Question 14 Marks
How many terms of the series $18 + 15 + 12 + ……..$ when added together will give $45?$
AnswerHere, we find that
$15 - 18 = 12 - 15 = -3$
Thus, the given series is an A.P. with first term 18 and common difference $-3.$
Let the number of term to be added be 'n'.
$S_n=\frac{n}{2}(2 a+(n-1) d)$
$\Rightarrow 45=\frac{n}{2}(2(18)+(n-1)(-3))$
$\Rightarrow 90 = n(36 - 3n + 3)$
$\Rightarrow 90 = n(39 - 3n)$
$\Rightarrow 90 = 3n(13 - n)$
$\Rightarrow 30 = 13n - n^2$
$\Rightarrow n^2 - 13n + 30 = 0$
$\Rightarrow n^2 - 10n - 3n + 30 = 0$
$\Rightarrow n(n - 10) - 3(n - 10) = 0$
$\Rightarrow (n - 10)(n - 3) = 0$
$\Rightarrow n - 10 = 0$ or $n - 3 = 0$
$\Rightarrow n = 10$ or $n = 3$
Thus, required number of term to be added is $3$ or $10.$
View full question & answer→Question 24 Marks
The first term of an A.P. is $20$ and the sum of its first seven terms is $2100;$ find the $31^{st}$ term of this A.P.
AnswerHere $a = 20$ and $S_7 = 2100$
We know that,
$S_n=\frac{n}{2}(2 a+(n-1) d)$
$\Rightarrow S_7=\frac{7}{2}(2 \times 20+(7-1) d)$
$\Rightarrow 2100=\frac{7}{2}(40+6 d)$
$\Rightarrow 4200=7(40+6 d)$
$\Rightarrow 600=40+6 d$
$\Rightarrow d=\frac{560}{6}$
To find: $t_{31} =?$
$t_n = a + (n - 1)d$
$\Rightarrow t_{31}=20+(31-1) \frac{560}{6}$
$=20+30 \times \frac{560}{6}$
$=20+5 \times x 50$
$=2820$
Therefore the $31^{st}$ term of the given A.P. is $2820.$
View full question & answer→Question 34 Marks
Find the sum of the A.P., $14, 21, 28, …, 168.$
AnswerHere $a = 14 , d = 7$ and $t_n = 168$
$t_n = a + (n - 1)d$
$\Rightarrow 168 = 14 + (n - 1)7$
$\Rightarrow 154 = 7n - 7$
$\Rightarrow 154 = 7n - 7$
$\Rightarrow 161 = 7n$
$\Rightarrow n = 23$
We know that,
$S_n=\frac{n}{2}(2 a+(n-1) d)$
$\Rightarrow S_{23}=\frac{23}{2}(2 \times 14+(23-1) 7)$
$=\frac{23}{2}(28+154)$
$=\frac{23}{2} \times 182$
$=2093$
Therefore the sum of the A.P., $14, 21, 28, …, 168$ is $2093.$
View full question & answer→Question 44 Marks
For an A.P., show that:
$(m + n)^{th} ~term + (m - n)^{th} ~term = 2 \times m^{th} ~term$
AnswerLet a and d be the first term and common difference respectively.
$\Rightarrow (m + n)^{th} term = a + (m + n - 1)d$ …. (i) and
$(m - n)^{th} term = a + (m - n - 1)d$ …. (ii)
From (i) + (ii), we get
$(m + n)^{th} term + (m - n)^{th} term$
$= a + (m + n - 1)d + a + (m - n - 1)d$
$= a + md + nd - d + a + md - nd - d$
$= 2a + 2md - 2d$
$= 2a + (m - 1)2d$
$= 2[ a + (m - 1)d]$
$= 2 \times m^{th}term$
Hence proved.
View full question & answer→Question 54 Marks
The sum of first $14$ terms of an A.P. is $1050$ and its $14^{th}$ term is $140$. Find the $20^{th}$ term.
AnswerLet 'a' be the first term and 'd' be the common difference of the given A.P.
Given,
$S_{14} = 1050$
$\Rightarrow \frac{14}{2}(2 a+(14-1) d)=1050$
$\Rightarrow 7(2a + 13d) = 1050$
$\Rightarrow 2a + 13d = 150$
$\Rightarrow a + 6.5d = 75$ ….(i)
And, $t_{14} = 140$
$\Rightarrow a + 13d = 140$ ….(ii)
Subtracting (i) from (ii), we get
$6.5d = 65$
$\Rightarrow d = 10$
$\Rightarrow a + 13(10) = 140$
$\Rightarrow a = 10$
Thus, $20^{th}$ term $= t_{20} = 10 + 19d = 10 + 19(10) = 200$
View full question & answer→Question 64 Marks
The sum of the $4^{th}$ and the $8^{th}$ terms of an A.P. is $24$ and the sum of the sixth term and the tenth is $44.$ Find the first three terms of the A.P.
