (8, 5$ and $2)$.
View full question & answer→Question 164 Marks
Find how many intergers between $200$ and $500$ are divisible by $8$.
AnswerThe first term between $200$ and $500$ divisible by $8$ is $208$, and the last term is $496$.
So, first term $(a) = 208$
Common difference $(d) = 8$
$a_n= a + (n - 1)d = 496$
$\Rightarrow 208 + (n - 1)8 = 496$
$\Rightarrow (n - 1)8 = 288$
$\Rightarrow n - 1 = 36 \Rightarrow n = 37$
Hence, there are $37$ integers between $200$ and $500$ which are divisible by $8$.
View full question & answer→Question 174 Marks
The first three terms of an AP are respectively (3y - 1), (3y + 5) and (5y + 1), find the value of y.
AnswerThe terms (3y - 1), (3y + 5) and (5y + 1) are in AP.
$\therefore$ (3y + 5) - (3y - 1) = (5y + 1) - (3y + 5)
⇒ 3y + 5 - 3y + 1 = 5y + 1 - 3y - 5
⇒ 6 = 2y - 4
⇒ 2y = 10
⇒ y = 5
Hence, the value of y is 5.
View full question & answer→Question 184 Marks
Find the sum of first n natural numbers.
AnswerThe first n natural numbers are 1, 2, 3, 4, 5, ..., n.
Here, a = 1 and d = (2 - 1) = 1
Sum of n terms of an AP is given by
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$=\big(\frac{\text{n}}{2}\big)\times\big[2\times1+(\text{n}-1)\times1\big]$
$=\big(\frac{\text{n}}{2}\big)\times\big[2+\text{n}-1\big]=\big(\frac{\text{n}}{2}\big)\times(\text{n}+1)=\frac{\text{n}(\text{n}+1)}{2}$
View full question & answer→Question 194 Marks
The sum of the first 9 terms of an AP is 81 and that of its first 20 terms is 400. Find the first and the common difference of the AP.
AnswerLet a be the first term and d be the common difference of the given A.P.
Then,
$\text{S}_9=81$
$\Rightarrow\frac{9}{2}\big[2\text{a}+8\text{d}\big]=81$
$\Rightarrow\frac{9\times2}{2}\big[\text{a}+4\text{d}\big]=81$
$\Rightarrow\text{a}+4\text{d}=9\dots(\text{i})$
Also, $\text{S}_{20}=400$
$\Rightarrow\frac{20}{2}\big[2\text{a}+19\text{d}\big]=400$
$\Rightarrow10\big[2\text{a}+19\text{d}\big]=400$
$\Rightarrow2\text{a}+19\text{d}=40\dots(\text{ii})$
Multiplying equation (i) by 2, we get
$2\text{a}+8\text{d}=18\dots(\text{iii})$
Subtracting (iii) from (ii), we get
$11\text{d}=22$
$\Rightarrow\text{d}=2$
$\Rightarrow\text{a}=9-4(2)=9-8=1$
Thus, the first term is 1 and the common difference is 2.
View full question & answer→Question 204 Marks
Find the sum of the following arithmetic series:
$34 + 32 + 30 + .... + 10.$
Answerwe havea $= 34,$
$d = 32 - 34 = -2$
Let the total number of terms be n.
Then $T_n = 10$
$\Rightarrow \text{a} + (\text{n} - 1)\text{d} = 10$
$\Rightarrow34+(\text{n}-1)(-2)=10$
$\Rightarrow(\text{n}-1)(-2)=-24$
$\Rightarrow\text{n}-1=\frac{-24}{-2}$
$\Rightarrow\text{n}-1=12$
$\Rightarrow\text{n}=13$
Now, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[\text{a}+\text{l}\big]$
$\therefore\text{S}_\text{13}=\frac{\text{13}}{2}\big[\text{34}+10\big]$
$=\frac{13}{2}\times44=286$
View full question & answer→Question 214 Marks
In a flower bed, there are $43$ rose plants in the first row, $41$ in the second, $39$ in the third, and so on. There are $11$ rose plants in the last row. How many rows are there flower bed?
