Question 514 Marks
The $10^{\text {th }}$ and $18^{\text {th }}$ terms of an $A.P.$ are $41$ and $73$ respectively. Find $26^{\text {th }}$ term.
AnswerIn the given problem, we are given $10^{\text {th }}$ and $18^{\text {th }}$ term of an $A.P.$ We need to find the $26^{\text {th }}$ term
Here,
$a_{10}=41$
$a_{18}=73$
Now, we will find $a_{10}$ and $a_{18}$ using the formula $a_n=a+(n-1) d$
So,
$a_{10}=a+(10-1) d$
$41=a+9 d \ldots . . .(i)$
Also,
$a_{18}=a+(18-1) d$
$73=a+17 d \ldots . .(i i)$
So, to solve for a and d
On susbtracting $(1)$ from $(2),$ we get
$8 d=32$
$d=\frac{32}{8}$
$d=4$
Substituting $d=4$ in $(1)$, we get
$41=a+9(4)$
$41-36=a$
$a=5$
Thus,
$a=5$
$d=4$
$n=26$
Substituting the above values in the formula, $a_n=a+(n-1) d$
$a_{26}=5+(26-1) 4$
$a_{26}=5+100$
$a_{26}=105$
Therefore, $a_{26}=105
View full question & answer→Question 524 Marks
A sum of Rs. $700$ is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs. $20$ less than its preceding prize, find the value of each prize.
AnswerIn the given problem,
Total amount of money $(S_n) = Rs\ 700$
There are a total of 7 prizes and each prize is Rs $20$ less than the previous prize. So let us take the first prize as Rs a.
So, the second prize will be Rs a - $20$, third prize will be Rs a $- 20 - 20.$
Therefore, the prize money will form an A.P. with first term a and common difference$ -20.$
So, using the formula for the sum of n terms,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
We get,
$700=\frac{7}{2}\Big[2\text{a}+(7-1)(-20)\Big]$
$700=\frac{7}{2}[2\text{a}+(6)(-20)]$
$700=\frac{7}{2}(2\text{a}-120)$
$700=7(\text{a}-60)$
On Further simplification, we get,
$\frac{700}{7}=\text{a}-60$
$100+60=\text{a}$
$\text{a}=160$
Therefore, the value of first prize is $Rs. 160.$
Second prize $= Rs. 140$
Third prize $= Rs. 120$
Fourth prize $= Rs. 100$
Fifth prize $= Rs. 80$
Sixth prize $= Rs. 60$
Seventh prize $= Rs. 40$
So the values of prizes are $Rs. 160, Rs. 140, Rs. 120, Rs. 100, Rs. 80, Rs. 60, Rs. 40.$
View full question & answer→Question 534 Marks
Let there be an $A.P$. with first term 'a', common difference ' $d$ '. If $a_n$ denotes in $n^{\text {th }}$ term and $S_n$ the sum of first $n$ terms, find.
d, if $\mathrm{a}=3, \mathrm{n}=8$ and $\mathrm{S}_{\mathrm{n}}=192$.
AnswerHere, we have an $A.P.$ whose first term $(a)$, Sum of first n terms $(S_n)$ and the number of terms $(n)$ are given. We need to find common difference $(d).$
Here,
First term $(a) = 3$
Sum of n terms $(S_n) = 192$
Number of terms $(n) = 8$
So here we will find the value of n using the formula, $a_n= a + (n - 1)d$
So. to find the common difference of this $A.P.$, we use the following formula for the sum of n terms of an $A.P.$
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
Where; a = first term for the given $A.P.$
$d =$ common difference of the given $A.P.$
$n =$ number of terms
So, using the formula for $n = 8$, we get,
$\text{S}_8=\frac{8}{2}[2(3)+(8-1)(\text{d})]$
$192 = 4[6 + (7)(d)]$
$192 = 24 + 28d$
$28d = 192 - 24$
Furhter solving for d,
$\text{d}=\frac{168}{28}$
$d = 6$
Therefore, the common difference of the given $A.P$. is $d = 6.$
View full question & answer→Question 544 Marks
Find the sum:
$\text{7}+10\frac{1}{2}+14+\ .....\ + 84$
AnswerGiven,
$\text{A.P., }7+10\frac{1}{2}+14+\ .....\ + 84$
Here,
First term, $a = 7$
Difference, $\text{d}=\frac{21}{2}-7=\frac{7}{2}$
and Last term, $a_n= 84$
We know, $a_n= a + (n - 1)d$
$\Rightarrow\ 84= 7 + (\text{n}-1)\frac{7}{2}$
$\Rightarrow\ 84=7+\frac{7\text{n}}{2}-\frac{7}{2}$
$\Rightarrow\ 84=\frac{14+7\text{n}-7}{2}$
$\Rightarrow\ 84\times2=7+7\text{n}$
$\Rightarrow\ 168=7+7\text{n}$
$\Rightarrow\ 7\text{n}=168-7$
$\Rightarrow\ \text{n}=\frac{161}{7}=23$
We know, Sum of n terms,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ \text{S}_{23}=\frac{23}{2}\Big[2(7)+(23-1)\frac{7}{2}\Big]$
$\Rightarrow\ \text{S}_{23}=\frac{23}{2}\Big[14+22\times\frac{7}{2}\Big]$
$\Rightarrow\ \text{S}_{23}=\frac{23}{2}[14+77]$
$\Rightarrow\ \text{S}_{23}=\frac{23}{2}\times91=\frac{2093}{2}$
Hecne, Sum of given A.P. is $\frac{2093}{2}$.
