Question 14 Marks
How many terms are there in the A.P.?
$-1,\frac{5}{6},\frac{2}{3},\frac{1}{2}, .....\frac{10}{3}.$
AnswerHere, A.P. is $-1,\frac{5}{6},\frac{2}{3},\frac{1}{2}, .....\frac{10}{3}.$
The first term $(a) = -1$
The last term $(\text{a}_\text{n})=\frac{10}{3}$
Now,
Common difference $(d) = a_1 - a$
$=-\frac{5}{6}-(-1)$
$=-\frac{5}{6}+1$
$=\frac{-5+6}{6}$
$=\frac{1}{6}$
Thus, using the above mentioned formula, we get,
$\frac{10}{3}=-1+(\text{n}-1)\frac{1}{6}$
$\frac{10}{3}+1=\frac{1}{6}\text{n}-\frac{1}{6}$
$\frac{13}{3}+\frac{1}{6}=\frac{1}{6}\text{n}$
Further solving for n, we get
$\frac{26+1}{6}=\frac{1}{6}\text{n}$
$\text{n}=\frac{27}{6}(6)$
$\text{n}=27$
Thus, $n = 27$
Therefore, the number of terms present in the given A.P. is $27$.
View full question & answer→Question 24 Marks
The $24^{\text {th }}$ term of an A.P. is twice its $10^{\text {th }}$ term. Show that its $72^{\text {nd }}$ term is $4$ times its $15^{\text {th }}$ term.
AnswerLet $a$ be the first term and $d$ be the common difference.
We know that, $n^{\text {th }}$ term $=a_n=a+(n-1) d$
According to the question,
$a_{24}=2 a_{10}$
$\Rightarrow a+(24-1) d=2(a+(10-1) d)$
$\Rightarrow a+23 d=2 a+18 d$
$\Rightarrow 23 d-18 d=2 a-a$
$\Rightarrow 5 d=a$
$\Rightarrow a=5 d$
Also,
$a_{72}=a+(72-1) d$
$a_{72}=5 d=71 d[a=5 d \text { From (i)] }$
$=76 d \text {.....(ii) }$
$\text { and } a_{15}=a+(15-1) d$
$=5 d+14 d[\text { from (i)] }$
$=19 d \text {....(iii) }$
On comparing (ii) and (iii), we get
$76 d=4 \times 19 d$
$\Rightarrow a_{72}=4 \times a_{15}$
Thus, $72^{\text {nd }}$ term of the given A.P. is 4 times its $15^{\text {th }}$ term.
View full question & answer→Question 34 Marks
All integers from $1$ to $500$ which are multiplies of $2$ or $5$.
AnswerSince, multiples of $2$ ot $5=$ Multiple of $2+$ Multiples of $5$ - Multiples of LCM $(2,5)$ i.e., $10$.
$\therefore$ MUltiples of $2$ or $5$ from $1$ to $500$
= List of multiples of $2$ from $1$ to $500+$ list of multiple of $5$ from $1$ to $500$ - list of multiple of $10$ from $1$ to $500$ $=(2,4,6, \ldots . ., $500$)+(5,10,15, \ldots . ., $500$)-(10,20, \ldots . ., $500$)$
All of these list from an A.P.
Now, number of terms in first list,
$500=2+\left(n_1-1\right) 2 \Rightarrow 498=\left(n_1-1\right) 2$
$\Rightarrow n_1-1=249 \Rightarrow n_1=250$
Number of terms in second list,
$500=5+\left(n_2-1\right) 5 \Rightarrow 495=\left(n_2-1\right) 5$
$\Rightarrow 99=\left(n_2-1\right) \Rightarrow n_2=100$
and number of terms in third list,
$500=10+\left(n_3-1\right) 10 \Rightarrow 490=\left(n_3-1\right) 10$
$\Rightarrow n_3-1=49 \Rightarrow n_3=50$
From Eq. (i), Sum of multiples of $2$ or $5$ from $1$ to $500$
$=\text { Sum of }(2,4,6, \ldots ., 500)+\operatorname{Sum} \text { of }(5,10, \ldots . .500)-\text { Sum of }(10,20, \ldots ., 500)$
$=\frac{n_1}{2}[2+500]+\frac{n_2}{2}[5+500]-\frac{n_3}{2}[10+500]$
${\left[\because S_{n}=\frac{n}{2}(a+1)\right]}$
$=\frac{250}{2} \times 502=\frac{100}{2} \times 505-\frac{50}{2} \times 510$
$=250 \times 251+505 \times 50-25 \times 510$
$=62750+25250-12750$
$=88000-12750$
$=75250$
View full question & answer→Question 44 Marks
The $26^{th}, 11^{th}$ and last term of an A.P. are $0, 3$ and $-\frac{1}{5},$ respectively. Find the common difference and the number of terms.
AnswerLet the first term, common difference and number of terms of an A.P. are a, d and n, respectively.
We know that, if last term of an A.P. is known, then $l = a + (n – 1) d ……(i)$ and
$n^{th}$ term of an A.P is $T_n = a + (n - 1)d .....(ii)$
Given that, $26^{th}$ term of an A.P. $= 0$
$\Rightarrow T_{26} = a + (26 - 1)d = 0 [$From eq. $(i)] $
$\Rightarrow a + 25d = 0 .....(iii)$
$11^{th}$ term of an A.P. $= 3$
$\Rightarrow T_{11} = a + (11 - 1)d = 3 [$From eq. $(ii)]$
$\Rightarrow a + 10d = 3 .....(iv)$ and last term of an A.P. $=-\frac{1}{5}$
$\Rightarrow l = a + (n - 1)d [$From eq. $(i)]$
$\Rightarrow\ -\frac{1}{5}=\text{a}+(\text{n}-1)\text{d}$
Now, subtracting Eq. (iv) from Eq. (iii),
$\Rightarrow\ \text{d}=-\frac{1}{5}$ Put the value of d in Eq. (iii),
we get $\text{a}+25\Big(-\frac{1}{5}\Big)=0$
$\Rightarrow a - 5 = 0 $
$\Rightarrow a = 5$
Now, put the value of a, d in Eq. (v),
we get $-\frac{1}{5}=5+(\text{n}-1)\Big(-\frac{1}{5}\Big)$
$\Rightarrow -1 = 25 - (n - 1) $
$\Rightarrow -1 = 25 - n + 1 $
$\Rightarrow n = 25 + 2 = 27$
Hencem the common difference and number of terms are $-\frac{1}{5}$ and $27$, respectively. View full question & answer→Question 54 Marks
Find the sum of the following arithmetic progressions:$\frac{\text{x}-\text{y}}{\text{x}+\text{y}}\frac{3\text{x}-2\text{y}}{\text{x}+\text{y}}\frac{5\text{x}-3\text{y}}{\text{x}+\text{y}}, .....\text{ to n terms.}$
AnswerIn an A.P. let first term = a, common difference = d, and there are n terms. Then, sum of n terms is,
$\text{S}_\text{n}=\frac{\text{n}}{2}\{2\text{a} + (\text{n} - 1)\text{d}\}$
Given, $\frac{\text{x}-\text{y}}{\text{x}+\text{y}}\frac{3\text{x}-2\text{y}}{\text{x}+\text{y}}\frac{5\text{x}-3\text{y}}{\text{x}+\text{y}}, .....$
Here,
First term $\text{a}=\frac{\text{x}-\text{y}}{\text{x}+\text{y}}$
Difference $\text{d}=\frac{3\text{x}-2\text{y}}{\text{x}+\text{y}}-\frac{\text{x}-\text{y}}{\text{x}+\text{y}}$
$\Rightarrow\ \text{d}=\frac{3\text{x}-2\text{y}-(\text{x}-\text{y})}{\text{x}+\text{y}}$
$\Rightarrow\ \text{d}=\frac{3\text{x}-2\text{y}-\text{x}+\text{y}}{\text{x}+\text{y}}$
$\Rightarrow\ \text{d}=\frac{2\text{x}-\text{y}}{\text{x}+\text{y}}$
No of terms n = n
We know, $\text{S}_\text{n}=\frac{\text{n}}{2}\{2\text{a} + (\text{n} - 1)\text{d}\}$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}\bigg[2\Big(\frac{\text{x}-\text{y}}{\text{x}+\text{y}}+(\text{n}-1)\frac{(2\text{x}-\text{y})}{(\text{x}+\text{y})}\bigg]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}\bigg[\frac{2\text{x}-2\text{y}}{\text{x}+\text{y}}+\frac{\text{n}(2\text{x}-\text{y})-1(2\text{x}-\text{y})}{\text{x}+\text{y}}\bigg]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}\bigg[\frac{2\text{x}-2\text{y}+\text{n}(2\text{x}-\text{y})-2\text{x}+\text{y}}{\text{x}+\text{y}}\bigg]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}\bigg[\frac{\text{n}(2\text{x}-\text{y})-\text{y}}{\text{x}+\text{y}}\bigg]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2(\text{x}+\text{y})}[\text{n}(2\text{x}-\text{y})-\text{y}]$
Hence, sum of n terms is $\frac{\text{n}}{2(\text{x}+\text{y})}[\text{n}(2\text{x}-\text{y})-\text{y}]$.
View full question & answer→Question 64 Marks
The sum of first n terms of an A.P is $5n^2 + 3n$. If its $m^{th}$ terms is $168$, find the value of m. Also, find the $20^{th}$ term of this A.P.
Answer$S_n = 5n^2 + 3n, T_m = 168$
$S_1 = 5(1)^2 + 3 \times 1 = 5 + 3 = 8$
$S_2 = 5(2)^2 + 3 \times 2 = 20 + 6 = 26$
$\therefore$ $T_2 = 26 - 8 = 18$ {$\because$ $T_2 = (S_2 - S_1)$}
$d = T_2 - T_1$
$d = 18 - 8 = 10, a = 8$
$\therefore$ $T_m = a + (m - 1)d$
$168 = 8 + (m - 1) \times 10$
$\Rightarrow 168 - 8 = (m - 1) \times 10$
$160 = 10(m - 1)$
$\frac{160}{10}=\text{m}-1$
$\Rightarrow m - 1 = 16$
$m = 16 + 1 = 17$
$\therefore$ $m = 17$
Now,$T_{20} = a + (20 - 1)d$
$= 8 + 19 \times 10 = 8 + 190 = 198$.