AnswerNumber of rose plants in first, second, third rows ....are $43, 41, 39 ....$ respectively.
There are $11$ rose plants in the last row
So, it is an AP. viz. $43, 41, 39 ....11$
$a = 43, d = 41 - 43 = -2, l = 11$
Let $n^{th}$ term be the last term
$\therefore$ $l = a + (n - 1)d$
$\Rightarrow 11 = 43 + (n - 1) \times (n - 1) \times (-2)$
$43 - 2n + 2 = 11 or 2n = 45 - 11 = 34$
$\therefore\text{n}=\frac{34}{2}=17$
Hence, there are $17$ rows in the flower bed.
View full question & answer→Question 224 Marks
If the numbers (2n - 1), (3n + 2) and (6n - 1) are in AP, find the value of n and the numbers.
AnswerIt is given that the numbers (2n - 1), (3n + 2) and (6n - 1) are in AP.
$\therefore$ (3n + 2) - (2n - 1) = (6n - 1) - (3n + 2)
⇒ 3n + 2 - 2n + 1 = 6n - 1 - 3n - 2
⇒ n + 3 = 3n - 3
⇒ 2n = 6
⇒ n = 3
When n = 3
2n - 1 = 2 × 3 - 1 = 6 - 1 = 5
3n + 2 = 3 × 3 + 2 = 9 + 2 = 11
6n - 1 = 6 × 3 - 1 = 18 - 1 = 17
Hence, the required value of n is 3 and the numbers are 5, 11 and 17.
View full question & answer→Question 234 Marks
Find an AP whose $4^{\text {th }}$ term is $9$ and the sum of its $6^{\text {th }}$ and $13^{\text {th }}$ terms is $40$ .
AnswerLet $a$ be the term and $d$ be the common difference of the AP. Then,
$a_4=9$
$\Rightarrow a+(4-1) d=9\left[a_n=a+(n-1) d\right]$
$\Rightarrow a+3 d=9 \ldots . .(1)$
Now,
$a_6+a_{13}=40 \text { (given) }$
$\Rightarrow(a+5 d)+(a+12 d)=40$
$\Rightarrow 2 a+17 d=40 \ldots . .(2)$
From (1) and (2), we get
$2(9-3 d)+17 d=40$
$\Rightarrow 18-6 d+17 d=40$
$\Rightarrow 11 d=40-18=22$
$\Rightarrow d=2$
Putting $d=2$ in (1), we get
$a+3 \times 2=9$
$\Rightarrow a=9-6=3$
Hence, the AP is $3,5,9,11, \ldots .$.
View full question & answer→Question 244 Marks
What is the common difference of an AP in which $a_{27} - a_7 = 84$?
AnswerLet 'a' is the first term and d is the common difference of the AP
Given:
$a_{27} - a_7 = 84$
$a_n= a + (n - 1)d$
$a_{27}= a + (27 - 1)d$
$a_7= a + (7 - 1)d$
$a_{27 -} a_7= 84$
$a + 26d - (a + 6d) = 84$
$a + 26d - a - 6d = 84$
$a - a + 26d - 6d = 84$
$26d - 6d = 84$
$20d = 84$
$\text{d} = \frac{84}{20} = \frac{21}{5}$
$d = 4.2$
View full question & answer→Question 254 Marks
Find four numbers in AP whose sum is $28$ and the sum of whose squares is $216$.
Answer
- Their sum
$(a - 3d) + (a - d) + (a + d) + (a + 3d) = 28$
$4a = 28$
$a = 7.$
- Their product
$(a - 3d)^2 + (a - d)^2 + (a + d)^2 + (a + 3d)^2 = 216$
$4a^2+ 20d^2 = 216$
$a^2 + 5d^2 = 54$
Putting the value of a
$49 + 5d^2 = 54$
$5d^2 = 5$
$d^2 = 1$
$\text{d}=\pm1.$
Hence, the numbers are $4, 6, 8$ & $10$. View full question & answer→Question 264 Marks
The sum of the first $n$ terms of an AP is $\left(3 n^2+6 n\right)$. Find the $n^{\text {th }}$ term and the $15^{\text {th }}$ term of this AP.