View full question & answer→Question 554 Marks
Find:
Which term in the A.P. $121, 117, 113, .....$ is its first negative term?
AnswerIn the given problem, we are given an A.P. and the value of one of its term. We need to find which term it is (n).
So here we will find the value of n using the formula, $a_n= a + (n - 1)d.$
Given,
A.P. $121, 117, 113, .....$
Here,
First term $= 121$
Difference $= 117 - 121 = -4$
We have to find which term of A.P. is its first negative term is then,
$a_n < 0$
We know, $n^{th}$ term of A.P.
$a + (n - 1)d < 0$
$⇒ 121 + (n - 1) -4 < 0$
$⇒ 121 + (-4n + 4) < 0$
$⇒ 121 - 4n + 4 < 0$
$⇒ 125 - 4n < 0$
$⇒ 4n > 125$
$\Rightarrow\ \text{n}>31\frac{1}{4}$
$\because\ \text{n}>31\frac{1}{4}$
$⇒ n = 32$
Hence, $32^{th}$ term of the given A.P. for getting first negative term.
View full question & answer→Question 564 Marks
Let there be an A.P. with first term ' $a$ ', common difference ' $d$ '. If $a_n$ denotes in $n^{\text {th }}$ term and $S_n$ the sum of first $n$ terms, find. $n$ and $a$, if $a_n=4, d=2$ and $S_n=-14$.
Answer$a_n=4, d=2, S_n=-14$
$\therefore a+(n-1) 2=4 \Rightarrow a+(n-1)^2=4$
$\Rightarrow a+2 n-2=4 \Rightarrow a+2 n=4+2$
$a+2 n=6 \ldots . .(\text { i })$
$S_n=\frac{n}{2}[a+1] \Rightarrow-14=\frac{n}{2}[a+4]$
$n(a+4)=-28 \ldots . . \text { (ii) }$
From (i) $a=6-2 n$
$\therefore \text { in (ii) }$
$n(6-2 n+4)=-28 \Rightarrow n(10-2 n)=-28$
$\Rightarrow 10 n-2 n^2=-28 \Rightarrow 2 n^2-10 n-28=0$
$\left.\Rightarrow n^2-5 n-14=0 \text { (Divisible by } 2\right)$
$\Rightarrow n^2-7 n+2 n-14=0$
$-14=-7 \times 2$
$-5=-7=2 \Rightarrow n(n-7)+2(n-7)=0$
$\Rightarrow(n-7)(n+2)=0$
Either $n-7=0$, then $n=7$
of $n +2=0$, then $n =-2$ but it is not possible being negative
$\therefore n=7$
Now $a =6-2 n =6-2 \times 7=6-14=-8$
Hence $a=-8, n=7$
View full question & answer→Question 574 Marks
Find the second term and $n ^{\text {th }}$ term of an $A.P.$ whose $6^{\text {th }}$ term is $12$ and $8^{\text {th }}$ term is $22 .$
AnswerIn the given problem, we are given $6^{\text {th }}$ and $8^{\text {th }}$ term of an A.P.