AnswerGiven
$t_4 + t_8 = 24$
$\Rightarrow(a+3 d)+(a+7 d)=24$
$\Rightarrow 2 a+10 d=24$
$\Rightarrow a+5 d=12 \ldots \ldots \ldots . .(i)$
And,
$t_6 + t_{10} = 44$
$\Rightarrow(a+5 d)+(a+9 d)=44$
$\Rightarrow 2 a+14 d=44$
$\Rightarrow a+7 d=22 \ldots \ldots \ldots(\text { iii) }$
Subtracting (i) from (ii), we get
$2d = 10$
$\Rightarrow d=5$
Substituting value of d in (i), we get
a + 5 x 5 = 12
$\Rightarrow a+25=12$
$\Rightarrow a=-13=1^{\text {st }} \text { term }$
$a+d=-13+5=-8=2^{\text {nd }} \text { term }$
$a+2 d=-13+2 \times 5=-13+10=-3=3^{\text {rd }} \text { tem }$
Hence, the first three terms of an A.P are $-13, -8$ and $-5$
View full question & answer→Question 74 Marks
A manufacturer of TV sets produces $600$ units in the third year and $700$ units in the $7^{th}$ year. Assuming that the production increases uniformly by a fixed number every year find:
(i) the production in the first year.
(ii) the production in the $10^{th}$ year.
(iii) the total production in $7$ years.
AnswerSince the Production increases uniformly by a fixed number every year, he sequence formed by the production in differnt years in an A.P.
Let the Production in the first year $= a$
Common differance $=$ number of unit by which the production increases every year $= d$
We have,
$t_3=600$
$\Rightarrow a+2 d=600$.
$t_7=700$
$\Rightarrow a+6 d=700$
Subtracting (1) from (2) we get
$4 d=100 \Rightarrow d=25$
$\Rightarrow a+2 \times 25=600$
$\Rightarrow a=550$
(1) The Producton in the first year $=550 Tv$ sets
(2) Production in the $10^{wedge}$ (th) year $=t_{10}=550++9 \times 25=775$ Tvsets
(3) Production in $7$ years $=S_7=\frac{7}{2}(2 \times 550+6 \times 25)$
$=\frac{7}{2}(1100+150)$
$=\frac{7}{2} \times 1250$
$=4375 \text { Tv sets }$
View full question & answer→Question 84 Marks
An article can be bought by paying $Rs. 28,000$ at once or by making $12$ monthly installments. If the first installment paid is $Rs. 3,000$ and every other installments is $Rs. 100$ less than the previous one, find :
(i) amount of installments paid in the $9^{th}$ month
(ii) total amount paid in the installment scheme
AnswerNumber of installments $=n=12$
First istalment=a $=R s .3000$
Depeciation in instalment $=d=-100$
$(1)$ Amount of installment paid in the $9^{th}$ month
$=t_9 $
$ =a+8 d$
$=3000+8 \times(-100)$
$ =300-800$
$=\text { Es. } 2200$
Total amount paid in the instakkment scheme
$=S_{12} $
$ =\frac{12}{2}(2 \times 3000+11 \times(-100))$
$ =6(6000-1100)$
$ =6 \times 4900 $
$ =\text { Rs. } 29,400$
View full question & answer→Question 94 Marks
A sum of $Rs. 700$ is to be paid to give seven cash prizes to the students of a school for their overall academic performance. If the cost of each prize is $Rs. 20$ less than its preceding prize; find the value of each of the prizes.
AnswerTotal amount of Prize- $s_n=R s .700$
Let the value of the first prize be Rs.a
$\text { Number of prizes }=n=7$
Let the value of first prize be Rs. a.
Depreciation in next prize $=R s .20$
We have,
$s_n=\frac{n}{2}(2 a+(n-1) d) $
$ \Rightarrow 700=\frac{7}{2}(2 a+6(-20))$
$ \Rightarrow 700=\frac{7}{2}(2 a=120)$
$ \Rightarrow 1400=14 a-840$
$ \Rightarrow 14 a=2240 $
$ \Rightarrow a=160$
$\Rightarrow$ Valueof $1^{st}$ Prize $= Rs. 160$
Value of $2^{\text {nd }} \operatorname{Priz}=R s(160-20) R s .140$
Value of $3^{\text {nd }}$ Prize $= Rs. (140-20)=R s .120$
Value of $4^{\text {th }}$ Prize $=R s .(120-20)=R s .100$
Value of $^{\text {th }}$ Prize $= Rs. (100-20) R s .80$
Value of $6^{\text {th }}$ Prize $=R s . .(80-20)=R s .60$
Value of $7^{\text {th }}$ Prize $=R s .(60-20)=R s .40$
View full question & answer→Question 104 Marks
Find four numbers in A.P. whose sum is $20$ and the sum of whose squares is $120.$
AnswerLet the four numbers in A.P be $(a-3 d),(a-d),(a+d)$ and $(a+3 d)$.