Answer$\text { we have, } S_n=3 n^2+6 n$
$\Rightarrow S_{n-1}=3(n-1)^2+6(n-1)$
$=3\left(n^2-2 n+1\right)+6 n-6$
$\Rightarrow 3 n^2-6 n+3+6 n-6$
$=3 n^2-3$
Now, $n ^{\text {th }}$ term, $T _{ n }= S _{ n }- S _{ n -1}$
$=3 n^2+6 n-3 n^2+3=6 n+3$
And, $15^{\text {th }}$ term $= T _{15}=6(15)+3$
$=90+3=93$
View full question & answer→Question 274 Marks
Is $-150$ a term of the AP $11, 8, 5, 2, ....?$
AnswerThe given AP is $11, 8, 5, 2, ...,$
Common difference $= 8 - 11 = -3$
The given term is $-150$.
The general term of an AP is given by
$a_n = a + (n - 1)d$
$\Rightarrow -150 = a + (n - 1)d$
$\Rightarrow -150 = 11 + (n - 1)(-3)$
$\Rightarrow -150 = 11 - 3n + 3$
$\Rightarrow 3n - 3 = 161$
$\Rightarrow 3n = 164$
$\Rightarrow\text{n}=\frac{164}{3}$
The number of terms cannot be a fraction.
So clearly, $-150$ is not a term of the AP.
View full question & answer→Question 284 Marks
The first and the last terms of an AP are $17$ and $350$ respectively. If the common difference is $9$, how many terms are there and what is their sum?
AnswerLet theire be n terms in the given A.P.
First terms, $a = 17$
Last term, $l = 350$
Common defference, $d = 9$
Now, $T_n = 350$
$\Rightarrow a + (n - 1)d = 350$
$\Rightarrow 17 + (n -1)9 = 350$
$\Rightarrow (n - 1)9 = 333$
$\Rightarrow n - 1 = 37$
$\Rightarrow n = 38$
Now, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[\text{a}+\text{l}\big]$
$\Rightarrow\text{S}_\text{38}=\frac{\text{38}}{2}\big[\text{17}+\text{350}\big]=19\times367=6973$
View full question & answer→Question 294 Marks
Sum of the first $14$ terms of an AP is $1505$ and its first term is $10$ . Find its $25^{\text {th }}$ term.
Answer$\text{S}_{14}=1505$ (Given)
$\Rightarrow\frac{14}{2}\big[2\times10+(14-\text{l})\times\text{d}\big]=1505$
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\Rightarrow 7(20 + 13d) = 1505$
$\Rightarrow 20 + 13d = 215$
$\Rightarrow 13d = 215 - 20 = 195$
$\Rightarrow d = 15$
$\therefore$ $25^{th}$ term of the AP, $a_{25}$
$= 10 + (25 - l) \times 15$
$[a_n = a + (n - l)d]$
$= 10 + 360$
$= 370$
Hence the required term is $370$.
View full question & answer→Question 304 Marks
The sum of first three terms of an AP is 48. If the product of first and second terms exceeds 4 times the third term by 12. Find the AP.
Hint: Let these terms be (a - b), a, (a + d).
AnswerLet the first three terms of the AP be (a - d), a and (a + d),
Then,
(a - d) + a + (a + d) = 48
⇒ 3a = 48
⇒ a = 16
Now,
(a - d) × a = 4(a + d) + 12
⇒ (16 - d) × 16 = 4 (16 + d) + 12
⇒ 256 - 16d = 64 + 4d + 12
⇒ 16d + 4d = 256 - 76
⇒ 20d = 180
⇒ d = 9
When a = 16 and d = 9,
a - d = 16 - 9 = 7
a + d = 16 + 9 = 25
Hence, the first three terms of the AP are 7, 16 and 25.