We need to find the $2^{\text {nd }}$ and $n ^{\text {th }}$ term,
Here, let us take the first term as a and the common difference as $d$,
We are given,
$a_6=12$
$a_8=22$
Now, we will find $a_6$ and $a_8$ using the formula $a_n=a+(n-1) d$
So,
$a_6=a+(6-1) d$
$12=a+5 d \ldots . .(i)$
Also,
$a_8=a+(8-1) d$
$22=a+7 d \ldots . . .(\text { (ii) }$
So, to solve for a and d
On subtracting (1) from (2), we get
$22-12=(a+7 d)-(a+5 d)$
$10=a+7 d-a-5 d$
$10=2 d$
$d=\frac{10}{2}$
$d=5 \ldots . . . \text { (iii) }$
Susbtituting (iii) in (i), we get
$12=a+5(5)$
$a=12-25$
$a=-13$
Thus,
$a=-13$
$d=5$
So, for the $2^{\text {nd }}$ term $(n=2)$,
$a_2=-13+(2-1) 5$
$=-13+(1) 5$
$=-13+5$
$=-8$
For the $n ^{\text {th }}$ term,
$a_n=-13+(n-1) 5$
$=-13+5 n-5$
$=-18+5 n$
Therefore, $a_2=-8, a_n=5 n-18$.
View full question & answer→Question 584 Marks
Show that the sequence defined by $a_n=5 n-7$ is an $A.P$, find its common difference.
AnswerIn the given problem, we need to show that the given sequence is an $A.P$. and then find its common difference. Here,
$a_n=5 n-7$
Now, to show that it is an $A.P$. Will find its few terms by substituting $n =1,2,3,4,5$
So,
Substituting $n =1$, we get
$a_1=5(1)-7$
$a_1=-2$
Subsituting $n =2$, we get
$a_2=5(2)-7$
$a_2=3$
Subsituting $n =3$, we get
$a_3 =5(3)-7$
$a_3 =8$
Subsituting $n =4$, we get
$a_4=5(4)-7$
$a_4=13$
Subsituting $n =5$, we get
$a_5=5(5)-7$
$a_5=18$
Further, for the given sequence to be an $A.P,$
We find the common difference $(d)$
$a=a_2-a_1=a_3-a_2$
thus,
$a_2-a_1=3-(-2)=5$
Since $a_2-a_1=a_3-a_2$
Hence, the given sequence is an $A.P$, and its common difference is $d=5$.
View full question & answer→Question 594 Marks
Find the sum:
$(-5) + (-8) + (-11) + ..... + (-230).$
AnswerGiven,
A.P. $(-5) + (-8) + ..... (-230)$
Here,
First term $a = -5$
Difference $d = -8 - (-5) = -8 + 5 = -3$
and Last term $a_n= -230$
We know $a_n= a + (n - 1)d$
$⇒ -230 = -5 + (n - 1)(-3)$
$⇒ -230 = -5 - 3n + 3$
$⇒ -230 = -2 - 3n$
$⇒ -228 = -3n$
$\Rightarrow\ \text{n}=\frac{-228}{-3}=76$
We know, sum of n terms,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ \text{S}_{76}=\frac{76}{2}[2(-5)+(76-1)(-3)]$
$ \Rightarrow S_{76}=38[-10+75 \times-3] $
$ \Rightarrow S_{76}=38[-10-225] $
$ \Rightarrow S_{76}=38[-235] $
$ \Rightarrow S_{76}=-8930$
Hence, Sum of given A.P. is -8930.
View full question & answer→Question 604 Marks
The sum of $4^{\text {th }}$ and $8^{\text {th }}$ terms of an $A.P.$ is $24$ and the sum of the $6^{\text {th }}$ and $10^{\text {th }}$ terms is $34 .$ Find the first term and the common difference of the $A.P.$
Answerlet a be the first term and d be the common difference, then
$a_4=a+(4-1) d=a+3 d$
$a_8=a+(8-1) d=a+7 d$
$\therefore a_4+a_8=a+3 d+a+7 d=24$
$\Rightarrow 2 a+10 d=24 \Rightarrow a+5 d=12 \ldots .(i)$
Similarly,
$a_6=a+5 d \text { and } a_{10}=a+9 d$
$\therefore a+5 d+a+9 d=34 \Rightarrow 2 a+14 d=34$
$\Rightarrow a+7 d=17 \ldots . \text { (ii) }$
Substracting $(i)$ from $(ii)$
$2\text{d}=5\Rightarrow\ \text{d}=\frac{5}{2}$
$\therefore\ \text{a}+5\text{d}=12$
$\Rightarrow \ \text{a}+\frac{5\times5}{2}=12$
$\Rightarrow\ \text{a}+\frac{25}{2}=12$
$\Rightarrow\ \text{a}=12-\frac{25}{2}=\frac{24-25}{2}=\frac{-1}{2}$
$\therefore\ \text{a}=\frac{-1}{2}\text{ and d}=\frac{5}{2}$
Hence first term $=\frac{-1}{2}$
and common difference $=\frac{5}{2}$
View full question & answer→Question 614 Marks
Which term of the arithmetic progression $8, 14, 20, 26, ...$ will be $72$ more than its $41^{st}$ term.