View full question & answer→Question 74 Marks
Write the first five terms of the following sequences whose $n^{th}$ terms are:
$a_n = n^2 - n + 1.$
Answer$a_n = n^2 - n + 1$.Here, the $n^{th}$ term is givne by the above expression. So, to find the first term we use $n = 1$, we get,
$a_1 = (1)^2 - (1) + 1$
$= 1 - 1 + 1$
$= 1$
Similarly, we find the other four terms,
Second term $(n = 2),$
$a_2 = (2)^3 - (2) + 1$
$= 4 - 2 + 1$
$= 3$
Third term $(n = 3),$
$a_3 = (3)^2 - (3) + 1$
$= 9 - 3 + 1$
$= 7$
Fourth term $(n = 4),$
$a_4 = (4)^2 - (4) + 1$
$= 16 - 4 + 1$
$= 13$
Fifth term $(n = 5),$
$a_5 = (5)^2 - (5) + 1$
$= 25 - 5 + 1$
$= 21$
Therefore, the first terms for the given sequence are $a_1 = 1, a_2 = 3, a_3 = 7, a_4 = 13, a_5 = 21$. View full question & answer→Question 84 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 d , if $a =8, a _{ n }=62$ and $S _{ n }=210$.
AnswerHere, we have an A.P. Whose $n^{th}$ term $(a_n)$, Sum of first n terms $(S_n)$ and first term $(a)$ are given. We need to find the number of terms $(n)$ and the common difference $(d)$.
Here,
First term $(a) = 8$
Last term $(a_n) = 62$
Sum of n terms $(S_n) = 210$
Now, here the sum of the n terms is given by 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 given A.P. on substituting the values in the formula for the sum of n terms of an A.P., we get,
$210=\Big(\frac{\text{n}}{2}\Big)[8+62]$
$210(2)=\text{n}(70)$
$\text{n}=\frac{420}{70}$
$\text{n}=6$
Also, here we will find the value of d using the formula.
$a_n = a + (n - 1)d$
So, Substituting the values in the above mentioned formula
$62 = 8 + (6 - 1)d$
$62 - 8 = (5)d$
$\frac{54}{5}=\text{d}$
$\text{d}=\frac{54}{5}$
Therefore, for the given A.P. $n = 6$ and $\text{d}=\frac{54}{5}$.
View full question & answer→Question 94 Marks
The first term of an A.P. is $5$ and its $100^{th}$ term is $-292$. Find the $50^{th}$ term of this A.P.
AnswerGiven
First term $a = 5$
and $100^{th}$ term $a_{100} = -292$
We know $n^{th}$ term of A.P.
$a_n = a + (n - 1)d$
then,
$\Rightarrow a_{100} = 5 + (100 - 1)d$
$\Rightarrow -292 = 5 + 99d$
$\Rightarrow 99d = -297$
$\Rightarrow\ \text{d}=\frac{-297}{99}=\frac{-27}{9}$
$\Rightarrow d = -3$
Now, we have to find $50^{th}$ term, then
$\Rightarrow a_{50} = a + (50 - 1)d$
$\Rightarrow a_{50} = 5 + 49 \times (-3)$
$\Rightarrow a_{50} = 5 - 147$
$\Rightarrow a_{50} = 142$
Hence, $50^{th}$ term of given A.P. is $-142$.
View full question & answer→Question 104 Marks
The first term of an A.P. is $2$ and the last term is $50$. The sum of all these terms is $442$. Find the common difference.
AnswerIn the given problem, we have the first and the last term of an A.P. along with the sum of all the terms of A.P. Here, we need to find the common difference of the A.P.
Here,
The first term of the A.P $(a) = 2$
The last term of the A.P $(l) = 50$
Sum of all the terms $S_n = 442$
Let the common difference of the A.P. be d.
So, let us first find the number of the terms (n) using the formula,
$442=\Big(\frac{\text{n}}{2}\Big)(2+50)$
$442=\Big(\frac{\text{n}}{2}\Big)(52)$
$442=({\text{n}})(26)$
$\text{n}=\frac{442}{26}$
$\text{n}=17$
Now, to find the common difference of the A.P. we use the following formula,
$l = a + (n - 1)d$
We get,
$50 = 2 + (17 - 1)d$
$50 = 2 + 17d - d$
$50 = 2 + 16d$
$\frac{50-2}{16}=\text{d}$
Further, solving For d,
$\text{d}=\frac{48}{16}$
$\text{d}=3$
Therefore the common difference of the A.P. $d = 3.$
View full question & answer→Question 114 Marks
Three numbers are in A.P. If the sum of these numbers be $27$ and the product $648$, find the numbers.
AnswerLet,
Three numbers are $(a - d), a, (a + d)$
Sum of three numbers,
$\Rightarrow (a - d) + a + (a + d) = 27$
$\Rightarrow a - d + a + a + d = 27$
$\Rightarrow 3a = 27$
$\Rightarrow a = 9$
Product of three numbers,
$\Rightarrow (a - d) a(a + d) = 648$
$\Rightarrow (a^2 - d^2)a = 648$
$\Rightarrow [(9)^2 - d^2]9 = 648$
$\Rightarrow\ 81-\text{d}^2=\frac{648}{9}$
$\Rightarrow 81 - d^2 = 72$
$\Rightarrow 81 - 72 = d^2$
$\Rightarrow 9 = d^2$
$\Rightarrow\ \text{d}=\sqrt{9}$
$\Rightarrow d = 3$
Now, the numbers are,
$(a - d), a, (a + d)$
$\Rightarrow (9 - 3), 9, (9 + 3)$
$\Rightarrow 6, 9, 12$
Hence, numbers are $6, 9, 12.$
View full question & answer→Question 124 Marks
Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.
AnswerHere, we are given that four number are in A.P., such that there sum is 50 and the greater number is 4 times the smallest.
So let us take the four terms as a - d, a, a + d, a + 2d.
Now, we are given that sum of these numbers is 50, so we get,
(a - d) + (a) + (a + d) + (a + 2d) = 50
a - d + a + a + d + a + 2d = 50
4a + 2d = 50
2a + d = 25 .....(i)
Also, the greater numbets is 4 times the smallest, so we get,
a + 2d = 4(a - d)
a + 2d = 4a - 4d
4d + 2d = 4a - a
6d = 3a
$\text{d}=\frac{3}{6}\text{a}\ .....\text{(ii)}$
Now using (ii) in (i), we get,
$2\text{a}+\frac{3}{6}\text{a}=25$
$\frac{12\text{a}+3\text{a}}{6}=25$
$15\text{a}=150$
$\text{a}=\frac{150}{15}$
$\text{a}=10$
Now, using the value of a in (ii), we get
$\text{d}=\frac{3}{6}(10)$
$\text{d}=\frac{10}{2}$
$\text{d}=5$
So, first term is given by,
a - d = 10 - 5
= 5
Second term is given by,
a = 10
Third term is given by,
a + d = 10 + 5
= 15
Fourth term is given by,
a + 2d = 10 + (2)(5)
= 10 + 10
= 20
Therefore, the four terms are 5, 10, 15, 20.
View full question & answer→Question 134 Marks
A man saved $Rs. 16500$ in ten years. In each year after the first he saved $Rs. 100$ more than he did in the preceding year.
How much did he save in the first year?
AnswerHere,
We are given that the total saving of a man is $Rs. 16500$ and every year he saved $Rs. 100$ more than the previous year.
So, let us take the first installment as a.
Second installment $= a + 100$
Third installment $= a + 100 + 100$
So, there installment will from an A.P. with the common difference $(d) = 100$
The sum of his savings every year $S_n = 16500$
Number of years $(n) = 10$
So, to find the first installment, 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 = 10, we get,
$\text{S}_{10}=\frac{10}{2}[2(\text{a})+(10-1)(100)]$
$16500 = 5[2a + (9)(100)]$
$16500 = 10a + 4500$
$16500 - 4500 = 10a$
Further solving for a,
$10a = 12000$
$a = Rs. 1200$
Therefore, man saved $Rs. 1200$ in the first year.
View full question & answer→Question 144 Marks
Write the expression $a_n - a_k$ for the A.P. $a, a + d, a + 2d, ...$
Hence, find the common difference of the A.P. for which,
$20^{th}$ term is $10$ more than the $18^{th}$ term.
AnswerWe know,
$a_n = a + (n - 1)d$
Let,
$n^{th}$ term $a_n = a + (n - 1)d$
$\Rightarrow a_n = a + nd - d$
$k^{th}$ term, $a_k = a + (k - 1)d$
$\Rightarrow a_k = a + kd - d$
Now,
$\Rightarrow a_n - a_k = (a + nd - d) - (a + kd - d)$
$\Rightarrow = a + nd - d - a - kd + d$
$\Rightarrow = nd - kd$
$\Rightarrow = d(n - k)$
Given,
$a_{20} = 10 + a_{18} .....(i)$
We know, $a_n = a+ (n - 1)d$
Then, $20^{th}$ term, $a_{20} = a + (20 - 1)d$
$\Rightarrow a_{20} = a + 19d$
$18^{th}$ term, $a_{18} = a + (18 - 1)d$
$\Rightarrow a_{18} = a + 17d$
By putting value of $a20$ and $a18$
$\Rightarrow a + 19d = 10 + a + 17d$
$\Rightarrow 19d = 17d + 10 + a - a$
$\Rightarrow 19d - 17d = 10$
$\Rightarrow 2d = 10$
$\Rightarrow d = 5$
Hence, Common differecne is $5$.
View full question & answer→Question 154 Marks
A man saved $Rs. 32$ during the first year, $Rs. 36$ in the second year and in this way he increases his saving by $Rs. 4$ every year. Find in what time his saving will be $Rs. 200.$
AnswerSavings for the first year $= Rs. 32$
For the second year $= Rs. 36$
$\therefore$ $a = 32 and d = 36 - 32 = 4$
$S_n = 200$
$\Rightarrow\ 200=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow 400 = n[2 \times 32 + (n - 1)4]$
$\Rightarrow 400 = n[64 + 4n - 4]$
$\Rightarrow 400 = n(60 + 4n) = 60n + 4n^2$
$\Rightarrow 4n^2 + 60n - 400 = 0 (Dividing by 4)$
$\Rightarrow 4(n^2 + 15n - 100 = 0$
$\Rightarrow n^2 + 15n - 100 = 0$
$\Rightarrow n^2 + 20n - 5n - 100 = 0$
$\Rightarrow n(n + 20) - 5(n + 20) = 0$
$\Rightarrow n = -20, n = 5.$
View full question & answer→Question 164 Marks
Find the sum of all odd numbers between:
$100$ and $200$.
AnswerIn this problem, we need to find the sum of all odd numbers lying between $100$ and $200$.
So, we know that the first odd number after $0$ is $101$ and the last odd number before $200$ is $199$.
Also, all these terms will from an A.P. with the common difference of $2$.
So here,
First term $(a) = 101$
Last term $(l) = 199$
Common difference $(d) = 2$
So, here the first step is to find the 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,
$199 = 101 + (n - 1)2$
$199 = 101 + 2n - 2$
$199 = 99 + 2n$
$199 - 99 = 2n$
Further simplifying,
$100 = 2n$
$\text{n}=\frac{100}{2}$
$n = 50$
Now, using the formula for the sum of n terms.