View full question & answer→Question 314 Marks
The first term of an AP is $p$ and its common difference is $q$. Find its $10^{th}$ term.
AnswerHere, $a = p$ and $d = q$
Now, $T_n = a + (n - 1)d$
$\Rightarrow T_n = p + (n - 1)q$
$\therefore$ $T_{10}= p + 9q$
View full question & answer→Question 324 Marks
Divide 32 into four parts which are the four terms of an AP such that the product of the first and the fourth terms is to the product of the second and third terms as 7 : 15
Hint: Let these parts be (a - 3d), (a - d), (a + d) and (a + 3d).
AnswerLet the four parts in AP be (a - 3d), (a - d), (a + d) and (a + 3d).
Then,
(a - 3d) + (a - d) + (a + d) + (a + 3d) = 32
⇒ 4a = 32
⇒ a = 8 ....(1)
Also,
(a - 3d)(a + 3d) : (a - d)(a + d) = 7 : 15
$\Rightarrow\frac{(8-3\text{d})(8+3\text{d})}{(8-\text{d})(8+\text{d})}=\frac{7}{15}$ [From (1)]
$\Rightarrow\frac{64-9\text{d}^2}{64-\text{d}^2}=\frac{7}{15}$
$\Rightarrow15(64-9\text{d}^2)=7(64-\text{d}^2)$
$\Rightarrow960-135\text{d}^2=448-7\text{d}^2$
$\Rightarrow135\text{d}^2-7\text{d}^2=960-448$
$\Rightarrow128\text{d}^2=512$
$\Rightarrow\text{d}^2=4$
$\Rightarrow\text{d}=\pm2$
When a = 8 and d = 2,
$\text{a}-3\text{d}=8-3\times2=8-6=2$
$\text{a}-\text{d}=8-2=6$
$\text{a}+\text{d}=8+2=10$
$\text{a}+3\text{d}=8+3\times2=8+6=14$
When a = 8 and d = -2,
$\text{a}-3\text{d}=8-3\times(-2)=8+6=14$
$\text{a}-\text{d}=8-(-2)=8+2=10$
$\text{a}+\text{d}=8-2=6$
$\text{a}+3\text{d}=8+3\times(-2)=8-6=2$
Hence, the four parts are 2, 6, 10 and 14.
View full question & answer→Question 334 Marks
What is the sum of first n terms of the AP $a, 3a, 5a, ....$
AnswerThe given AP is $a, 3a, 5a, ...$
Here
First term, $A = a$
Common difference, $D = 3a - a = 2a$
$\therefore$ sum of first $n$ terms, $S_n$
$=\frac{\text{n}}{2}\big[2\times\text{a} + (\text{n}-1)\times2\text{a}\big]$ $\Big\{\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{A} + (\text{n}-1)\text{D}\big]\Big\}$
$=\frac{\text{n}}{2}(2\text{a}+2\text{an}-2\text{a})$
$=\frac{\text{n}}{2}\times2\text{an}$
$=\text{an}^2$
Hence, the required sum is $an^2$.
View full question & answer→Question 344 Marks
If $10$ times the $10^{th}$ term of an AP is equal to $15$ times the $15^{th}$ term, show that its $25^{th}$ term is zero.
AnswerLet a be the first term and d be the common difference
$T_{10} = a + 9d, T_{15}= a + 14d$
$10T_{10} = 15T_{15}$
$\Rightarrow 10 (a + 9d) = 15(a + 14d)$
or $2(a + 9d) = 3(a + 14d)$
$\Rightarrow a + 24d = 0$
$\therefore$ $T_{25} = 0$
View full question & answer→Question 354 Marks
The sum of three numbers in AP is $3$ and their product is $-35$. Find the numbers.
AnswerLet the required numbers be $(a - d), a$ and $(a + d)$.