AnswerIn the given problem. let us first find the $41^{st}$ term of the given $A.P.$
A.P. is $8, 14, 20, 26, .....$
Here,
First term $(a) = 8$
Common difference of the A.P. $(d) = 14 - 8 = 6$
Now, as we know,
$a_n= a + (n - 1)d$
So, for $41^{st}$ term $(n = 41),$
$a_{41}= 8 + (41 - 1)(6)$
$= 8 + 40(6)$
$= 8 + 240$
$= 248$
Let us take the term which is $72$ more than the $41^{st}$ term as an. So,
$a_n= 72 + a_{41}$
$= 72 + 248$
$= 320$
Also, $a_n= a + (n - 1)d$
$320 = 8 + (n - 1)6$
$320 = 8 +6n - 6$
$320 = 2 + 6n$
$320 - 2 = 6n$
Further simplifying, we get,
$318 = 6n$
$\text{n}=\frac{318}{6}$
$n = 53$
Therefore, the $53^{nd}$ term of the given $A.P.$ is $72$ more than the $41^{st}$ term.
View full question & answer→Question 624 Marks
All integers between $100$ and $550$ which are not divisible by $9.$
AnswerThe sum of the integers between $100$ and $550$ which is not divisible by $9 =$ (Sum of total numbers between $100$ and $550) -$ (Sum of totel numbers between $100$ and $550$ which is divisible by $9)$
Here,
$a = 101, d = 102 - 101 = 1$ and $a_n = l = 549$
$\because$ $a_n = l = a + (n - 1)1$
$\Rightarrow 549 = 101 + (n - 1)1$
$\Rightarrow (n - 1) = 448 $
$\Rightarrow n = 449$
$\therefore$ Sum of terms between 100 and 550
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ \text{S}_{449}=\frac{449}{2}[2\times101+(449-1)1]$
$=\frac{449}{2}[202+448]=\frac{449}{2}\times650$
$= 449 \times325 =145,925$
No. below $100$ and $550$ which are divisible by $9$
$108, 117, 126, 135 ...... 540$
here $a=108, d=9, a_n=540$
Therefore,
$a_n=a+(n-1) d$
549 = 108 + (n - 1)9
549 = 108 = (n - 1)9
$=\frac{441}{9}=\text{n}-1$
$49 = n - 1$
$n = 50$
$\text{S}_{50}=\frac{50}{2}(108+549)\Big[\text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}+1)\Big]$
$\text{S}_{50}=\frac{50}{2}(657)$
$\text{S}_{50}=25\times657$
$\text{S}_{50}=16425$
So that from conditior
$= 145,925 - 16,425 = 129, 500$
Hence, the required sum is $129,500.$
View full question & answer→Question 634 Marks
The first and the last terms of an $A.P$. are $7$ and $49$ respectively. If sum of all its terms is $420$, find its common difference.
Answer$a = 7$
$a_n \text { or } I=49 \Rightarrow a_n=49 $
$ S_n=420$
Let d be the common difference
Now $I\left(a_n\right)=a+(n-1) d$
$⇒ 49 = 7 + (n - 1)d$
$⇒ (n - 1)d = 49 - 7 = 42 .....(i)$
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$420=\frac{\text{n}}{2}[2\times7+42]\ [\text{from (i)}]$
$420=\frac{\text{n}}{2}(14+42)$
$420=\frac{\text{n}}{2}(56)$
$\Rightarrow\ \text{n}=\frac{420\times2}{56}=15$
$\therefore\ \text{d}=\frac{42}{\text{n}-1}\ [\text{from (i)}]$
$=\frac{42}{15-1}=\frac{42}{14}=3$
$\therefore\ \text{d}=3$
View full question & answer→Question 644 Marks
Find the sum to n term of the $A.P.$ $5, 2, -1, -4, -7, ...,$
AnswerIn the given problem, we need to find the sum of the n terms of the given $A.P.$ $"5, 2, -1, -4, -7, ....."$
So, here we use the following formula for the sum of n terms of an $A.P.$
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
Where; $a =$ first term for the given $A.P.$
$d =$ common difference of the given $A.P.$
$n =$ number of terms
For the given $A.P.$ $(5, 2, -1, -4, -7, .....)$
Common difference of the $A.P.$ $(d) = a_2- a_1$
$= 2 - 5$
$= -3$
Number of terms $(n) = n$
First term for the given $A.P.$ $(a) = 5$
So, using the formula we get,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2(5)+(\text{n}-1)(-3)]$
$=\frac{\text{n}}{2}[10+(-3\text{n}+3)]$
$=\frac{\text{n}}{2}[10-3\text{n}+3]$
$=\frac{\text{n}}{2}[13-3\text{n}]$
Therefore, the sum of first n terms for the given $A.P.$ is $=\frac{\text{n}}{2}[13-3\text{n}]$.