Then, $(a-3 d)+(a-d)+(a+d)+(a+3 d)=20$
$\Rightarrow 4 a=20$
$ \Rightarrow a=5$
It is given that
$(a-3 d)^2+(a-d)^2+(a+d)^2+(a+3 d)^2=120$
$\Rightarrow a^2+9 d^2-6 a d+a^2+d^2-2 a d+a^2+d^2+2 a d+a^2+9 d^2+6 a d=120$
$\Rightarrow 4 a^2+20 d 2=120 $
$\Rightarrow a^2+5 d^2=30$
$ \Rightarrow 5^2+5 d^2=30$
$ \Rightarrow 25+5 d^2=30 $
$\Rightarrow 5 d^2=5 $
$\Rightarrow d^2=1$
$ \Rightarrow d= \pm 1$
$\text { When } a=5, d=1 $
$a-3 d=5-3(1)=2$
$a-d=5-1=4$
$ a-d=5-1=4 $
$a+d=5+1=6$
$ a+3 d=5+3(1)=8$
When $a=5, d=1$
$a-3 d=5-3(-1)=8$
$a-d=5-(-1)=6$
$a+d=5-(-1)=6$
$a+3 d=5+3(-1)=2$
Thus, the four parts are $(2,4,6,8)$ or $(8,6,4,2)$.
View full question & answer→Question 114 Marks
The sum of three numbers in A.P.is $15$ the sum of the square of the exterme is $58.$ find the numbers.
AnswerLet the three numberss in A.P. Be $(a-d),$ a and $(a+d).$
Then, $(a-d)+a+(a+d)=15$
$\Rightarrow 3 a=15 $
$ \Rightarrow a=5$
It is given that
$(a-d)^2+(a+d)^2=58 $
$ \Rightarrow a^2+d^2-2 a d+a^2+d^2+2 a d=58$
$\Rightarrow 2 a^2+2 d^2=58$
$\Rightarrow 2\left(a^2+d^2\right)=58$
$\Rightarrow a^2+d^2=29 $
$\Rightarrow 5^2+d^2=29 $
$ \Rightarrow 25+d^2=29$
$ \Rightarrow d^2=4 $
$ \Rightarrow d= \pm 2$
When $a=5$ and $d=2$
$a-d=5-2=3$
$a=5 $
$a+d=5+2=7$
When $a=5$ and $d=-2$
$a-d=5-(-2)=7 $
$a=5$
$a+d=5+(-2)=3$
Thus, the tree numbers in A.P are $(3,5,7)$ or $(7,5,3)$
View full question & answer→Question 124 Marks
Find three numbers in A.P whose sum is $24$ and whoseproduct is $440$
AnswerLet the three number in AP. Be a-d, a and $a+d$.
$ \therefore(a-d)+a+(a+d)=24$
$\Rightarrow 3 a=24$
$\Rightarrow a=8 \ldots \ldots \ldots \ldots . .(1)$
$\text { Also, }(a-d) \times a \times(a+d)=440$
$\Rightarrow\left(a^2-d^2\right) \times a=440$
$\Rightarrow\left(8^2-d^2\right) \times 8=440 \ldots \ldots . . \ldots \ldots . . .[\text { from (1)] }$
$\Rightarrow 64-d^2=55$
$\Rightarrow d^2=9$
$\Rightarrow d= \pm 3$
When $a=8$ and $d=3$
Required terms $=a-d, a$ and $a+d$
$=8-3,8,8+3 $
$ =5,8,11$
When $a=8$ and $d=-3$
$ \begin{aligned} \text { Required term } & =a-d, a \text { and } a+d \\ & =8-(-3), 8,8+(-3) \\ & =11,8,5 \end{aligned} $
View full question & answer→Question 134 Marks
The angles of a polygon are in A.P. with common differance $5^\circ.$ if the smallest angle in $120^\circ,$ find the number ofsides of the polygoa.
AnswerLet the number of sides of a polygon be n.
The smallest angle $= 120^\circ = a$
Common difference in angles $= d = 5^\circ$
Now, in a polygon of n sides, the sum of interior angles $= (2n - 4) x 90^\circ$
$\Rightarrow \frac{n}{2}\left(2 \times 120^{\circ}+(n-1) \times 5^{\circ}\right)=(2 n-4) \times 90^{\circ}$
$\Rightarrow \frac{n}{2}\left(240^{\circ}+5 n-5^{\circ}\right)=180 n-360^{\circ}$
$\Rightarrow n\left(235^{\circ}+5 n\right)=360 n-720^{\circ}$
$\Rightarrow 5 n^2-125 n+720=0$
$\Rightarrow n^2-25 n+144=0$
$\Rightarrow n^2-16 n-9 n+144=0$
$\Rightarrow n(n-16)-9(n-16)=0$
$\Rightarrow(n-16)(n-9)=0$
$\Rightarrow n =16 \text { or } n =9$
View full question & answer→Question 144 Marks
The first term of an A.P is $5,$ the last term is $45$ and the sum of its terms is $1000.$ Find the number of terms and the common difference of the A.P.