Then $(a - d) + a + (a + d) = 3$
$\Rightarrow 3a = 3$
$\Rightarrow a = 1$
Also, $(a - d) × a × (a + d) = -35$
$\Rightarrow a(a^2 - d^2) = -35$
$\Rightarrow 1 \times (1 - d^2) = -35$
$\Rightarrow d^2 = 36$
$\Rightarrow\text{d}=\pm6$
Thus, $a = 1$ and $\text{d}=\pm6$
Hence, the required numbers are $(-5, 1$ and $7)$ or $(7, 1$ and $-5).$
View full question & answer→Question 364 Marks
Which term of the AP $21, 18, 15, ....$ is zero?
AnswerIn the given AP, first term, $a = 21$ and common difference, $d = (18 - 21) = -3$
Let's its $n^{th}$ term be $0$.
Then $T_n = 0$
$\Rightarrow a + (n - 1)d = 0$
$\Rightarrow 21 + (n - 1) \times (-3) = 0$
$\Rightarrow 24 - 3n = 0$
$\Rightarrow 3n = 24$
$\Rightarrow n = 8$
Hence, the $8^{th}$ term of the given AP is $0$.
View full question & answer→Question 374 Marks
Write the next term of the $\text{AP}\sqrt{8},\sqrt{18},\sqrt{32},....$
AnswerThe given AP is $\sqrt{8},\sqrt{18},\sqrt{32},....$
On simplifying the terms, we get :
$2\sqrt{2},3\sqrt{2},4\sqrt{2},....$
Here, $\text{a}=2\sqrt{2}$ and $\text{d}=(3\sqrt{2}-2\sqrt{2})=\sqrt{2}$
$\therefore$ Next term, $\text{T}_4=\text{a}+3\text{d}=2\sqrt{2}+3\sqrt{2}=5\sqrt{2}=\sqrt{50}$
View full question & answer→Question 384 Marks
The sum of first m term of an AP is $(4m^2 - m)$. If its $n^{th}$ term is $107$, find the value of n. Also, find the $21^{st}$ term of this AP.
AnswerLet be the sum of the first m terms of the AP.
$S_m = 4m^2_- m$
$\Rightarrow S_{m - 1} = 4(m - 1)^2 - (m - 1)$
$\Rightarrow S_{m - 1} = 4(m^2 - 2m + 1) - m + 1$
$\Rightarrow S_{m - 1} = 4m^2 - 8m + 4 - m + 1$
$\Rightarrow S_{m - 1}= 4m^2 - 9m + 5$
Let $a_m$ be the $m^{th}$ term of the AP.
$\therefore$ $a_m = S_m = S_{m-1}$
$\therefore$ $a_m = (4m^2 - m) - (4m^2 - 9m + 5)$
$\therefore$ $a_m = 4m^2 - m - 4m^2+ 9m - 5$
$\therefore$ $a_m = 8m - 5 ....(i)$
Given that $a_n = 107.$
$\Rightarrow 8n - 5 = 107$
$\Rightarrow 8n = 112$
$\Rightarrow n = 14$
The $21^{st}$ term $= a_{21} = 8(21) - 5 = 163$
View full question & answer→Question 394 Marks
A sum of $₹ 700$ is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is $₹ 20$ less than its precending prize, find the value of each prize.