View full question & answer→Question 654 Marks
Find the common difference of the $A.P$. and write the next two terms:
$75, 67, 59, 51, .....$
AnswerGiven,
$75,67,59,51, \ldots \ldots$
Here, $a_1-75, a_2=67, a_3=59, a_4=51$
Difference between terms
$d_1=a_2-a_1=67-75=-8$
$d_2=a_3-a_2=59-67=-8$
$\text { and } d_3=a_4-a_3=51-59=-8$
So, common difference is -8
Now, next two terms:
$a_n=a_{n-1}+d$
$5^{\text {th }} \text { term }$
$a_5=a_{5-1}+(-8)$
$=a_4-8$
$=51-8$
$\Rightarrow a_5=43$
$6^{\text {th }} \text { terms, }$
$a_6=a_{6-1}+(-8)$
$=a_5-8$
$=43-8$
$\Rightarrow a_6=35$
Hence, common differnce is $-8$ and next two terms $43$ and $35 .$
View full question & answer→Question 664 Marks
There are $25$ trees at equal distances of $5$ metres in a line with a well, the distance of the well from the nearest tree being $10$ metres. A gardener water all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to eater all the trees.
AnswerNumber of trees = 25
Distance between one to other tree $= 5m$
Distance between first near and the well $= 10m$
Now in order to water the first tree, the gardener has to cover $10m + 10m = 20m$
And to water the second tree, the distance to covered is $15 + 15 = 30 m$
To water the third tree, the distance to cover is $= 20 + 20 = 40 m$
The series will be $20, 30, 40, .....$
where a $= 20, d = 30 - 20 = 10$ and $n = 25$
Total distance was covered
$\text{S}_{25}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$=\frac{25}{2}[2\times20+(25-1)\times10]$
$=\frac{25}{2}[40+24+10]=\frac{25}{2}[40+240]$
$=\frac{25}{2}\times280=25\times140=3500\text{m}$ View full question & answer→Question 674 Marks
In an $A.P$., the first term is $22$, $n^{th}$ term is $-11$ and the sum to first n terms is $66$. Find n and d, the common difference.
AnswerIn the given problem, we have the first and the $n^{th}$ term of an $A.P$. along with the sum of the n terms of $$A.P.$$ Here, we need to find the number of terms and the common difference of the $A.P.$
Here,
The first term of the $A.P$ $(a) = 22$
The $n^{th}$ term of the $A.P$ $(l) = -11$
Sum of all the terms $s_n= 66$
Let the common difference of the $A.P.$ be d.
So, let us first find the number of the terms (n) using the formula,
$66=\Big(\frac{\text{n}}{2}\Big)[22+(-11)]$
$66=\Big(\frac{\text{n}}{2}\Big)(22-11)$
$(66)(2)=(\text{n})(11)$
Further, Solving for n
$\text{n}=\frac{(66)(2)}{11}$
$n = (6)(2)$
$n = 12$
Now, to find the common difference of the $A.P.$ we use the following formula,
$l = a + (n - 1)d$
We get
$-11 = 22 + (12 - 1)d$
$-11 = 22 + (11)d$
$\frac{-11-22}{11}=\text{d}$
Further, Solvinf for d,
$\text{d}=\frac{-33}{11}$
$\text{d}=-3$
Therefore, the number of terms is $n = 12$ and the common difference of the $A.P.$ $d = -3.$
View full question & answer→Question 684 Marks
If the sum of first n terms of an is $\frac{1}{2}(3\text{n}^2+7\text{n})$, then find its $n^{th}$ term. Hence write its $20^{th}$ term.
Answer$\text{S}_\text{n}=\frac{1}{2}[3\text{n}^2+7\text{n}]$
Then, $\text{S}_{\text{n}-1}=\frac{1}{2}[3(\text{n}-1)^2+7(\text{n}-1)]$
$\text{T}_\text{n}=\text{S}_\text{n}-\text{S}_{\text{n}-1}$
$=\frac{1}{2}[3\text{n}^2+7\text{n})-\frac{1}{2}[3(\text{n}-1)^2+7(\text{n}-1)]$
$=\frac{1}{2}[(3\text{n}^2+7\text{n})-[3(\text{n}^2-2\text{n}+1)+7\text{n}-7]$
$=\frac{1}{2}[3\text{n}^2+7\text{n}-3\text{n}^2+6\text{n}-3-7\text{n} +7]$
$=\frac{1}{2}[6\text{n}+4]=3\text{n}+2$
and $\text{T}_{20}=3(20)+2=60+2=62$
View full question & answer→Question 694 Marks
Find the sum of,
All $3$ - digit natural numbers which are divisible by $13.$
AnswerSo, we know that the first $3$ digit multiple of $13$ is $104$ and the last $3$ digit multiple of $13$ is $988.$
Also, all these terms will from an $A.P.$ with the common difference of $13$.