AnswerFirst-term $a = 5$
Last term $l = 45$
Sum of terms $= 1000$
Let there be n terms in this A.P
Now sum of first $n$ terms $=\frac{n}{2}[a+l]$
$\Rightarrow 1000=\frac{n}{2}[5+45] $
$ \Rightarrow 2000=n \times 50$
$ \Rightarrow n=40$
$l = a + (n - 1)d$
$\Rightarrow 45 = 5 + (40 - 1)d$
$\Rightarrow 40 = 39d$
$\Rightarrow d=\frac{40}{39}$
Hence numbers of terms are $40$ and common difference is $\frac{40}{39}$
View full question & answer→Question 154 Marks
Find the sum of all odd natural numbers less than $50$
AnswerOdd natural numbers less than $50$ are as follows
$1, 3, 5, 7, 9, ........, 49$
Now, $3 - 1 = 2, 5 - 3 = 2$ and so on
Thus, this forms an A.P. with first term $a = 1,$ common difference $d = 2$ and last term $l = 49$
Now,$ l = a + (n -1)d$
>$\Rightarrow 49 = 1 + (n - 1) xx 2$
$\Rightarrow 48=(n-1) \times 2 $
$\Rightarrow 24=n-1 $
$\Rightarrow n=25$
Sum of first $n$ terms $=S=\frac{n}{2}[a+1]$
$\ \Rightarrow $ Sum of odd natural numbers less than $50=\frac{25}{2}[1+49]$
$=\frac{25}{2} \times 50 $
$ =625$
View full question & answer→Question 164 Marks
How many terms of the A.P: $24, 21, 18, ………$ must be taken so that their sum is $78?$
AnswerLet the number of terms taken be n.
The given A.P is $24, 21, 18, .....$
Here $a = 24$ and $d = 21 - 24= -3$
$S=\frac{n}{2}[2 a+(n-1) d] $
$\Rightarrow 78=\frac{n}{2}[2 \times 24+(n-1) \times(-3)]$
$ \Rightarrow 78=\frac{n}{2}[48-3 n+3]$
$\Rightarrow156= n [51-3 n ] $
$ \Rightarrow 156=51 n-3 n^2 $
$\Rightarrow 3 n^2-51 n+156=0$
$\Rightarrow n^2-17 n+52=0$
$ \Rightarrow n^2-13 n-4 n+52=0 $
$ \Rightarrow n(n-13)-4(n-13)=0 $
$ \Rightarrow n =13 \text { or } n =4$
$\therefore$ Required number of term $= 4$ or $13$
View full question & answer→Question 174 Marks
Find the sum of all multiples of $7$ between $300$ and $700.$
AnswerNumbers between $300$ and $700$ which are multiples of $7$ are as follows:
$301, 308, 315, 322, ....., 693$
Clearly, this forms an A.P with first term $a = 301,$ common difference $d = 7$ and last term $l = 693$
$l = a + (n - 1)d$
$\Rightarrow 693=301+(n-1) \times 7$
$ \Rightarrow 392=(n-1) \times 7 $
$ \Rightarrow n-1=56$
$\Rightarrow n=57$
Sum of first $n$ terms $=S=\frac{n}{2}(a+l)$
$\Rightarrow $ Required sum $=\frac{57}{2}(301+693)$
$=\frac{57}{2} \times 994 $
$=57 \times 497$
$=28329$
View full question & answer→Question 184 Marks
If the 8th term of an A.P is $37$ and the $15$th term is $15$ more than the $12$th term, find the A.P. Also, find the sum of first $20$ terms of A.P.
AnswerFor an A.P
$t_8=37$
$\Rightarrow a+7 d=37 \ldots \text {. (i) }$
$\text { Also } t_{15}-t_{12}=15$
$\Rightarrow(a+14 d)-(a+11 d)=15$
$\Rightarrow a+14 d-a-11 d=15$
$\Rightarrow 3 d =15$
$\Rightarrow d =5$
Subsituting $d = 5$ in (i) we get
$a+7 \times 5=37$
$\Rightarrow a+35=37$
$\Rightarrow a=2$ ∴ Required A.P $= a, a +d, a + 2d, a + 3d, .....$
$= 2, 7, 12, 17, .....$
Sum of the first $20$ terms of this A.P $=\frac{20}{2}(2 \times 2+19 \times 5)$
$= 10(4 + 95)$
$= 10 \times 99$
$= 990$
View full question & answer→Question 194 Marks
In an A.P, the first term is $25,$ nth term is $-17$ and the sum of n terms is $132.$ Find n and the common difference.
AnswerFirst-term $a = 25$
nth term $= -17 \Rightarrow$ last term $l = -17$
Sum of n terms = 132
$\Rightarrow \frac{n}{2}[a+l]=132$
$ \Rightarrow n (25-17)=264 $
$ \Rightarrow n \times 8=264 $
$\Rightarrow n=33$
Now $l = -17$
$\Rightarrow a+(n-1) d=-17$
$\Rightarrow 25+32 d=-17$
$\Rightarrow 32 d=-42$
$\Rightarrow d=-\frac{42}{32}$
$\Rightarrow d=-\frac{21}{16}$
View full question & answer→Question 204 Marks
If number n - 2, 4n - 1 and 5n + 2 are in A.P find the value of n and its next two terms.
AnswerSince (n - 2), (4n - 1) and (5n + 2) are in A.P we have
(4n - 1) - (n - 2) = (5n + 2)- (4n -1)
⇒ 4n - 1 - n + 2 = 5n + 2 - 4n + 1
⇒ 3n + 1 = n + 3
⇒ 2n = 2
⇒ n = 1
∴ (n - 2), (4n - 1) and (5n + 2)
∴ (1 - 2), (4(1) - 1) and (5(1) + 2)
So, the given numbers are -1, 3, 7
⇒ a = - 1 and d = 3 - (-1) = 4
hence the next two terms are (7 + 4) and (7 + 2 × 4)
i.e 11 and 15.