AnswerLet the $1^{st}$ prize be a
So the $2^{nd}$ prize $= a - 20$
The $3^{rd}$ prize $= (a - 20 - 20 = a - 40$
And so on
So, the series will be
$a, a - 20, a - 40, ....$
Since difference is same, it is an AP
Here, sum of $Rs 700$ is given,
So, $S_n= 700$
Also,
$a = a$
$d = (a - 20) - a = a - 20 - a = -20$
We know that
$\text{S}_\text{n}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})$
Putting the values
$\text{S}_\text{n}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})$
$700=\frac{7}{2}(2\text{a}+(7-1)\times-20)$
$700=\frac{7}{2}(2\text{a}+6\times-20)$
$700=\frac{7}{2}(2\text{a}-120)$
$700\times\frac{7}{2}=2\text{a}-120$
$200=2\text{a}-120$
$200+120=2\text{a}$
$320=2\text{a}$
$\frac{320}{2}=\text{a}$
$160=\text{a}$
$\text{a}=160$
So,
$1^{st}$ prize $= a = 160$
$2^{nd}$ prize $= a - 20 = 160 - 20 = 140$
$3^{rd}$ prieze $= a - 40 = 160 - 40 = 120$
$4^{th}$ prize $= a - 60 = 160 - 60 =100$
$5^{th}$ prize $= a - 80 = 160 - 80 = 80$
$6^{th}$ prize $= a - 100 = 160 - 100 = 60$
$7^{th}$ prize $= a - 120 = 160 - 120 = 40$
View full question & answer→Question 404 Marks
Show that $(a - b)^2, (a^2 + b^2)$ and $(a + b)^2$ are in AP.
AnswerThe given number are $(a - b)^2, (a^2 + b^2)$ and $(a + b)^2$.
Now,
$(a^2 + b^2) - (a - b)^2 = a^2 + b^2 - (a^2 - 2ab + b^2) = a^2 + b^2 - a^2 + 2ab - b^2 = 2ab$
$(a + b)^2 - (a^2+ b^2) = a^2 + 2ab + b^2 - a^2 - b^2 = 2ab$
So, $(a^2 + b^2) - (a - b)^2 = (a + b)^2 - (a^2 + b^2) = 2ab$ (Constant)
Since each term differs from its preceding term by a constant, therefore, the given numbers are in AP.
View full question & answer→Question 414 Marks
How many terms of the AP 21, 18, 15, ..... must be added to get the sum 0?
AnswerHere a = 21, d = (18 - 21) = -3
Let the required number of terms be n, then
$\text{S}_\text{n}=0\Rightarrow\frac{\text{}\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]=0$
$\Rightarrow\frac{\text{n}}{2}\big[2\times21+(\text{n}-1)(-3)\big]=0$
$\Rightarrow\frac{\text{n}}{2}(45-3\text{n})=0$
$\Rightarrow\text{n}(45-3\text{n})=0$
$\Rightarrow45-3\text{n}=0\Rightarrow3\text{n}=45$
$\Rightarrow\text{n}=15$
$\therefore$ Sum of first 15 terms = 0
View full question & answer→Question 424 Marks
The sum of the first n terms of an AP is $\Big(\frac{3\text{n}^2}{2}+\frac{5\text{n}}{2}\Big).$ Find its $n^{th}$ term and the $25^{th}$ term.
Answerwe have, $\text{S}_\text{n}=\frac{3\text{n}^2}{2}+\frac{5\text{n}}{2}=\frac{3\text{n}^2+5\text{n}}{2}$
$\Rightarrow\text{S}_{\text{n}-1}=\frac{3(\text{n}-1)^2+5(\text{n}-1)}{2}$
$=\frac{3(\text{n}^2-2\text{n}+1)+5\text{n}-5}{2}$
$=\frac{3\text{n}^2-6\text{n}+3+5\text{n}-5}{2}$
$=\frac{3\text{n}^2-\text{n}-2}{2}$
Now, $n^{th}$ term
$ \text{T}_\text{n} = \text{S}_\text{n} - \text{S}_{\text{n} - 1}$
$=\frac{3\text{n}^2+5\text{n}}{2} - \frac{3\text{n}^2-\text{n}-2}{2}$
$=\frac{3\text{n}^2+5\text{n}-3\text{n}^2+\text{n}+2}{2}$
$=\frac{6\text{n}+2}{2}=3\text{n}+1$
$25^{th}$ term $= T_{25} = 3(25) + 1 = 75 + 1 = 76$
View full question & answer→Question 434 Marks
The sum of the first $7$ terms of an AP is 182. If its $4^{th}$ and $17^{th}$ terms are in the ratio $1 : 5$, find the AP.