So here,
First term $(a) = 104$
Last term $(l) = 988$
Common difference $(d) = 13$
So, here the first step is to find total number of terms. Let us take the number of terms as n.
Now, as we know,
$a_n= a + (n - 1)d$
So, for the last term,
$988 = 104 + (n - 1)13$
$988 = 104 + 13n - 13$
$988 = 91 + 13n$
Further simplifying,
$\text{n}=\frac{988-91}{13}$
$\text{n}=\frac{897}{13}$
$\text{n}=69$
Now using the formula for the sum of n terms, we get
Now, using the formula for the sum of n terms, we get
$\text{S}_\text{n}=\frac{69}{2}[2(104)+(69-1)13]$
$=\frac{69}{2}[208+(68)13]$
$=\frac{69}{2}(208+884)$
On further simplification, we get
$\text{S}_\text{n}=\frac{69}{2}(1092)$
$=69(546)$
$=37674$
Therefore, the sum of all $3$ digit multiples of $13$ is $S_n= 37674.$
View full question & answer→Question 704 Marks
Let there be an $A.P$. with first term 'a', common difference ' $d$ '. If $a_n$ denotes in $n^{\text {th }}$ term and $S_n$ the sum of first $n$ terms, find.
$n$ and $S_n$, if $a=5, d=3$ and $a_n=50$.
AnswerHere, we have an $A.P$. whose $n ^{\text {th }}$ term ( $a _{ n }$ ), first term $(a)$ and common difference $(d)$ are given, We need to find the number of terms $(n)$ and the sum of first $n$ terms $\left(S_n\right)$.
Here,
First term $(a)=25$
Last term $\left(a_n\right)=50$
Common difference $( d )=3$
So here we will find the value of $n$ using the formula, an $=a+(n-1) d$
So, substituting the values in the above mentioned formula
$50=5+(n-1) 3$
$50=5+3 n-3$
$50=2+3 n$
$3 n=50-2$
Furhter simplifying for n ,
3n = 48
$\text{n}=\frac{48}{3}$
$n = 16$
Now, here we can find the sum of the n terms of the given $A.P$., using the formula,
$\text{S}_\text{n}=\Big(\frac{\text{n}}{2}\Big)(\text{a}+\text{l})$
Where, a = the first term
l = the last term
So, for the givne $A.P$, on substituting the values in the formula for the sum of n terms of an $A.P$., we get,
$\text{S}_{16}=\Big(\frac{16}{2}\Big)[5+50]$
$= 8(55)$
$= 440$
Therefore, for the given $A.P. n = 16$ and $Sn = 440$
View full question & answer→Question 714 Marks
Find the term of the arithmetic progression $9, 12, 15, 18, .....$ which is $39$ more than its $36^{th}$ term.
AnswerGiven,
A.P. $9, 12, 15, 18$
Here,
First term $a = 9$
Difference $d = 12 - 9 = 3$
and Last term $a_n$
We know, $a_n= a + (n - 1)d$
then $a_n= 9 + (n - 1)3$
$\Rightarrow a_n= 9 + 3n - 3$
$\Rightarrow a_n= 6 + 3n .....(i)$
Let
$36^{th}$ term $a_{36}= 9 + (36 - 1)3$
$= 9 + 35 \times 3$
$= 9 + 105$
$\Rightarrow a_{36}= 114$
Now, term is 39 more then $36^{th}$ term
$\Rightarrow a_n= 39 + a_{36}$
$\Rightarrow a_n= 39 + 114$
$\Rightarrow a_n= 153.$
Put in eq. (i)
$\Rightarrow 153 = 6 + 3n$
$\Rightarrow 3n = 153 - 6$
$\Rightarrow 3n = 147$
$\Rightarrow\ \text{n}=\frac{147}{3}$
$\Rightarrow n = 49$
Hence, $49^{th}$ term of given $A.P$. is $39$ more than its $36^{th}$ term.
View full question & answer→Question 724 Marks
Which term of the $A.P. 3, 10, 17, ..$. will be $84$ more than its $13^{th}$ term?