View full question & answer→Question 214 Marks
If the $3$rd and the $9$th terms of an arithmetic progression are $4$ and $-8$ respectively, Which term of it is zero?
AnswerFor an A.P
$t_3=4$
$\Rightarrow a +2d = 4 .....(i)$
$t_9=-8$
$\rightarrow a + 8d = -8 ....(ii)$
Substractinjg (i) from (ii) we get
$6d = -12$
$\Rightarrow d = -2$
Substituting $d = -2$ in (i) we get
$a + 2(-2) = 4$
$\Rightarrow a - 4 = 4$
$\Rightarrow a = 8$
$\Rightarrow$ general term $=t_n=8+(n-1)(-2)$
Let pth term of this A.P be 0
$\Rightarrow 8+(p-1) \times(-2)=0$
$\Rightarrow 8-2 p+2=0$
$\Rightarrow 10-2 p=0$
$\Rightarrow 2 p=10$
$\Rightarrow p=5$
Thus 5th term of this A.P is $0.$
View full question & answer→Question 224 Marks
How many two-digit numbers are divisible by 3?
AnswerThe two digit numbers divisible by 3 are as follows:
12, 15, 18, 21, ...., 99
Clearly, this form an A.P with first term a = 12 and common difference d = 3
Last term = nth term = 99
The general term of an A.P is given by
$t_n=a+(n-1) d$
$\Rightarrow$ 99 12 + (n - 1)(3)
$\Rightarrow$ 99 = 12 + 3n - 3
$\Rightarrow$ 90 = 3n
$\Rightarrow$ n = 30
Thus 30 two-digit numbers are divisible by 3.
View full question & answer→Question 234 Marks
In an A.P, if mth term is n and nth term is m, show that its rth term is (m + n – r)
AnswerFor an A.P.
$t_m=n$
$\Rightarrow a+(m-1) d=n \ldots . .( i )$
And $t_n=m$
$\Rightarrow$ a + (n - 1)d = m .....(ii)
Substracting (i) from (ii) we get
(n - 1)d - (m - 1)d = m - n
$\Rightarrow$ d(n - m) = m - n
$\Rightarrow$ -d(n - m) = m - n
$\Rightarrow$ d = -1
Substituting d = -1 in equation (i) we get
a + (m - 1)(-1) = n
$\Rightarrow$ a - m + 1 = n
$\Rightarrow$ a = m + n -1
Now $t_r=a+(r-1) d$
= (m + n - 1) + (r - 1)(-1)
= m + n - 1 - r + 1
= m + n - r
View full question & answer→Question 244 Marks
The sum of the $2$nd term and the $7$th term of an A.P is $30.$ If its $15$th term is $1$ less than twice of its $8$th term, find the A.P
AnswerThe general term of an AP is given by $t_n=a+(n-1) d$
Now $t_2+t_7=30$
$\Rightarrow (a + d) + (a + 6d) = 30$
$\Rightarrow 2a + 7d = 30 ...(i)$
Next $2 \times t_8-t_{15}=1$
$\Rightarrow 2 \times(a+7 d)-(a+14 d)=1$
$\Rightarrow 2a + 14d - a - 14d = 1$
$\Rightarrow a = 1$
Substituting the value of a in (i) we get
$2 \times 1+7 d=30$
$\Rightarrow 7 d=28$
$\Rightarrow d=4$ Thus required A.P $= a, a + d, a + 2d, a + 3d$
$= 1, 5, 9, 13, ....$
View full question & answer→Question 254 Marks
If a, b and c are in A.P show that: $a + 4, b + 4$ and $c + 4$ are in A.P.
Answera, b and c are in A.P
$\Rightarrow b - a = c - b$
$\Rightarrow 2b = a + c$
Given terms are $(a + 4), (b + 4)$ and $(c + 4)$
Now (b + 4) - (a + 4)= b - a
$=\frac{a+c}{2}-a$
$=\frac{a+c-2 a}{2}$
$=\frac{c-a}{2}$
And $(c + 4) - (b + 4) = c - b$
$=c-\frac{a+c}{2}$
$=\frac{2 c-a-c}{2}$
$=\frac{c-a}{2}$
Since $(b + 4) - (a + 4) = (c + 4) - (b + 4),$ then given terms are in A.P
View full question & answer→Question 264 Marks
Is $402$ a term of the sequence: $8, 13, 18, 23.....?$
AnswerThe given sequence is $8, 13, 18, 23.....$
Now
$13 - 8 = 5$
$18 - 13 = 5$
$23 - 18 = 5,$ etc
hence the given sequence is an A,P with first term $a = 8$ and common difference $d= 5.$
The general term of an A.P is given by
$t_n=a+(n-1)(5)$
$\Rightarrow402=8+(n-1)(5)$
$\Rightarrow394=5 n-5$
$\Rightarrow399=5 n$
$\Rightarrow n=\frac{399}{5}$
The number of terms cannot be a fraction.
So clearly, $402$ is not the term of the given sequence.
View full question & answer→Question 274 Marks
If the third term of an A.P. is $5$ and the seventh term is $9$ , find the $17^{\text {th }}$ term.