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{S}_\text{n}=\frac{50}{2}[2(101)+(50-1)2]$
$= 25[202 + (49)2]$
$= 25(202 + 98)$
$= 25(300)$
$= 7500$
therefore, the sum of all the odd numbers lying between $100$ and $200$ is $S_n = 7500.$
View full question & answer→Question 174 Marks
Find the sum of all odd numbers between:
$0$ and $50$
AnswerIn this problem, we need to find the sum of all odd numbers lying between $0$ and $50.$
So, we know that the first odd number after $0$ is $1$ and the last odd number before $50$ is $49$.
Also. all these terms will form an A.P. with the common difference of $2$.
So here,
First term $(a) = 1$
Last term $(l) = 49$
Common difference $(d) = 2$
So, here the first step is to find the 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,
$49 = 1 + (n - 1)2$
$49 = 1 + 2n - 2$
$49 = 2n - 1$
$49 + 1 = 2n$
Further simplifying,
$50 = 2n$
$\text{n}=\frac{50}{2}$
$n = 25$
Now, using the formula for the sum of n terms,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
For $n = 25$, we get
$\text{S}_{25}=\frac{25}{2}\Big[2\times1+(25-1)2\Big]$
$=\frac{25}{2}[2+48]$
$=\frac{25}{2}(50)=25\times25$
$=625$
Therefore, the sum of all the odd numberes lying between $0$ and $50$ is $S_n = 625.$
View full question & answer→Question 184 Marks
If the sum of the first $n$ terms of an A.P is $4 n-n^2$, what is the first term? What is the sum of first two terms? What is the second term? Similarly, find the third, the tenth and the $n ^{\text {th }}$ terms.
AnswerIn the given problem, the sum of n terns of an A.P. is given by the expression,
$S_n=4 n-n^2$
So here, we can find the first term by susbstituting $n =1$,
$S_n=4 n-n^2$
$S_1=4(1)-(1)^2$
$=4-1$
$=3$
Similarly, the sun of first two terms can be given by,
$S^2=4(2)-(2)^2$
$=8-4$
$=4$
Now, as we know,
$a_n=S_n-S_{n-1}$
$\text { So, }$
$a_2=S_2-S_1$
$=4-3$
$=1$
Now, using the same method we have to find the third, tenth and $n^{th}$ term of the A.P.
So, for the third term
$a_3 = S_3 - S_2$
$= [4(3) - (3)^2] - [4(2) - (2)^2]$
$= (12 - 9) - (8 - 4)$
$= 3 - 4$
$= 1$
Also, for the tenth term,
$a_{10} = S_{10} - S_9$
$= [4(10) - (10)^2] - [4(9) - (9)^2]$
$= (40 - 100) - (36 - 81)$
$= -60 + 45$
$= -15$
So, for the nth term,
$a_n = S_n - S_{n-1}$
$= [4(n) - (n)^2] - [4(n - 1) - (n - 1)2]$
$= (4n - n^2) - (4n - 4 -n_2 - 1 + 2n)$
$= 4n - n_2 - 4n + 4 + n_2 + 1 - 2n$
$= 5 - 2n$
Therefore, $a = 3, S_2 = 4, a_2 = 1, a_3 = -1, a_{10} = -15$.
View full question & answer→Question 194 Marks
Find the sum of all even integers between $101$ and $999.$
AnswerIn this problem, we need to find the sum of all the even numbers lying between $101$ and $999$.
So, we know that the first even number after $101$ is $102$ and the last even number before $999$ is $998$.
Also, all these terms will form an A.P. with the common difference of $2$.
So here,
First term $(a) = 102$
Last term $(l) = 998$
Common difference $(d) = 2$
So, here the first step is to find the 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,
$998 = 102 + (n - 1)2$
$998 = 102 + 2n - 2$
$998 = 100 + 2n$
$998 - 100 = 2n$
Further Simplifying,
$898 = 2n$
$\text{n}=\frac{898}{2}$
$n = 449$
Now, using the formula for the sum of n terms,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
For n = 449, we get,
$\text{S}_\text{449}=\frac{449}{2}[2(102)+(449-1)2]$
$=\frac{449}{2}[204+(448)2]$
$=\frac{449}{2}(204+896)$
$=\frac{449}{2}(1100)$
On further simplification, we get,
$S_n = 449(550)$
$= 246950$
Therefore, the sum of all the even numbers lying between $101$ and $999$ is $S_n = 246950.$
View full question & answer→Question 204 Marks
The sum of the first $7$ terms of an A.P. is $63$ and the sum of its next $7$ terms is $161$. Find the $28^{th}$ term of this A.P.
AnswerSum of first $7$ terms $(S_7) = 63$ Sum of next $7$ terms $= 161$
$\therefore$ $S_{14} = 63 + 161 = 224$
Let a be first term and d be the common difference, then $\therefore\ \text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{S}_7=\frac{7}{2}[2\text{a}+(7-1)\text{d]}$
$=\frac{7}{2}[2\text{a}+6\text{d}]$
$\Rightarrow\ 63=7\text{a}+21\text{d}$
$\therefore\ \text{a}+3\text{d}=9\ .....(\text{i})$ and $\text{S}_{14}=\frac{14}{2}[2\text{a}+(14-1)\text{d}]$
$224=7[2\text{a}+13\text{d}]$ $32=2\text{a}+13\text{d}$
$\Rightarrow\ 2\text{a}+13\text{d}=32$ Multiply (i) by $2$ and (ii) by $1$
$\therefore\ \text{a}=9-3\text{d}=9-3\times2=9-6=3$
$\therefore\ \text{a}=3,\text{d}=2$
Now $\text{T}_{28}=\text{a}+(\text{n}-1)\text{d}=3+(28-1)\times2$
$=3+27\times2=3+54=57$ View full question & answer→Question 214 Marks
Find the common difference and write the next four terms of the following arithmetic progressions:
$-1,\frac{5}{6},\frac{2}{3},\ .....$
AnswerHere, first term $(a_1) = -1$
Common difference $(d) = a_2 - a_1$
$=-\frac{5}{6}-(-1)$
$=\frac{-5+6}{6}$
$=\frac{1}{6}$
Now, we need to find the next four terms of the given A.P
That is we need to find $a_4, a_5, a_6, a_7.$
So, using the formula $a_n = a + (n - 1)d$
Substituting $n = 4, 5, 6, 7$ in the above formula
Substituting $n = 4$, we get
$\text{a}_4=-1+(4-1)\Big(\frac{1}{6}\Big)$
$\text{a}_4=-1+\Big(\frac{1}{2}\Big)$
$\text{a}_4=\frac{-2+1}{2}=\frac{-1}{2}$
Substituting $n = 5$, we get
$\text{a}_5=-1+(5-1)\Big(\frac{1}{6}\Big)$
$\text{a}_5=-1+\frac{2}{3}$
$\text{a}_5=\frac{-3+2}{3}$
$\text{a}_5=-\frac{1}{3}$
Substituting $n = 6$, we get
$\text{a}_6=-1+(6-1)\Big(\frac{1}{6}\Big)$
$\text{a}_6=-1+\frac{5}{6}$
$\text{a}_6=\frac{-6+5}{6}$
$\text{a}_6=-\frac{1}{6}$
Substituting $n = 7$, we get
$\text{a}_7=-1+(7-1)\Big(\frac{1}{6}\Big)$
$\text{a}_7=-1+1$
$\text{a}_7=0$
Therefore, the common difference is $\text{d}=\frac{1}{6}$ and the next four terms are $-\frac{1}{2},-\frac{1}{3},-\frac{1}{6},0$.
View full question & answer→Question 224 Marks
Find the middle term of the A.P. $213, 205, 197, ...., 37.$
AnswerLet a be the first term and d be the common difference.
We know that, $n^{th}$ term $= a_n = a + (n - 1)d$
It is given that $a = 213, d= -8$ and $a_n = 37$
According to the question,
$\Rightarrow 37 = 213 + (n - 1)(-8)$
$\Rightarrow 37 = 213 - 8n + 8$
$\Rightarrow 8n = 221 - 37$
$\Rightarrow 8n = 184$
$\Rightarrow n = 23 .....(i)$
Therefore, Total number of terms is $23$.
Since, there are odd number of terms.
So, Middle term will be $\Big(\frac{23 + 1}{2}\Big)^{\text{th}}$ term, i.e., the $12^{th}$ term.
$\therefore$ $a_{12} = 213 + (12 - 1)(-8)$
$= 213 - 88$
$= 125$
Thus, the middle term of the A.P. $213, 205, 197, ....., 37$ is $125$.
View full question & answer→Question 234 Marks
Find the next five terms of the following sequences given:
$a_1 = 4, a_n = 4a_{n-1} + 3, n > 1.$
AnswerGiven,
First term $(a_1) = 4$
$n^{th}$ term $(a_n) = 4a_{n-1} + 3, n > 1$
To find next fice terms i.e., $a_2, a_3, a_4, a_5, a_6$_ we put $n = 2, 3, 4, 5, $6 is $a_n$
Then, we get
$a_2 = 4a_{2-1} + 3 = 4.a_1 + 3 = 4(4) + 3 = 19$ ($\therefore$ $a_1 = 4$)
$a_3 = 4a_{3-1} + 3 = 4.a_2 + 3 = 4(19) + 3 = 79$
$a_4 = 4a_{4-1} + 3 = 4.a_3 + 3 = 4(79) + 3 = 319$
$a_5 = 4a_{5-1} + 3 = 4.a_4 + 3 = 4(319) + 3 = 1279$
$a_6 = 4a_{6-1} + 3 = 4.a_5 + 3 = 4(1279) + 3 = 5119.$
$\therefore$ The required next five terms are,
$a_2 = 19, a_3 = 79, a_4 = 319, a_5 = 1279, a_6 = 5119.$
View full question & answer→Question 244 Marks
Find the next five terms of the following sequences given:
$a_1 = a_2 = 2, a_n = a_{n-1} - 3, n > 2.$
Answer$a_1=a_2=2$
$a_n=a_{n-1}-3$
Let $n=3,4,5,6,7$
$a_3=a_{3-1}-3=a_2-3$
$=2-3=-1$
$a_4=a_{4-1}-3=a_3-3$
$=-1-3=-4$
$a_5=a_{5-1}-3=a_4-3$
$=-4-3=-7$
$a_6=a_{6-1}-3=a_5-3$
$=-7-3=-10$
$a_7=a_{7-1}-3=a_6-3$
$=-10-3=-13 .$
View full question & answer→Question 254 Marks
If the $n ^{\text {th }}$ term of the A.P. $9,7,5, \ldots$ is same as the $n ^{\text {th }}$ term of the A.P. $15,12,9, \ldots$ find $n$.