AnswerGiven A.P., $3, 10, 7, .....$
First term $a = 3$
and Difference $d = 10 - 3 = 7$
We know, $a_n=a+(n-1) d$
Let,
$13^{\text {th }}$ term, $a_{13}=3+(13-1) 7$
$=3+12 \times 7$
$=3+84$
$\Rightarrow \mathrm{a}_{13}=87$
Now, $\mathrm{n}^{\text {th }}$ term is more than 84
$ \Rightarrow a_n=84+a_{13} $
$ \Rightarrow a_n=84+87 $
$ \Rightarrow a_n=171$
Now, we have to find term.
$a_n=a+(n-1) d $
$ \Rightarrow 171=3+(n-1) 7 $
$ \Rightarrow 171=3+7 n-7 $
$ \Rightarrow 7 n=175 $
$ \Rightarrow n=25$
Hence, $25^{\text {th }}$ term of the given A.P. is 84 more than its $13^{\text {th }}$ term.
View full question & answer→Question 734 Marks
A man is employed to count Rs. $10710$. He counts at the rate of Rs. $180$ per minute for half an hour. After this he counts at the rate of Rs. $3 $less every minute than the preceding minute. Find the time taken by him to count the entire amount.
AnswerTotal amount to be counted = Rs. $10710$
Amount count for first half an hour ($30$ minutes) at the rate of Rs. 180 per minute $= 180 \times 30 = Rs. 5310$
After half hour,
Let a be the first term and d be the common difference, then
$a = 180 - 3 = 177, d = -3$ and $S_n= 5310$
But $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ 5310=\frac{\text{n}}{2}[2\times177+(\text{n}-1)\times-3]$
$ 10620=n[354-3 n+3]$
$10620=n(357-3 n) \Rightarrow 10620=357 n-3 n^2 $
$ \Rightarrow 3 n^2-357 n+10620=0 $
$ \left.n^2-119 n+3540=0 \text { (Dividing by } 3\right) $
$ \Rightarrow n^2-59 n-60 n+3540=0$
$\begin{Bmatrix}\because3540 = -59\times(-60)-119 = -59 - 60 \end{Bmatrix}$
$\Rightarrow n(n - 59) - 60(n - 59) = 0$
$\Rightarrow (n - 59) (n - 60) = 0$
Either $n - 59 = 0$, then $n - 59$ or $n - 60 = 0,$ then $n = 60$
Total time $= 59 + 30 = 89$ minutes or $= 60 + 30 = 90$ minutes.
View full question & answer→Question 744 Marks
If the $10^{th}$ term of an $A.P.$ is $21$ and the sum of its first $10$ terms is $120$, find its $n^{th}$ term.
AnswerWe know that, sum of first n term $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
and $n^{th}$ term $= a_n= a + (n - 1)d$
Now,
$\text{S}_{10}=\frac{10}{2}[2\text{a}+(10- 1)\text{d}]$
$\Rightarrow 120 = 5(2a + 9d)$
$\Rightarrow 24 = 2a + 9d$
$\Rightarrow 2a + 9d = 24 .....(i)$
Also,
$a_{10} = a + (10 - 1)d$
$\Rightarrow 21 = a + 9d$
$\Rightarrow 2a + 18d = 42 .....(ii)$
Subtracting (i) from (ii), we get
$18d - 9d = 42 - 24$
$\Rightarrow 9d = 18$
$\Rightarrow d = 2$
$\Rightarrow 2a = 24 - 9d [$From$ (i)]$
$\Rightarrow 2a = 24 - 9 \times 2$
$\Rightarrow 2a = 24 - 18$
$\Rightarrow 2a = 6$
$\Rightarrow a = 3$
Also,
$a_n= a + (n - 1)d$
$= 3 + (n - 1)2$
$= 3 + 2n - 2$
$\Rightarrow 1 + 2n$
Thus, $n^{th}$ term if this A.P. is $1 + 2n.$
View full question & answer→Question 754 Marks
How many terms are there in the $A.P$. whose first and fifth term are $-14$ and $2$ respectively and the sum if the terms is $40?$
AnswerFirst term of an $A.P. = -14$
and Fifth term $= 2$
Sum of terms $= 40$
Let n be the number of terms, then
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$ a_5=2$
$ \Rightarrow a_5=a+(5-1) d $
$ \Rightarrow 2=-14+4 d \Rightarrow 4 d=14+2=16 $
$ \Rightarrow d=\frac{16}{4}=4$
$ \text { Now, } \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}] $
$ \Rightarrow 40=\frac{\mathrm{n}}{2}[2 \times(-14)+(\mathrm{n}-1) \times 4] $
$ \Rightarrow 80=\mathrm{n}(-28+4 \mathrm{n}-4)$
$ \Rightarrow 80=\mathrm{n}(-32+4 \mathrm{n}) $
$ \Rightarrow 80=-32 \mathrm{n}+4 \mathrm{n}^2 $
$ \Rightarrow 4 \mathrm{n}^2-32 \mathrm{n}-80=0 $
$ \left.\Rightarrow \mathrm{n}^2-8 \mathrm{n}-20=0 \text { (Dividing by } 4\right) $
$ \Rightarrow \mathrm{n}^2-10 \mathrm{n}+2 \mathrm{n}-20=0$
$\begin{Bmatrix}\therefore\ -20 = - 10 \times2 \\ -8 = -10 + 2 \end{Bmatrix}$
$⇒ n(n - 10) + 2(n - 10) = 0$
$⇒ (n - 10)(n + 2) = 0$
Either $n - 10 = 0,$ then $n = 10$
or $n + 2 = 0 ⇒ n = -2$ but it is not possible being negative
$\therefore$ Number of term $= 10.$
View full question & answer→Question 764 Marks
If $(m+1)^{\text {th }}$ term of an $A.P$ is twice the $(n+1)^{\text {th }}$ term, prove that $(3 m+1)^{\text {th }}$ term is twice the $(m+n+1)^{\text {th }}$ term.