AnswerLet 'a' be the first term and 'd' be the common difference of the given A.P.
Now, $t_3 = 5$ (given)
$\Rightarrow a + 2d = 5 ….(i)$
And,
$t_7 = 9$ (given)
$\Rightarrow a + 6d = 9 ….(ii)$
Subtracting (i) from (ii), we get
$4d = 4$
$\Rightarrow d = 1$
$\Rightarrow a + 2(1) = 5$
$\Rightarrow a = 3$
Hence, $17^{th}$ term $= t_{17} = a + 16d = 3 + 16(1) = 19$
View full question & answer→Question 284 Marks
Which term of the service $21, 18, 15 ......$ is $-81?$
Can any term of this series be zero? if yes find the number of terms.
AnswerThe given A.P is $21, 18, 15, .....$
Here, first term $a = 21$ and common difference $d = 18 - 21 = 3$
Let nth term of the given A.P be $-81$
$\Rightarrow -81 = a + (n - 1)d$
$\Rightarrow-81=21+(n-1) \times(-3)$
$\Rightarrow-102=(n-1) \times(-3)$
$\Rightarrow n-1=34$
$\Rightarrow n =35$
Thus the $35$th term of the given A.P is $-81.$
Let $p^{th}$ term of this A.P be $0$
$\Rightarrow 21+(p-1) \times(-3)=0$
$\Rightarrow 21-3 p+3=0$
$\Rightarrow p = 8$
Thus 8 th term of this A.P is $0$
View full question & answer→Question 294 Marks
Find the $31^{st}$ term of an A.P whose $10^{th}$ term is $38$ and the $16^{th}$ term is $74.$
AnswerThe general term of an A.P is given by
$t_n=a+(n-1) d$
Now $t_{10}=38$
$\Rightarrow a + 9d = 38... (i)$
And $t_{16}=74$
$\Rightarrow a + 15d = 74 .....(ii)$
Substracting (i) from (ii) we get
$6d = 36$
$\Rightarrow d=6$
Substiting $d = 6$ in (i) we get
$a+9 \times 6=38$
$\Rightarrow a + 54 = 38$
$\Rightarrow a = -16$
$\Rightarrow t_n=-16+(n-1)(6)$
$\Rightarrow t_{31}=-16+30 \times 6=-16+180=164$
View full question & answer→Question 304 Marks
If $5$th and $6$th terms of an A.P are respectively $6$ and $5.$ Find the 11th term of the A.P
AnswerThe general term of an A.P is given by
$t_n=a+(n-1) d$
Now $t_5=6$
$\Rightarrow a + (5 - 1)d = 6$
$\Rightarrow a + 4d = 6 ....(i)$
And $t_6=5$
$\Rightarrow$ a + (6 - 1)d = 5
$\Rightarrow a + 5d = 5 ...(ii)$
Substracting (ii) from (i) we get
$-d = 1$
$\Rightarrow d = -1$
Substituting $d = -1$ in (i) we get
$a + 4(-1) = 6$
$\Rightarrow a - 4 = 6$
$\Rightarrow a = 10$
$\Rightarrow t_n=10+(n-1)(-1)$
$\Rightarrow t_{11}=10+(11-1)(-1)=10-10=0$
View full question & answer→Question 314 Marks
Find five numbers in $A.P$ whose sum is $12 \frac{1}{2}$ and the ratio of the first to the last terms $2:3.$
AnswerLet the five numbers in $A.P$. be
$(a-2 d),(a-d), a,(a+d)$ and $(a+2 d)$.
Then, $(a-2 d)+(a-d)+a+(a+d)+(a+2 d)=12 \frac{1}{2}$
$\Rightarrow 5 a=\frac{25}{2}$
$\Rightarrow a=\frac{5}{2}$
It is given that
$\frac{a-2 d}{a+2 d}=\frac{2}{3}$
$\Rightarrow 3 a-6 d=2 a+4 d$
$\Rightarrow a=10 d$
$\Rightarrow \frac{5}{2}=10 d$
$\Rightarrow d=\frac{1}{4}$
$\Rightarrow a=\frac{5}{2}$ and $ d=\frac{1}{4}$
Thus, We have
$a-2 d=\frac{5}{2}-2 \times \frac{1}{4}=\frac{5}{2}-\frac{1}{2}=\frac{4}{2}=2$
$a-d=\frac{5}{2}-\frac{1}{4}=\frac{10-1}{4}=\frac{9}{4}$
$a=\frac{5}{2}$
$a+d=\frac{5}{2}+\frac{1}{4}=\frac{10+1}{4}=\frac{11}{4}$
$a+3 d=\frac{5}{2}+2 \times \frac{1}{4}=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3$
Thus, the five numbers in $A . P=2, \frac{9}{4}, \frac{5}{2}, \frac{11}{4}$ and $3$
$=2,2.25,2.5,2.75$ and $3$
View full question & answer→Question 324 Marks
Divide $96$ into four parts which are in $A.P$ and the ratio between product of their means to product of their extremes is $15:7.$
AnswerLet the four parts be $(a-3 d),(a-d),(a+d)$ and $(a+3 d)$
Then, $(a-3 d)+(a-d)+(a+d)+(a+3 d)=96$
$\Rightarrow 4 a=96$
$\Rightarrow a=24$
It is given that
$\frac{(a-d)(a+d)}{(a-3 d)(a+3 d)}=\frac{15}{7}$
$\Rightarrow \frac{a^2-d^2}{a^2-9 d^2}=\frac{15}{7}$
$\Rightarrow \frac{576-d^2}{576-9 d^2}=\frac{15}{7}$
$\Rightarrow 4032-7 d^2=8640-135 d^2$
$\Rightarrow 128 d^2=4608$
$\Rightarrow d^2=36$
$\Rightarrow d= \pm 6$
When $a=24, d=-6$
$a-3 d=24-3(6)=6$
$a-d=24-(-6)=18$
$a+d=24+6=30$
$a+3 d=24+3(6)=42$
When $a=24, d=-6$
$a-3 d=24-3(-6)=42$
$a-d=24-(-6)=30$
$a+d=24+(-6)=18$
$a+3 d=24+3(-6)=6$
Thus , the four parts are $(6,18,30,42)$ or $(42,30,18,6)$
View full question & answer→Question 334 Marks
The sum of three consecutive terms of an $A.P.$ is $21$ and the sum of their squares is $165.$ find these terms.