AnswerGiven,
$AP_1=9,7,5, \ldots . . n_1$
$AP_2=15,12,9, \ldots . . n_2$
$n^{\text {th }}$ term of both AP is equal then
$n_1=n_2=n$
we know, $n ^{\text {th }}$ term of A.P.
$a_n=a+(n-1) d$
$\text { For } A P _1, a_{n_1}= n$
$a=9$
$\text { and } d=(7-9)=-2$
$\text { then, } n=9+(n-1)-2$
$\Rightarrow n=9-2 n+2$
$\Rightarrow n=11-2 n \ldots . . \text { (i) }$
For $A P _2, a _{ n _2}= n$
$a=15$
and $d=(12-15)=-3$
then, $n=15+(n-1)-3$
$\Rightarrow n=15-3 n+3$
$\Rightarrow n=18-3 n \ldots . . .(\text { iii) }$
From eq. (i) and (ii), we get
$\Rightarrow 11-2 n=18-3 n$
$\Rightarrow-2 n+3 n=18-11$
$\Rightarrow n=7$
Hence, value of $n$ is $7$ .
View full question & answer→Question 264 Marks
In an A.P., if the $5^{\text {th }}$ and $12^{\text {th }}$ terms are $30$ and $65$ respectively, what is the sum of first $20$ terms?
AnswerGiven,
$5^{\text {th }}$ term $a_5=30$
$12^{\text {th }}$ term $a _{12}=65$
We know $a_n=a+(n-1) d$
$\Rightarrow 30=a+4 d \ldots$...(i)
$12^{\text {th }}$ term $a _{12}= a +(12-1) d$
$\Rightarrow 65=a+11 d \ldots . . . . \text { (ii) }$
Subtracting eq. (i) from Eq. (ii)
$\Rightarrow 65-30=a+11 d-(a+4 d)$
$\Rightarrow 35=a+11 d-a-4 d$
$\Rightarrow 35=7 d$
$\Rightarrow d=\frac{35}{7}$
$\Rightarrow d=5$
From eq. (i) $30=a+4 \times 5$
$\Rightarrow 30=a+20$
$\Rightarrow a=30-20$
$\Rightarrow a=10$
We know, Sum of $n$ terms
$S_n=\frac{n}{2}[2 a+( n -1) d]$
$\Rightarrow S_{20}=\frac{20}{2}[2 \times 10+(20-1) 5]$
$\Rightarrow S_{20}=10[20+19 \times 5]$
$\Rightarrow S_{20}=10[20+95]$
$\Rightarrow S_{20}=10 \times 115$
$\Rightarrow S_{20}=1150$
Hence, Sum of first $20$ terms is $1150$.
View full question & answer→Question 274 Marks
In an A.P. the first term is $8, n^{th}$ term is $33$ and the sum to first n terms is $123$. Find n and d, the common differences.
AnswerIn an A.P,
First term $(a) = 8$
$n^{th}$ term $(a_n) = 33$
$S_n = 123$
Let d the common difference an n be the number of terms
$\therefore$ $a_n = a + (n - 1)d$
$\Rightarrow 33 = 8 + (n - 1)d$
$\Rightarrow (n - 1)d = 33 - 8 = 25 .....(i)$
and $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ 123=\frac{\text{n}}{2}[2\times8+25]\ [\text{ from (i)}]$
$123\times2=\text{n}(16+25)=41\text{n}$
$\Rightarrow\ \text{n}=\frac{123\times2}{41}=6$
From (i) $(n - 1)d = 25$
$\Rightarrow\ (6-1)\text{d}=25\Rightarrow\ 5\text{d}=25$
$\Rightarrow\ \text{d}=\frac{25}{5}=5$
Hence $n = 6, d = 5.$
View full question & answer→Question 284 Marks
Find the sum:$1 + 3 + 5 + 7 + ..... + 199.$
AnswerGiven,
A.P. $1 + 3 + 5 + 7 + ..... + 199.$
Here,
First term, $a = 1$
Difference, $d = 3 - 1 = 2$
and Last term $a_n = 199$
We know, $a_n = a + (n - 1)d$
$\Rightarrow 199 = 1 + (n - 1)2$
$\Rightarrow 199 = 1 + 2n - 2$
$\Rightarrow 200 = 2n$
$\Rightarrow n = 100$
We know, Sum of n terms,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ \text{S}_{100}=\frac{100}{2}[2(1)+(100-1)2]$
$\Rightarrow S_{100} = 50[2 + 99 \times 2]$
$\Rightarrow S_{100} = 50[2 + 198]$
$\Rightarrow S_{100} = 50[200]$
$\Rightarrow S_{100} = 10000.$
Hecne, Sum of given A.P. is $10000.$
View full question & answer→Question 294 Marks
The first and the last terms of an A.P. are $5$ and $45$ respectively. If the sum of all its terms is $400$, find its common difference.
AnswerLet a be the first term and d be the common difference.
We know that, Sum of first n terms $=\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
Also, $n^{th}$ term $a_n = a + (n - 1)d$
According to the question,
$a = 5, a_n = 45$ and $S_n = 400$
Now,
$a_n = a (n - 1)d$
$\Rightarrow 45 = 5 + (n - 1)d$
$\Rightarrow 40 = nd - d$
$\Rightarrow nd - d = 40 .....(1)$
Also,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\times5+(\text{n}-1)\text{d}]$
$\Rightarrow\ 400=\frac{\text{n}}{2}[10+\text{nd}-\text{d}]$
$\Rightarrow 800 = n$$[10 + 40]$ [From (i)]
$\Rightarrow 50n = 800$
$\Rightarrow n = 16 .....(ii)$
On substituting (ii) in (i), we get
$nd - d = 40$
$\Rightarrow (16 - 1)d = 40$
$\Rightarrow 15d = 40$
$\Rightarrow\ \text{d}=\frac{8}{3}$
Thus, common difference of the given A.P. is $\frac{8}{3}$.
View full question & answer→Question 304 Marks
Write the expression $a_n - a_k$ for the A.P. $a, a + d, a + 2d, ...$
Hence, find the common difference of the A.P. for which,
$11^{th}$ term is $5$ and $13^{th}$ term is $79$.
AnswerWe know,
$a_n=a+(n-1) d$
$\text { Let, }$
$n^{\text {th }} \text { term } a_n=a+(n-1) d$
$\Rightarrow a_n=a+n d-d$
$k^{\text {th }} \text { term, } a_k=a+(k-1) d$
$\Rightarrow a_k=a+k d-d$
Now,
$\Rightarrow a_n-a_k=(a+n d-d)-(a+k d-d)$
$\Rightarrow=a+n d-d-a-k d+d$
$\Rightarrow=n d-k d$
$\Rightarrow=d(n-k)$
Given, $11^{\text {th }}$ term, $a_{11}=5$
and $13^{\text {th }}$ term, $a_{13}=79$
We know, an $=a+(n-1) d$
then,
$11^{\text {th }} \text { term, } a_{11}=a+(11-1) d$
$\Rightarrow 5=a+10 d \ldots . .(i)$
$13^{\text {th }} \text { term, } a_{13}=a+(13-1) d$
$\Rightarrow 79=a+12 d . . . . . \text { (ii) }$
By subtituting eq. (i) from eq. (ii)
$\Rightarrow 79-5=a+12 d-(a+10 d)$
$\Rightarrow 74=a+12 d-a-10 d$
$\Rightarrow 74=2 d$
$\Rightarrow d=37$
View full question & answer→Question 314 Marks
If $10$ times the $10^{\text {th }}$ term of an A.P. is equal to $15$ times the $15^{\text {th }}$ term, show that $25^{\text {th }}$ term of the A.P. is zero.
AnswerHere, let us take the first term of the A.P. as a and the common difference as d.
We are given that 10 times the $10^{\text {th }}$ term is equal to 15 times the $15^{\text {th }}$ term. We need to show that $25^{\text {th }}$ term is zero.
So, let us first find the two terms.
So, as we know,
$a_n=a+(n-1) d$
For $10^{\text {th }}$ term $(n=10)$
$a_{10}=a+(10-1) d$
$=a+9 d$
For $15^{\text {th }}$ term $(n=15)$,
$a_{15}=a+(15-1) d$
$=a+14 d$
Now, we are given
$10(a+9 d)=15(a+14 d)$
Solving this, we get,
$10 a+90 d=15 a+210 d$
$90 d-210 d=15 a-10 a$
$-120 d=5 a$
$-24 d=a \ldots . .(i)$
Next, we need to prove that the $25^{\text {th }}$ term of the A.P. is zero. For that, let us find the $25^{\text {th }}$ term using $n=25$,
$a_{25}=a+(25-1) d$
$=-24 d+24 d \text { (using } 1)$
$=0$
Thus, the $25^{\text {th }}$ term of the given A.P. is zero.
Hence proved.
View full question & answer→Question 324 Marks
The sum of first $9$ terms of an A.P. is $162$. The ratio of its $6^{th}$ term to its $13^{th}$ term is $1 : 2$. Find the first and $15^{th}$ term of the A.P?
AnswerSum of first $9$ terms $= 162$
Ratio of its $6^{th}$ term and $13^{th}$ term $= 1 : 2$
$\text{S}_9=\frac{9}{2}[2\text{a}+(9-1)\text{d}]=162\ .....(\text{i})$
$\text{T}_6=[\text{a}+(6-1)\text{d}]=(\text{a}+5\text{d})$
$=\text{a}+5\text{d}$
and $\text{T}_{13}=[\text{a}+(13-1)\text{d}]=(\text{a}+12\text{d})$
But $\text{T}_6:\text{T}_{13}=1:2\Rightarrow\ \frac{\text{T}_6}{\text{T}_{13}}=\frac{1}{2}$
$\Rightarrow\ \frac{\text{a}+5\text{d}}{\text{a}+12\text{d}}=\frac{1}{2}\Rightarrow\ 2\text{a}+10\text{d}=\text{a}+12\text{d}$
$\Rightarrow\ 2\text{a}-\text{a}=12\text{d}-10\text{d}\Rightarrow\ \text{a}=2\text{d}$
From (i),
$\because\ \frac{9}{2}[2\text{a}+(9-1)\text{d}]=162$
$\Rightarrow\ \frac{9}{2}(2\text{a}+8\text{d})=162$
$\Rightarrow\ \frac{9}{2}(4\text{d}+8\text{d})=162\ (\because\ \text{a}=2\text{d})$
$\Rightarrow\ \frac{9}{2}\times12\text{d}=162\Rightarrow\ 54\text{d}=162$
$\Rightarrow\ \text{d}=\frac{162}{54}=3$
$\text{a}=2\text{d}=2\times3=6$
Now $\text{T}_{15}=\text{a}+(15-1)\text{d}=\text{a}+14\text{d}$
$=6+14\times3=6+42=48$
Hence first term $= 6$ and $T_{15} = 48.$
View full question & answer→Question 334 Marks
If $12^{th}$ term of an A.P. is -$13$ and the sum of the first four terms is $24,$ what is the sum of first $10$ terms?