AnswerLet $a, a+d, a+2 d, a+3 d, \ldots .$. is an A.P.
$ \therefore(m+1)^{\text {th }} \text { term }=a+(m+1-1) d $
$ =a+m d $
$ a n d(n+1)^{\text {th }} \text { term }=a+(n+1-1) d $
$ =a+n d $
$ \because(m+1)^{\text {th }} \text { term }=2(n+1)^{\text {th }} \text { term } $
$ \therefore a+m d=2(a+n d) $
$ \Rightarrow a+m d=2 a+2 n d \Rightarrow 2 a-a=m d-2 n d $
$ a=d(m-2 n)=(m-2 n) d \ldots . .(\text { (i) }$
Now $(3 m+1)^{\text {th }}$ term $=a+(3 m+1-1) d$
$ =a+3 m d=(m-2 n) d+3 m d $
$ =(m-2 n+3 m) d=(4 m-2 n) d=2(2 m-n) d \ldots $
$ \text { and }(m+n+1)^{\text {th }} \text { term }=a+(m+n+1-1) d $
$ =a+(m+n) d $
$ =(m-2 n) d+(m+n) d $
$ =(m-2 n+m+n) d $
$ =(2 m-n) d \ldots . . .(i i i)$
From (ii) and (iii)
$(3 m+1)^{\text {th }}$ term $=2(m+n+1)^{\text {th }}$ term.
Hence proved.
View full question & answer→Question 774 Marks
Find the sum of n terms of the series $\Big(4-\frac{1}{\text{n}}\Big)+\Big(4-\frac{2}{\text{n}}\Big)+\Big(4-\frac{3}{\text{n}}\Big)+\ .....$
AnswerLet the given series be $\text{S}=\Big(4-\frac{1}{\text{n}}\Big)+\Big(4-\frac{2}{\text{n}}\Big)+\Big(4-\frac{3}{\text{n}}\Big)+\ .....$
$=[4+4+4+\ .....]-\Big[\frac{1}{\text{n}}+\frac{2}{\text{n}}+\frac{3}{\text{n}}+\ .....\Big]$
$=4[1+1+1+\ .....]-\frac{1}{\text{n}}[1+2+3+\ .....]$
$=\text{S}_1-\text{S}_2$
$\text{S}_1=4[1+1+1+\ .....]$
$\text{a}=1,\text{d}=0$
$\text{S}_2=\frac{1}{\text{n}}[2\times1+(\text{n}-1)\times0]$
$\Big(\text{S}_\text{n}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})\Big)$
$\Rightarrow\ \text{S}_1=4\text{n}$
$\text{S}_2=\frac{1}{\text{n}}[1+2+3+\ .....]$
$\text{a}=1,\text{d}=2-1=1$
$\text{S}_2=\frac{1}{\text{n}}\times\frac{\text{n}}{2}[2\times1+(\text{n}-1)\times1]$
$=\frac{1}{2}[2+\text{n}-1]$
$=\frac{1}{2}[1+\text{n}]$
Thus, $\text{S}=\text{S}_1-\text{S}_2=4\text{n}-\frac{1}{2}[1+\text{n}]$
$\text{S}=\frac{8\text{n}-1-\text{n}}{2}=\frac{7\text{n}-1}{2}$
Hence, the sum of n terms of the series is $\frac{7\text{n}-1}{2}$.
View full question & answer→