AnswerLet the treee consecutive terms in $A.P.$ be $a-d, a$ and $a+d.$
$\therefore(a-d)+a+(a+d)=21$
$\Rightarrow a=7 ........ (1)$
Also,$(a-d)^2+a^2+(a+d)^2=165$
$\Rightarrow a^2+d^2-2 a d+a^2+a^2+d^2+2 a d=165$
$\Rightarrow 3 a^2+2 d^2=165$
$\Rightarrow 3 \times(7)^2+2 d^2=165 ...... [$From $(1)]$
$\Rightarrow 3 \times 49+2 d^2=165$
$\Rightarrow 147+2 d^2=165$
$\Rightarrow 2 d^2=18$
$\Rightarrow d^2=9$
$\Rightarrow d= \pm 3$
when $a=7$ and $d=3$
Required terms $=a-d, a$ and $a+d$
$=7-3,7,7+3$
$=4,7,10$
When $a=7$ and $d=-3$
$\text { Required terms }=a-d, a \text { and } a+d$
$=7-(-3), 7,7+(-3)$
$=10,7,4$
View full question & answer→Question 344 Marks
If the sum of first $7$ terms of an $A.P.$ is $49$ and that of its first $17$ terms is $289,$ find the sum of first n terms of the $A.P.$
AnswerLet the first term and the common difference of the given $AP$ be a and d, respectively.
Sum of the first $7$ terms, $S_7 = 49$
We know
$S=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \frac{7}{2}(2 a+6 d)=49$
$\Rightarrow \frac{7}{2} \times 2(a+3 d)=49$
$\Rightarrow a+3 d=7 ....(1)$
Sum of the first $17$ terms, $S_{17} = 289$
$\Rightarrow \frac{17}{2}(2 a +16 d )=289$
$\Rightarrow \frac{17}{2} \times 2( a +8 d )=289$
$\Rightarrow a+8 d=\frac{289}{17}$
$\Rightarrow a+8 d=17....(2)$
Subtracting $(2)$ from $(1),$ we get
$5d = 10$
$ 5 d =10$
$ d=\frac{5}{10}$
$ \Rightarrow d =2$
Substituting the value of $d$ in $(1),$ we get
$a=1$
Now,
sum of the first $n$ terms is given by
$ S_n=\frac{n}{2}[2 a+(n-1) d]$
$ =\frac{n}{2}[2 \times 1+(n-1) \times 2]$
$ =\frac{n}{2}[2+2 n-2]$
$ =\frac{n}{2}[2 n]$
$ = n ^2$
Therefore, the sum of the first $n$ terms of the $AP$ is $n^2$.
View full question & answer→Question 354 Marks
The fourth term of an$ A.P.$ is $11$ and the term exceeds twice the fourth term by $5$ the $A.P$ and the sum of first $50$ terms.
AnswerFor an $A.P \ t_4=11$
$\Rightarrow a + 3d = 11 ....(1)$
Also ,$ t_8 - 2t_4 = 5$
$\Rightarrow (a + 7d) - 2 \times 11 = 5$
$\Rightarrow a + 7d - 22 = 5$
$\Rightarrow a + 7d = 27 ....(2)$
Substracting equation $1$ from equation $2,$ we get
$a+3 d=11$
$a+7 d=27$
$\frac{----}{-4 d=-16}$
$d=\frac{-16}{-4}$
$\Rightarrow d=4$
substituting $d = 4$ in equation $1,$ we get
$a + 3d = 11$
$a + 3 \times 4 = 11$
$\Rightarrow a + 12 = 11$
$\Rightarrow a = -1$
$\therefore$ required $A.P = a, a + d, a + 2d, a + 3d,....$
$\therefore a = - 1$
$\therefore a + d = - 1 + 4 = 3$
$\therefore a + 2d = - 1 + 2(4) = - 1 + 8 = 7$
$\therefore a + 3d = - 1 + 3(4) = - 1 + 12 = 11$
Sum of first $50$ terms of this $A.p =\frac{50}{2}(2 \times(-1)+49 \times 4)$
$= 25(- 2 + 196)$
$= 25 \times 194$
$= 4850$
View full question & answer→Question 364 Marks
$4^{th}$ term of an $A.P$ is equal to $3$ times its first term and $7^{th}$ term exceeds twice the $3^{rd}$ time by $I.$ Find the first term and the common difference.