AnswerGiven, $a_{12} = -13, a + a_2 + a_3 + a_4 = 24$
$\text{S}_4=\frac{4}{2}(2\text{a}+3\text{d})=24$
$2\text{a}+3\text{d}=\frac{24}{2}=12\ .....(\text{i})$
$\Rightarrow\ \text{a}+(12-1)\text{d}=-13$
$\text{a}+11\text{d}=-13\ .....(\text{ii})$ Subtract (i) from (ii) $\times 2$
$\text{d}=\frac{-38}{19}=-2$
put $d = -2$ in
(ii) $a + 11(-2) = -13 a = -13 + 22 a = 9$
Given to find sum of first $10$ terms. $\text{S}_{10}=\frac{10}{2}[2\text{a}+(10-1)\text{d}]$
$\text{S}_{10}=5\big[2\times9+9\times(-2)\big]$
$=5(18-18)$
$=0$
$\therefore\ \text{S}_{10}=0$ View full question & answer→Question 344 Marks
If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms.
AnswerLet a be the first term and d be the common difference of an A.P.
Sum of 7 terms = 49
and sum of 17 terms = 289
$\text{S}_7=\frac{7}{2}[2\text{a}+(7-1)\text{d}]=49$
$\Rightarrow\ \frac{7}{2}[2\text{a}+6\text{d}]=49$
$7\text{a}+21\text{d}=49\Rightarrow\ \text{a}+3\text{d}=7\ (\text{Dividing by }7)$
$\therefore\ \text{a}+3\text{d}=7\ .....(\text{i})$
Similarly $\text{S}_{17}=\frac{17}{2}[2\text{a}+(17-1)\text{d}]=289$
$\Rightarrow\ \frac{17}{2}[2\text{a}+16\text{d}]=289$
$\Rightarrow\ 17\text{a}+136\text{d}=289\ (\text{Dividing by }17)$
$\Rightarrow\ \text{a}+8\text{d}=17\ .....(\text{ii})$
Subtracting (i) from (ii)
$5\text{d}=10\Rightarrow\ \text{d}=\frac{10}{5}=2$
and $\text{a}+3\text{d}=7\Rightarrow\ \text{a}+2\times3=7$
$\Rightarrow\ \text{a}+6=7\Rightarrow\ \text{a}=7-6=1$
$\therefore\ \text{a}=1,\text{d}=2$
$\therefore\ \text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$=\frac{\text{n}}{2}[2\times1+(\text{n}-1)2]$
$=\frac{\text{n}}{2}[2+2\text{n}-2]=\frac{\text{n}}{2}\times2\text{n}=\text{n}^2$
View full question & answer→Question 354 Marks
If $S_n$ denotes the sum of first n terms of an A.P., prove that $S_{12} = 3(S_8 - S_4).$
AnswerLet a be the first term and d be the common difference
We know that, Sum of first n terms $=\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
Now,
$\text{S}_4=\frac{4}{2}[2\text{a}+(4-1)\text{d}]$
$=2(2\text{a}+3\text{d})$
$=4\text{a}+6\text{d} \ .....\text{(i)}$
$\text{S}_8=\frac{8}{2}[2\text{a}+(8-1)\text{d}]$
$=4(2\text{a}+7\text{d})$
$=8\text{a}+28\text{d}\ .....(\text{ii})$
$\text{S}_{12}=\frac{12}{2}[2\text{a}+(12-1)\text{d}]$
$=6(2\text{a}+11\text{d})$
$=12\text{a}+66\text{d}\ .....(\text{iii})$
On Subtracting (i) from (ii), we get
$S_8 - S_4 = 8a + 28d - (4a + 6d)$
$= 4a + 22d$
Multiplying both sides by $3$, we get
$3(S_8 - S_4) = 3(4a + 22d)$
$= 12a + 66d$
$= S_{12}$ [from (iii)]
Thus, $S_{12} = 3(S_8 - S_4).$
View full question & answer→Question 364 Marks
Find the $12^{th}$ term from the end of the following arithmetic progressions:
$1, 4, 7, 10, ....., 88.$
Answer$1,4,7,10, \ldots . . ., 88$.
Here, to find the $12^{\text {th }}$ term from the end let us first find the total number of terms. Let us take the total number of terms as $n$ .
So,
First term $(a) = 1$
Last term $\left(a_n\right)=88$
Common difference, $d=4-1=3$
Now, as we know
$a_n=a+(n-1) d$
So, for the last term,
$88=1+(n-1) 3$
$88=1+3 n-3$
$88=-2+3 n$
$88+2=3 n$
Further simplifying,
$90=3 n$
$n=\frac{90}{3}$
$n=30$
So, the $12^{\text {th }}$ term from the end means the $19^{\text {th }}$ term from the beginning.
So, for the $19^{\text {th }}$ term $( n =19)$
$a_{19}=1+(19-1) 3$
$=1+(18) 3$
$=1+54$
$=55$
Therefore, the $12^{\text {th }}$ term the end of the given A.P. is $55$ .
View full question & answer→Question 374 Marks
Find the next five terms of the following sequences given:
$\text{a}_1=-1,\text{a}_\text{n}=\frac{\text{a}_\text{n}-1}{\text{n}},\text{n}\geq2$
AnswerGiven, first term $(a_1) = -1$
$n^{th}$ term $(a_n)$ $=\frac{\text{a}_\text{n}-1}{\text{n}},\text{n}\geq2$
To find next five terms i.e., $a_2, a_3, a_4, a_5, a_6$ we put $n = 2, 3, 4, 5, 6$ is an
$\text{a}_2=\frac{\text{a}_{2-1}}{2}=\frac{\text{a}_1}{2}=\frac{-1}{2}$
$\text{a}_3=\frac{\text{a}_{3-1}}{3}=\frac{\text{a}_2}{3}=\frac{-\frac{1}{2}}{3}=\frac{-1}{6}$
$\text{a}_4=\frac{\text{a}_{4-1}}{4}=\frac{\text{a}_3}{4}=\frac{\frac{-1}{6}}{4}=\frac{-1}{24}$
$\text{a}_5=\frac{\text{a}_{5-1}}{5}=\frac{\text{a}_4}{5}=\frac{\frac{-1}{24}}{5}=\frac{-1}{120}$
$\therefore$ The next five terms are,
$\text{a}_2=\frac{-1}{2},\text{a}_3=\frac{-1}{6},\text{a}_4=\frac{-1}{24},\text{a}_5=\frac{-1}{120},\text{a}_6=\frac{-1}{720}$
View full question & answer→Question 384 Marks
Find the sum of all integers between $84$ and $719$, which are multiples of $5$.
AnswerIn this problem, we need to find the sum of all the multiples of $5$ lying between $84$ and $719.$
So, we know that the first multiple of $5$ after $84$ is $85$ and the last multiple of $5$ before $719$ is $715$.
Also, all these terms will form an A.P. with the common difference of $5$.
So here,
First term $(a) = 85$
Last term $(l) = 715$
Common difference $(d) = 5$
So, here the first step is to find the 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,
$715 = 85 + (n - 1)5$
$715 = 85 + 5n - 5$
$715 = 80 + 5n$
$715 - 80 = 5n$
Further simplifying,
$635 = 5n$
$\text{n}=\frac{635}{5}$
$n = 127$
Now, 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,
$\text{S}_\text{n}=\frac{127}{2}[2(85)+(127-1)5]$
$=\frac{127}{2}[170+(126)5]$
$=\frac{127}{2}(170+630)$
$=\frac{127(800)}{2}$
On further simplification, we get,
$S_n = 127(400)$
$= 50800$
Therefore, the sum of all the multiples of $5$ lying between $84$ and $719$ is $S_n = 50800.$
View full question & answer→Question 394 Marks
Find the sum of n terms of an A.P. whose nth terms is given by $an = 5 - 6n$.
AnswerGiven, $a_n = 5 - 6n$
By putting $n = 1, 2, 3, .....$
$a_1 = 5 - 6(1) = 5 - 6 = -1$
$a_2 = 5 - 6(2) = 5 - 12 = -7$
and $a_3 = 5 - 6(3) = 5 - 18 = -13$
So, A.P. is $-1, -7, -13, .....$
Here,
First term $a = -1$
and Difference $d = -7 - (-1) = -7 + 1 = -6$
we know, sum of n terms
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}[2(-1)+(\text{n}-1)-6]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}[-2+6\text{n}+6]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}[4-6\text{n}]$
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}\times2[2-3\text{n}]$
$\Rightarrow\ \text{S}_\text{n}=\text{n}[2-3\text{n}]$
Hence, Sum of n terms is $n[2 - 3n].$
View full question & answer→Question 404 Marks
The $4^{th}$ term of an A.P. is three times the first and the $7^{th}$ term exceeds twice the third term by $1$. Find the first term and the common difference.
AnswerGiven,
$4^{\text {th }}$ term of an A.P. is three times the times the first term
$a_4=3 . a$
$n^{\text {th }} \text { term of a sequ }$
$a+(4-1) d=3 a$
$a+3 d=3 a$
$3 d=2 a$
$a=\frac{3}{2} d \ldots . .(i)$
$n^{\text {th }} \text { term of a sequence } a_n=a+(n-1) \cdot d$
Seventh term exceeds twice the third term by $1$.
$a_7+1=2 \cdot a 3$
$a+(7-1) d+1=2(a+(3-1) d)$
$a+6 d+1=2 a+4 d$
$a=2 d+1 \ldots . . \text { (ii) }$
By equating (1), (2)
$\frac{3}{2} d=2 d+1$
$\frac{3}{2} d-2 d=1$
$-d=2$
$d=-2$
Put $d =-2$ in $a =\frac{3}{2} d$
$=\frac{3}{2} \cdot x$
$=-3$
$\therefore$ First term $a=-3$, common difference $d=-2$.
View full question & answer→Question 414 Marks
Which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.
$12, 2, -8, -18, .....$
AnswerIn the given problem, we are given various sequences.
We need to find out that the given sequences are an A.P. of not and then find its common difference $(d).$
$12, 2, -8, -18, .....$
Let $a_1 = 12, a_2 = 2, a_3 = -8, a_4 = -18$
Now $a_2 - a_1 = 2 - 12 = -10$
$a_3 - a_2 = -8 - 2 = -10$
$a_4 - a_3 = -18 - (-8) = -18 + 8 = -10$
$\therefore$ We see that common difference is $-10$
$\therefore$ It is an A.P.
View full question & answer→Question 424 Marks
The $10^{th}$ and $18^{th}$ terms of an A.P. are $41$ and $73$ respectively. Find $26^{th}$ term.