AnswerLet a be the first term and d be the common difference of the given A.P.
Then,
$T_4 = a + 3d$ and $T_{17} = a + 16d$
Now, $\frac{\text{T}_4}{\text{T}_{17}}=\frac{1}{5}$
$\Rightarrow\frac{\text{a}+3\text{d}}{\text{a}+16\text{d}}=\frac{1}{5}$
$\Rightarrow5\text{a}+15\text{d}=\text{a}+16\text{d}$
$\Rightarrow4\text{a}-\text{d}=0$
$\Rightarrow4\text{a}=\text{d}\dots(\text{i})$
Also, $\text{S}_7=182$
$\Rightarrow\frac{7}{2}\big[2\text{a}+6\text{d}\big]=182$
$\Rightarrow\frac{7\times2}{2}\big[\text{a}+3\text{d}\big]=182$
$\Rightarrow\text{a}+3\text{d}=26$
$\Rightarrow\text{a}+3\text{(4a})=26\dots\big[\text{From(i)}\big]$
$\Rightarrow13\text{a}=26$
$\Rightarrow\text{a}=2$
$\Rightarrow\text{d}=4(2)=8$
Thus, we have
$T_1 = 2$
$T_2= T_1 + d = 2 + 8 = 10$
$T_3= T_1 + 2d = 2 + 2(8) = 2 + 16 = 18$
$T_4 = T_1 + 3d = 2 + 3(8) = 2 + 24 = 26$
Thus, the given A.P. is $2, 10, 18, 26, .....$
View full question & answer→Question 444 Marks
In an AP, the first term is 2, the last term is 29 and the sum of all the terms is 155. Find the common difference.
AnswerFirst term 'a' of an AP = 2
The last term l = 29
Sum of n term $\text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}+\text{l})=155$
$\therefore\frac{\text{n}}{2}(\text{2}+29)=155$
$\text{n}=\frac{155\times2}{31}=10$
Also, l = a + (n - 1)d
or 29 = 2 + (10 - 2)d = 2 + 9d
⇒ 9d = 29 - 2 = 27 $\therefore\text{d}=\frac{27}{9}=3$
$\therefore$ common difference = 3
View full question & answer→Question 454 Marks
If (3y - 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value of y.
AnswerIt is given that (3y - 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP.
$\therefore$ (3y + 5) - (3y - 1) = (5y + 1) - (3y + 5)
⇒ 3y + 5 - 3y + 1 = 5y + 1 - 3y - 5
⇒ 6 = 2y - 4
⇒ 2y = 6 + 4 = 10
⇒ y = 5
Hence, the value of y is 5.
View full question & answer→Question 464 Marks
How many numbers are there between $101$ and $999$, which are divisible by both $2$ and $5$?
AnswerFor the number to be divisible by both 2 and 5, they have to be divisible by the LCM of $2$ and $5 = 10$.
The numbers divisible by 10 between 101 and 999
Are $110, 120, 130, ...., 990$
Here
$a = 110$
$d = 10$
$a_n = a + (n - 1)d$
$\Rightarrow 990 = 110 + (n - 1)(10)$
$\Rightarrow 990 = 110 + 10n - 10$
$\Rightarrow 890 = 10n$
$\Rightarrow n = 89$
Thus, $89$ numbers between $101$ and $999$ are divisible by both $2$ and $5$.
View full question & answer→Question 474 Marks
How many two-digit numbers are divisible by $6?$
AnswerThe two-digit numbers divisible by $6$ start from
$12, 18, 24, ..., 96$
Here,
$a = 12$
$d = 6$
$a_n = a + (n - 1)d$
$\Rightarrow 96 = 12 + (n - 1)(6)$
$\Rightarrow 96 = 12 + 6n - 6$
$\Rightarrow 90 = 6n$
$\Rightarrow n = 15$
This, $15$ two-digit number are divisible by $6$.
View full question & answer→