AnswerThe general term of an $AP$ is given by $t_n=a+(n-1) d$
Now $t_4=3 \times a$
$\Rightarrow a + 3d = 3a$
$\Rightarrow 2a - 3d = 0 ...(1)$
$\text { Next } t_7-2 \times t_3=1$
$\Rightarrow a+6 d-2(a+2 d)=1$
$\Rightarrow a+6 d-2 a-4 d=1$
$\Rightarrow -a+2 d=1 \ldots . \text { (ii) }$
multiplying $(ii)$ by $2$ we get
$-2a + 4d = 2 ....(iii)$
Adding equation $(i)$ and $(iii)$ we get
$d = 2$
Substituting the value of $d$ in $(ii)$ we get
$ -a+2 \times 2=1$
$\Rightarrow-a+4=1$
$= > a=3 $
Hence $a =3$ and $d =2$
View full question & answer→Question 374 Marks
An $A.P$ consists of $57$ terms of which $7^{th}$ term is $13$ and the last term is $108.$ Find the $45^{th}$ term of this $A.P.$
AnswerNumber of terms $= n = 57$
$t_7=13$
$\Rightarrow a + 6d = 13 ....(i)$
Last term $=t_{57}=108$
$\Rightarrow a + 56d = 108 ....(ii)$
Substracting $(i)$ from $(ii)$ we get
$50d = 95$
$\Rightarrow d=\frac{95}{50}$
$\Rightarrow d=\frac{19}{10}$
Substituting value of $d$ in $(i)$ we get
$a+6 \times \frac{19}{10}=13$
$\Rightarrow a+\frac{57}{5}=13$
$\Rightarrow a=13-\frac{57}{5}=\frac{65-57}{5}=\frac{8}{5}$
$\Rightarrow$ General term $=t_n=\frac{8}{5}+(n-1) \times \frac{19}{10}$
$\Rightarrow t_{45}=\frac{8}{5}+44 \times \frac{19}{10}$
$=\frac{8}{5}+\frac{418}{5}$
$=\frac{426}{5}$
$=85.2$
View full question & answer→Question 384 Marks
The sum of the $4^{th}$ and the $8^{th}$ terms of an $A.P.$ is $24$ and the sum of the $6^{th}$ and the $10^{th}$ terms of the same $A.P.$ is $34.$ Find the first three terms of the $A.P.$
AnswerLet 'a' be the first term and $ 'd\ ' $ be the common difference of the given $A.P.$
$t_4 + t_8 = 24 ($given$)$
$\Rightarrow (a + 3d) + (a + 7d) = 24$
$\Rightarrow 2a + 10d = 24$
$\Rightarrow a + 5d = 12 ….(i)$
And,
$t_6 + t_{10} = 34 ($given$)$
$\Rightarrow (a + 5d) + (a + 9d) = 34$
$\Rightarrow 2a + 14d = 34$
$\Rightarrow a + 7d = 17 ….(ii)$
Subtracting $(i)$ from $(ii),$ we get
$2d = 5$
$\Rightarrow d=\frac{5}{2}$
$\Rightarrow a+5 \times \frac{5}{2}=12$
$\Rightarrow a+\frac{25}{2}=12$
Thus we have,
$1^{\text {st }}$ term $=-\frac{1}{2}$
$2^{n d}=a+d=-\frac{1}{2}+\frac{5}{2}=2$
$3^{r d}$ term $=a+2 d=-\frac{1}{2}+2 \times \frac{5}{2}=-\frac{1}{2}+5=\frac{9}{2}$
View full question & answer→Question 394 Marks
If $t_n$ represents $n ^{\text {th }}$ term of an $A.P\ t_2+t_5-t_3=10$ and $t_2+t_9=17$ Find its first term and its common difference.
AnswerLet the first term of an $A.P$ be a and the common difference be $d.$
The general term of an $A.P$ is given by $t_n=a+(n-1) d$
Now $t_1+t_5-t_3=10$
$\Rightarrow(a+d)+(a+4 d)-(a+2 d)=10$
$\Rightarrow a+d+a+4 d-a-2 d=10$
$\Rightarrow a + 3d = 10 ...(i)$
Also $t_2+t_9=17$
$\Rightarrow (a + d) + (a + 8d) = 17$
$\Rightarrow 2a + 9d = 17 ....(ii)$
Multiplying equation $(i)$ by $2$ we get
$2a + 6d = 20 ....(iii)$
Subtracting $(ii)$ from $(iii)$ we get
$-3d= 3$
$\Rightarrow d = -1$
Substituting value of $d$ in $(i)$ we get
$a + 3(-1) = 10$
$\Rightarrow a - 3 = 10$
$\Rightarrow a = 13$
Hence $a = 13$ and $d = -1$
View full question & answer→