AnswerIn the given problem, we are given $10^{th}$ and $18^{th}$ term of an A.P.
We need to find the $26^{th}$ 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,
$a_{10} = a + (10 - 1)d$
$41 = a + 9d .....(i)$
Also,
$a_{18} = a + (18 - 1)d$
$73 = a + 17d .....(ii)$
So, to solve for a and d
On susbtracting (1) from (2), we get
$8d = 32$
$\text{d}=\frac{32}{8}$
$d = 4$
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 434 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 $\text {Rs}. 160.$
Second prize = $\text {Rs}. 140$
Third prize =$\text {Rs}. 120$
Fourth prize = $\text {Rs}. 100$
Fifth prize = $\text {Rs}. 80$
Sixth prize = $\text{Rs}. 60$
Seventh prize = $\text {Rs}. 40$
So the values of prizes are $\text {Rs}. 160.$, $\text {Rs}. 140$, $\text {Rs}. 120$, $\text {Rs}. 100$, $\text {Rs}. 80$, $\text{Rs}. 60$, $\text {Rs}. 40$
View full question & answer→Question 444 Marks
Let there be an A.P. with first term $'a'$, common difference $'d'$. If $a_n$ denotes in $n^{th}$ term and $S_n$ the sum of first $n$ terms, find.
d, if $a = 3, n = 8$ and $S_n = 192$.
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 454 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 464 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.
$a + (n - 1)d < 0$
$\Rightarrow 121 + (n - 1) -4 < 0$
$\Rightarrow 121 + (-4n + 4) < 0$
$\Rightarrow 121 - 4n + 4 < 0$
$\Rightarrow 125 - 4n < 0$
$\Rightarrow 4n > 125$
$\Rightarrow\ \text{n}>31\frac{1}{4}$
$\because\ \text{n}>31\frac{1}{4}$
$\Rightarrow n = 32$
Hence, $32^{th}$ term of the given A.P. for getting first negative term.
View full question & answer→Question 474 Marks
Let there be an A.P. with first term $'a'$, common difference $'d'$. If $a_n$ denotes in $n^{th}$ 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$a_n=4, d=2, S_n=-14$
$\therefore a+(n-1) 2=4 \Rightarrow a+(n-1)^2=4$
$\Rightarrow a+2 n-2=4 \Rightarrow a+2 n=4+2$
$a+2 n=6 \ldots . . \text { (i) }$
$S_n=\frac{n}{2}[a+1] \Rightarrow-14=\frac{n}{2}[a+4]$
$n(a+4)=-28 \text {.....(ii) }$
From (i) $a=6-2 n$
$\therefore \text { in (ii) }$
$n(6-2 n+4)=-28 \Rightarrow n(10-2 n)=-28$
$\Rightarrow 10 n-2 n^2=-28 \Rightarrow 2 n^2-10 n-28=0$
$\left.\Rightarrow n^2-5 n-14=0 \text { (Divisible by } 2\right)$
$\Rightarrow n^2-7 n+2 n-14=0$
$\begin{Bmatrix} \therefore\ -14=-7\times2\\ -5=-7=2 \end{Bmatrix}$
$\Rightarrow n(n - 7) + 2(n - 7) = 0$
$\Rightarrow (n - 7)(n + 2) = 0$
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 484 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,
$a_8=a+(8-1) d$
$22=a+7 d \ldots . . .(i i)$
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 $2^{nd}$ term $(n = 2),$
$a_2 = -13 + (2 - 1)5$
$= -13 + (1)5$
$= -13 + 5$
$= -8$
For the $n^{th}$ term,
$a_n = -13 + (n - 1)5$
$= -13 + 5n - 5$
$= -18 + 5n$
Therefore, $a_2 = -8, a_n = 5n - 18.$
View full question & answer→Question 494 Marks
Show that the sequence defined by $a_n = 5n - 7$ 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,
$a_n = 5n - 7$
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 504 Marks
Find the sum:
$(-5) + (-8) + (-11) + ..... + (-230).$
AnswerGiven,
$\text { A.P. }(-5)+(-8)+\ldots . . .(-230)$
Here,
First term $a=-5$
Difference $d=-8-(-5)=-8+5=-3$
and Last term $a_n=-230$
We know $a_n=a+(n-1) d$
$\Rightarrow-230=-5+(n-1)(-3)$
$\Rightarrow-230=-5-3 n+3$
$\Rightarrow-230=-2-3 n$
$\Rightarrow-228=-3 n$
$\Rightarrow n=\frac{-228}{-3}=76$
We know, sum of $n$ terms,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow S_{76}=\frac{76}{2}[2(-5)+(76-1)(-3)]$
$\Rightarrow S_{76}=38[-10+75 \times-3]$
$\Rightarrow S_{76}=38[-10-225]$
$\Rightarrow S_{76}=38[-235]$
$\Rightarrow S_{76}=-8930$
Hence, Sum of given A.P. is $-8930$.
View full question & answer→Question 514 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
$a_4 = a + (4 - 1)d = a + 3d$
$a_8 = a + (8 - 1)d = a + 7d$
$\therefore$ $a_4 + a_8 = a + 3d + a + 7d = 24$
$\Rightarrow 2a + 10d = 24 \Rightarrow a + 5d = 12 .....(i)$
Similarly,
$a_6 = a + 5d$ and $a_{10} = a + 9d$
$\therefore a + 5d + a + 9d = 34 \Rightarrow 2a + 14d = 34$
$\Rightarrow a + 7d = 17 .....(ii)$
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 524 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 $41^{\text {st }}$ term of the given A.P.
A.P. is $8,14,20,26, \ldots .$.
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 $41^{\text {st }}$ 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 $41^{\text {st }}$ term as an. So,
$a_n=72+a_{41}$
$=72+248$
$=320$
Also, $a_n=a+(n-1) d$
$320=8+(n-1) 6$
$320=8+6 n-6$
$320=2+6 n$
$320-2=6 n$
Further simplifying, we get,
$318=6 n$
$n=\frac{318}{6}$
$n=53$
Therefore, the $53^{\text {nd }}$ term of the given A.P. is $72$ more than the $41^{\text {st }}$ term.
View full question & answer→Question 534 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 544 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$a = 7$
$a_n or l = 49 \Rightarrow a_n = 49$
$S_n = 420$
Let d be the common difference
Now $l(a_n) = a + (n - 1)d$
$\Rightarrow 49 = 7 + (n - 1)d$
$\Rightarrow (n - 1)d = 49 - 7 = 42 .....(i)$
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$420=\frac{\text{n}}{2}[2\times7+42]\ [\text{from (i)}]$
$420=\frac{\text{n}}{2}(14+42)$
$420=\frac{\text{n}}{2}(56)$
$\Rightarrow\ \text{n}=\frac{420\times2}{56}=15$
$\therefore\ \text{d}=\frac{42}{\text{n}-1}\ [\text{from (i)}]$
$=\frac{42}{15-1}=\frac{42}{14}=3$
$\therefore\ \text{d}=3$
View full question & answer→Question 554 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 564 Marks
Find the common difference of the A.P. and write the next two terms:
$75,67,59,51, \ldots . .$
AnswerGiven,
$75,67,59,51, \ldots . .$
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 574 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 584 Marks
In an A.P., the first term is $22, n ^{\text {th }}$ 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 $n ^{\text {th }}$ 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 $n ^{\text {th }}$ term of the A.P $(1)=-11$
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,
$66=\left(\frac{n}{2}\right)[22+(-11)]$
$66=\left(\frac{n}{2}\right)(22-11)$
$(66)(2)=(n)(11)$
Further, Solving for n
$n =\frac{(66)(2)}{11}$
$n=(6)(2)$
$n=12$
Now, to find the common difference of the A.P. we use the following formula,
$I=a+(n-1) d$
We get
$-11=22+(12-1) d$
$-11=22+(11) d$
$\frac{-11-22}{11}=d$
Further, Solvinf for $d$,
$d=\frac{-33}{11}$
$d=-3$
Therefore, the number of terms is $n=12$ and the common difference of the A.P. $d=-3$.
View full question & answer→Question 594 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 604 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 $( I )=988$ 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,
$988=104+(n-1) 13988$
$=104+13 n-13988=91+13 n$
Further simplifying,
$n =\frac{988-91}{13}$
$n =\frac{897}{13}$
$n=69$
Now using the formula for the sum of n terms, we get Now,
using the formula for the sum of n terms, we get
$S_{n}=\frac{69}{2}[2(104)+(69-1) 13]$
$=\frac{69}{2}[208+(68) 13]$
$=\frac{69}{2}(208+884)$
On further simplification,
$\text { we get } S_n=\frac{69}{2}(1092)=69(546)$
$=37674$
Therefore, the sum of all $3$ digit multiples of $13$ is $S_n=37674$.
View full question & answer→Question 614 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 ( $S _{ n }$ ).
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 $n$,
$3 n=48$
$n=\frac{48}{3}$
$n=16$
Now, here we can find the sum of the n terms of the given A.P., using the formula,
$S_n=\left(\frac{n}{2}\right)(a+1)$
Where, $a=$ the first term
I = 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,
$S_{16}=\left(\frac{16}{2}\right)[5+50]$
$=8(55)$
$=440$
Therefore, for the given A.P. $n=16$ and $S_n=440$
View full question & answer→Question 624 Marks
Find the term of the arithmetic progression $9,12,15,18, \ldots .$. which is 39 more than its $36^{\text {th }}$ 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 $a_n=9+(n-1) 3$
$\Rightarrow a_n=9+3 n-3$
$\Rightarrow a_n=6+3 n$
Let
$36^{\text {th }}$ term $a_{36}=9+(36-1) 3$
$=9+35 \times 3$
$=9+105$
$\Rightarrow a_{36}=114$
Now, term is $39$ more then $36^{\text {th }}$ term
$\Rightarrow a_n=39+a_{36}$
$\Rightarrow a_n=39+114$
$\Rightarrow a_n=153$
Put in eq. (i)
$\Rightarrow 153=6+3 n$
$\Rightarrow 3 n=153-6$
$\Rightarrow 3 n=147$
$\Rightarrow n=\frac{147}{3}$
$\Rightarrow n=49$
Hence, $49^{\text {th }}$ term of given A.P. is $39$ more than its $36^{\text {th }}$ term.
View full question & answer→Question 634 Marks
Which term of the A.P. $3,10,17, \ldots$ will be 84 more than its $13^{\text {th }}$ 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, $a_{13}=3+(13-1) 7$
$=3+12 \times 7$
$=3+84$
$\Rightarrow a_{13}=87$
Now, $n ^{\text {th }}$ 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 644 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 $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ 5310=\frac{\text{n}}{2}[2\times177+(\text{n}-1)\times-3]$
$10620 = n[354 - 3n + 3]$
$10620 = n(357 - 3n) \Rightarrow 10620 = 357n - 3n^2$
$\Rightarrow 3n^2 - 357n + 10620 = 0$
$n^2 - 119n + 3540 = 0$ (Dividing by $3$)
$\Rightarrow n^2 - 59n - 60n + 3540 = 0$
$\begin{Bmatrix}\because3540 = -59\times(-60) \\ -119 = -59 - 60 \end{Bmatrix}$
$\Rightarrow n(n - 59) - 60(n - 59) = 0$
$\Rightarrow (n - 59) (n - 60) = 0$
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 654 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 $S _{ n }=\frac{ n }{2}[2 a +( n - 1 ) d ]$ and $n^{\text {th }}$ term $=a_n=a+(n-1) d$
Now,
$S_{10}=\frac{10}{2}[2 a+(10-1) d]$
$\Rightarrow 120=5(2 a+9 d)$
$\Rightarrow 24=2 a+9 d$
$\Rightarrow 2 a+9 d=24 \ldots . . .(i)$
Also,
$a_{10}=a+(10-1) d$
$\Rightarrow 21=a+9 d$
$\Rightarrow 2 a+18 d=42 \ldots . . .(\text { ii) }$
Subtracting (i) from (ii), we get
$18 d-9 d=42-24$
$\Rightarrow 9 d=18$
$\Rightarrow d=2$
$\Rightarrow 2 a=24-9 d[\text { From (i) }]$
$\Rightarrow 2 a=24-9 \times 2$
$\Rightarrow 2 a=24-18$
$\Rightarrow 2 a=6$
$\Rightarrow a=3$
Also,
$a_n=a+(n-1) d$
$=3+(n-1) 2$
$=3+2 n-2$
$\Rightarrow 1+2 n$
Thus, $n ^{\text {th }}$ term if this A.P. is $1+2 n$.
View full question & answer→Question 664 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
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$a_5 = 2$
$\Rightarrow a_5 = a + (5 - 1)d$
$\Rightarrow 2 = -14 + 4d \Rightarrow 4d = 14 + 2 = 16$
$\Rightarrow\ \text{d}=\frac{16}{4}=4$
Now, $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ 40=\frac{\text{n}}{2}[2\times(-14)+(\text{n}-1)\times4]$
$\Rightarrow 80 = n(-28 + 4n - 4)$
$\Rightarrow 80 = n(-32 + 4n)$
$\Rightarrow 80 = -32n + 4n^2$
$\Rightarrow 4n^2 - 32n - 80 = 0$
$\Rightarrow n^2 - 8n - 20 = 0$ (Dividing by $4$)
$\Rightarrow n^2 - 10n + 2n - 20 = 0$
$\begin{Bmatrix}\therefore\ -20 = - 10 \times2 \\ -8 = -10 + 2 \end{Bmatrix}$
$\Rightarrow n(n - 10) + 2(n - 10) = 0$
$\Rightarrow (n - 10)(n + 2) = 0$
Either $n - 10 = 0,$ then $n = 10$
or $n + 2 = 0 \Rightarrow n = -2$ but it is not possible being negative
$\therefore$ Number of term $= 10$.
View full question & answer→Question 674 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.
$\therefore(m+1)^{\text {th }} \text { term }=a+(m+1-1) d$
$=a+m d$
$\text { and }(n+1)^{\text {th }} \text { term }=a+(n+1-1) d$
$=a+n d$
$\because(m+1)^{\text {th }} \text { term }=2(n+1)^{\text {th }} \text { term }$
$\therefore a+m d=2(a+n d)$
$\Rightarrow a+m d=2 a+2 n d \Rightarrow 2 a-a=m d-2 n d$
$a=d(m-2 n)=(m-2 n) d \ldots . . .(i)$
Now $(3 m+1)^{\text {th }}$ term $=a+(3 m+1-1) d$
$=a+3 m d=(m-2 n) d+3 m d$
$=(m-2 n+3 m) d=(4 m-2 n) d=2(2 m-n) d \ldots$
$\text { and }(m+n+1)^{\text {th }} \text { term }=a+(m+n+1-1) d$
$=a+(m+n) d$
$=(m-2 n) d+(m+n) d$
$=(m-2 n+m+n) d$
$=(2 m-n) d \ldots . . .(\text { iii })$
From (ii) and (iii)
$(3 m+1)^{\text {th }}$ term $=2(m+n+1)^{\text {th }}$ term.
Hence proved.
View full question & answer→Question 684 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→Question 694 Marks
The $17^{\text {th }}$ term of an A.P. is 5 more than twice its $8^{\text {th }}$ term. If the $11^{\text {th }}$ term of the A.P. is 43 . find the $n ^{\text {th }}$ term.
AnswerGiven,
$a_{17}=5+2\left(a_8\right) \ldots . .(i)$
$\text { and } a_{11}=43$
$\text { We know, } a_n=a+(n-1) d$
$8^{\text {th }} \text { term, } a_8=a+(8-1) d$
$\Rightarrow a_8=a+7 d$
$11^{\text {th }} \text { term, } a_{11}=a+(11-1) d$
$\Rightarrow 43=a+10 d$
$\Rightarrow a=43-10 d \ldots . .(2)$
$17^{\text {th }} \text { term, } a_{17}=a+(17-1) d$
$\Rightarrow a_{17}=a+16 d$
$\Rightarrow a_{17}=43-10 d+16 d$
$\Rightarrow a_{17}=43+6 d$
By putting value of $a_8$ and $a_{17}$ in eq. (i)
$\Rightarrow 43+6 d=5+2(a+7 d)$
$\Rightarrow 43+6 d=5+2 a+14 d$
$\Rightarrow 43-5=2 a+14 d-6 d$
$\Rightarrow 38=2 a+8 d$
$\Rightarrow 38=2(43-10 d)+8 d[\text { by eq. (ii) }]$
$\Rightarrow 38=86-20 d+8 d$
$\Rightarrow 38=86-12 d$
$\Rightarrow 12 d=86-38$
$\Rightarrow d=\frac{48}{12}=4$
From eq. (ii) $a=43-10 d$
$=43-10 \times 4$
$=43-40=3$
We know, $n ^{\text {th }}$ term of A.P.
$a_n=a+(n-1) d$
$=3+(n-1) 4$
$=3+4 n-4$
$\Rightarrow a_n=4 n-1$
Hence, $n ^{\text {th }}$ term $4 n -1$.
View full question & answer→Question 704 Marks
How many terms of the A.P. $63,60,57, \ldots .$. must be taken so that their sum is $693$ ?
AnswerGiven,
A.P. $63,60,57, \ldots .$.
and Sum of terms, $S_n=693$
Here, First term a $=63$
and Difference $d=60-63=-3$
We know, $S _{ n }=\frac{ n }{2}[2 a +( n -1) d ]$
$\Rightarrow 693=\frac{ n }{2}[2(63)+( n -1)(-3)]$
$\Rightarrow 693 \times 2=n[126-3 n+3]$
$\Rightarrow 1386=n[129-3 n]$
$\Rightarrow 1386=129 n-3 n^2$
$\Rightarrow 3 n^2-129 n+1386=0$
$\Rightarrow 3\left(n^2-43 n+462\right)=0$
$\Rightarrow n^2-43 n+462=0$
$\Rightarrow n ^2-22 n -21 n +462=0$
$\Rightarrow n ( n -22)-21( n -22)=0$
$\Rightarrow(n-22)(n-21)=0$
Now,
$n-22=0 \text { and } n-21=0$
$n=22, n=21$
Hence, no of terms are $21$ and $22$.
View full question & answer→Question 714 Marks
The sum $4^{\text {th }}$ and $8^{\text {th }}$ terms of an A.P. is $24$ and the sum of $6^{\text {th }}$ and $10^{\text {th }}$ terms is $44$ . Find the A.P.?
Answer$\text { Given, } a_4+a_8=24$
$a_6+a_{10}=44 \ldots . . . .(\text { (ii) }$
We know, $a_n=a+(n-1) d$
Then,
$4^{\text {th }} \text { term, } a_4=a+(4-1) d$
$\Rightarrow a 4=a+3 d$
$6^{\text {th }} \text { term, } a_6=a+(6-1) d$
$\Rightarrow a_6=a+5 d$
$8^{\text {th }} \text { term, } a_8=a+(8-1) d$
$\Rightarrow a_8=a+7 d$
$10^{\text {th }} \text { term, } a_{10}=a+(8-1) d$
$\Rightarrow a_{10}=a+9 d$
Put the value of $a_4$ and $a_8$ in eq. (i)
$\Rightarrow a+3 d+a+7 d=24$
$\Rightarrow 2 a+10 d=21 \ldots . . .(\text { iii) }$
Put the value of $a_6$ and $a_{10}$ in eq (ii)
$\Rightarrow a+5 d+a+9 d=44$
$\Rightarrow 2 a+14 d=44 \ldots \ldots(iv)$
By substituting eq. (iii) from (iv)
$\Rightarrow 2 a+14 d-(2 a+10 d)=44-24$
$\Rightarrow 2 a+14 d-2 a-10 d=20$
$\Rightarrow 4 d=20$
$\Rightarrow d=5$
By putting value of $d$ in eq. (iii)
$\Rightarrow 2 a+10(5)=24$
$\Rightarrow 2 a+50=24$
$\Rightarrow 2 a=24-50$
$\Rightarrow a=-13$
We know, A.P. is
$a, a+d, a+2 d, \ldots \ldots$
$\Rightarrow-13,-13+5,-13+2(5), \ldots \ldots$
$\Rightarrow-13,-8,-13+10, \ldots \ldots$
$\Rightarrow-13,-8,-3, \ldots \ldots$
Hence, A.P. is $-13,-8,-3, \ldots .$.
View full question & answer→Question 724 Marks
The sum of first seven terms of an A.P. is $182$. If its $4^{th}$ and the $17^{th}$ terms are in the ratio $1 : 5$, find the A.P.
AnswerLet a be the first term and d be the common difference.
We know that, sum of first n terms $=\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
According to the question,
$S_7 = 182$
$\Rightarrow\ \frac{7}{2}[2\text{a}+(7-1)\text{d}]=182$
$\Rightarrow\ \frac{1}{2}(2\text{a}+6\text{d})=26$
$\Rightarrow\ \text{a}+3\text{d}=26$
$\Rightarrow\ \text{a}=26-3\text{d}\ .....(\text{i})$
Also,
$\frac{\text{a}_4}{\text{a}_{17}}=\frac{1}{5}$
$\Rightarrow\ \frac{\text{a}+(4-1)\text{d}}{\text{a}+(17-1)\text{d}}=\frac{1}{5}$
$\Rightarrow\ \frac{\text{a}+3\text{d}}{\text{a}+16\text{d}}=\frac{1}{5}$
$\Rightarrow\ 5(\text{a}+3\text{d})=\text{a}+16\text{d}$
$\Rightarrow\ 5\text{a}+15\text{d}=\text{a}+16\text{d}$
$\Rightarrow\ 5\text{a}-\text{a}=16\text{d}-15\text{d}$
$\Rightarrow\ 4\text{a}=\text{d}\ .....(\text{ii})$
On Substituting (ii) in (i), we get
$a = 26 - 3(4a)$
$\Rightarrow a = 26 - 12a$
$\Rightarrow 12a + a = 26$
$\Rightarrow 13a = 26$
$\Rightarrow a = 2$
$\Rightarrow d = 4 \times 2$ [From (ii)]
$\Rightarrow d = 8$
Thus, the A.P. is $2, 10, 18, 26, .....$
View full question & answer→Question 734 Marks
The $6^{th}$ and $17^{th}$ terms of an A.P. are $19$ and $41$ respectively, find the $40^{th}$ term?
AnswerGiven,
$6^{\text {th }}$ term $=a_6=19$
and $17^{\text {th }}$ term $=a_{17}=41$
We know, $n ^{\text {th }}$ term of A.P.
$a n=a+(n-1) d$
Then, $6^{\text {th }}$ term,
$\Rightarrow a_6=a+(6-1) d$
$\Rightarrow 19=a+5 d \ldots . . .(i)$
and $17^{\text {th }}$ term,
$\Rightarrow a_{17}=a+(17-1) d$
$\Rightarrow 41=a+16 d \ldots . . .(\text { ii) }$
by substuting eq. (i) from (ii)
$\Rightarrow 41-19=(a+16 d)-(a+5 d)$
$\Rightarrow 22=a+16 d-a-5 d$
$\Rightarrow 22=11 d$
$\Rightarrow d=\frac{22}{11}$
$\Rightarrow d=2$
Now from eq. (i)
$\Rightarrow 19=a+5 d$
$\Rightarrow 19=a+5 \times 2$
$\Rightarrow a=19-10$
$\Rightarrow a=9$
We have to find $40^{\text {th }}$ term then putting $n=40$
$a_{40}=a+(40-1) d$
$=9+39 \times 2$
$\Rightarrow a_{40}=87$
Hence, $40^{\text {th }}$ term is $87$ .
View full question & answer→Question 744 Marks
Find the common difference of the A.P. and write the next two terms:
$0,\frac{1}{4},\frac{1}{2},\frac{3}{4}, .....$
AnswerGiven,
$0,\frac{1}{4},\frac{1}{2},\frac{3}{4}, .....$
$\text{a}_1=0,\text{a}_2=\frac{1}{4},\text{a}_3=\frac{1}{2},\text{a}_4=\frac{3}{4}$
Difference between terms
$\text{d}_1=\text{a}_2-\text{a}_1=\frac{1}{4}-0=\frac{1}{4}$
$\text{d}_2=\text{a}_3-\text{a}_2=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$
and $\text{d}_3=\text{a}_4-\text{a}_3=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}$
So, common difference is $\frac{1}{4}$
New, next two terms:
$a_n = a_{n-1} + d$
$5^{th}$ term,
$\text{a}_5=\text{a}_{5-1}+\frac{1}{4}$
$=\text{a}_4+\frac{1}{4}$
$\Rightarrow\ \text{a}_5=\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1$
$6^{th}$ term.
$\text{a}_6=\text{a}_{6-1}+\frac{1}{4}$
$=\text{a}_5+\frac{1}{4}$
$\Rightarrow\ \text{a}_6=1+\frac{1}{4}=\frac{4+1}{4}=\frac{5}{4}$
Hence, common difference is $\frac{1}{4}$ and next two terms $1$ and $\frac{5}{4}$.
View full question & answer→Question 754 Marks
The students of a school decided to beautify the school on the annual day by fixing colourful on the straight passage of the school. They have $27$ flags to be fixed at intervals of every $2$ metre. The flags are stored at the position of the middle most flag Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag?
AnswerGiven that, the students of a school decided to beautify the school on the annual day by fixing colourful flags on the straight passage of the school.
Given that, the number of flags $=27$
and distance between each flag $=2 m$.
Also, the flags are stored at the position of the middle most flag i.e., $14^{\text {th }}$ flag and Ruchi was given the responsibility of placing the flags.
Ruchi kept her books, where the flags were stored i.e., $14^{\text {th }}$ flag and she could carry only one flag at a time.
Let she placed $13$ flags into her left position from middle most flag i.e., $14^{\text {th }}$ flag.
For placing second flag and return his initial position distance travelled $=2+2=4 m$.
Similarly, for placing third flag and return his initial position, distance travelled $=4+4=8 m$.
For placing fourth flag and return his initial position, distance travelled $=6+6=12 m$.
For placing fourteenth flag and return his initial position, distance travelled $=26+26=52 m$.
Proceed same manner into her right position from middle most flag i.e., $14^{\text {th }}$ flag.
Total distance travelled in that case $=52 m$.
Also, when Ruchi placed the last flag she return her middle position and collect her books.
This distance also included in placed the last flag.
So, these distances from a series.
$4+8+12+16+\ldots \ldots+52$ [for left]
and $4+8+12+16+\ldots . . .+52$ [for right].
Total distance covered by Ruchi for placing these flags
$=2 \times(4+8+12+\ldots . .+52)$
$=2 \times\left[\frac{13}{2}\{2 \times 4+(13-1) \times(8-4)\}\right]$
$\left\{\because \frac{\text { Sumpf n terms of an A.P. }}{S_{n}=\frac{n}{2}[2 a+(n-1) d]}\right\}$
$=2 \times\left[\frac{13}{2}(8+12 \times 4)\right]$
[ $\because$ both sides of Ruchi number of flags i.e., $n=13]$
$=2 \times[13(4+12 \times 2)]=2 \times 13(4+24)$
$=2 \times 13 \times 28=728 m$
Hence, the required is $728\ m$ in which she did cover in completing this job and returning back to collect her books.
Now, the maximum distance she travelled carrying a flag = Distance travelled by Ruchi during placing the $14^{th}$ flag in her left position or $27^{th}$ flag in her right position
$= (2 + 2 + 2 + ..... + 13 times)$
$= 2 \times 13 = 26m.$
Hence, the required maximum distance she travelled carrying a flag is $26\ m.$
View full question & answer→Question 764 Marks
A man arranges to pay off a debt of Rs 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of all debt unpaid, find the value of the first instalment.
AnswerIn the given problem,
Total amount of debt to be paid in 40 installments Rs. 3600
After 30 installments one-third of his debt is left unpaid. This means that he paid two third of the debt in 30 installments. So,
Amount he paid in 30 installments $=\frac{2}{3}(3600)$
= 2(1200)
= 2400
Let us take the first installment as a and common difference as d.
So, using the formula for the sum of n terms of an A.P,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
Let us find a and d, for 30 installments.
$\text{S}_{30}=\frac{30}{2}[2\text{a}+(30-1)\text{d}]$
$2400=15[2\text{a}+(29)\text{d}]$
$\frac{2400}{15}=2\text{a}+29\text{d}$
$160=2\text{a}+29\text{d}$
$\text{a}=\frac{160-29\text{d}}{2}\ .....\text{(i)}$
Similarly, we find a and d for 40 installments.
$\text{S}_{40}=\frac{40}{2}[2\text{a}+(40-1)\text{d}]$
$3600=20[2\text{a}+(39)\text{d}]$
$\frac{3600}{20}=2\text{a}+39\text{d}$
$180=2\text{a}+39\text{d}$
$\text{a}=\frac{180-39\text{d}}{2}\ .....\text{(ii)}$
Subtracting (i) from (ii), we get,
$\text{a}-\text{a}=\bigg(\frac{180-39\text{d}}{2}\bigg)-\bigg(\frac{160-29\text{d}}{2}\bigg)$
$0=\frac{180-39\text{d}-160+29\text{d}}{2}$
$0=20-10\text{d}$
Furhter solving for d,
$10\text{d}=20$
$\text{d}=\frac{20}{10}$
$\text{d}=2$
Substituting the value of d in (i), we get,
$\text{a}=\frac{160-29(2)}{2}$
$=\frac{160-58}{2}$
$=\frac{102}{2}$
$=51$
Therefore, the first installment is Rs. 51
View full question & answer→Question 774 Marks
In an A.P., the sum of first ten terms is $-150$ and the sum of its next ten terms is $-550$. Find the A.P.
AnswerHere, we are given $S _{10}=-150$ and sum of the next ten terms is $-550$ .
Let us take the first term of the A.P. as a and the common difference as d .
So, let us first find $a_{10}$. For the sum of first $10$ terms of this A.P,
First term $=a$
Last term $=a_{10}$
So, we know,
$a_n=a+(n-1) d$
For the $10^{\text {th }}$ term $( n =10)$,
$a_{10}=a+(10-1) d$
$=a+9 d$
So, here we can find the sum of the $n$ terms of the given A.P.,
$S_{n}=\left(\frac{n}{2}\right)(a+l)$
using the formula,
Where, $a=$ the first term
I = the last term
So, for the given A.P.
$S_{10}=\left(\frac{10}{2}\right)(a+a+9 d)$
$-150=5(2 a+9 d)$
$-150=10 a+45 d$
$a=\frac{-150-45 d}{10} \ldots (i)$
Similarly, for the sum of next $10$ terms $\left( S _{10}\right)$,
First term $= a_{11}$
Last term $= a_{20}$
For the $11^{th}$ term $(n = 11),$
$a_{11} = a + (11 - 1)d$
$= a + 10d$
For the $20^{th}$ term $(n = 20),$
$a_{20} = a + (20 - 1)d$
$= a + 19d$
So, for the given A.P,
$\text{S}_{10}=\frac{10}{2}[\text{a}+10\text{d}+\text{a}+19\text{d}]$
$-550=5(2\text{a}+29\text{d})$
$-550=10\text{a}+145\text{d}$
$\text{a}=\frac{-550-145\text{d}}{10}\ .....\text{(ii)}$
Now, subtracting (i) from (ii),
$\text{a}-\text{a}=\Big(\frac{-550-145\text{d}}{10}\Big)-\Big(\frac{-150-45\text{d}}{10}\Big)$
$0=\frac{-550-145\text{d}+150+45\text{d}}{10}$
$0=-400-100\text{d}$
$100\text{d}=-400$
$\text{d}=-4$
Substituting the value of d in (i)
$\text{a}=\frac{-150-45(-4)}{10}$
$=\frac{-150+180}{10}$
$=\frac{30}{10}$
$=3$
So, the A.P. is $3, -1, -5, -9, .....$ with $a = 3, d = -4.$
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