Question 515 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 525 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 535 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 545 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 555 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 565 Marks
Find the common difference of the A.P. and write the next two terms:
75, 67, 59, 51, .....
AnswerGiven,
75, 67, 59, 51, .....
Here, $a_1 - 75, a_2 = 67, a_3 = 59, a_4 = 51$
Difference between terms
$d_1 = a_2 - a_1 = 67 - 75 = -8,$
$d_2 = a_3 - a_2 = 59 - 67 = -8,$
and $d_3 = a_4 - a_3 = 51 - 59 = -8$
So, common difference is -8
Now, next two terms:
$a_n = a_{n-1} + d$
$5^{th}$ term
$a_5 = a_{5-1} + (-8)$
$= a_4 - 8$
$= 51 - 8$
$\Rightarrow a_5 = 43$
$6^{th}$ 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 575 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 585 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 595 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 605 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, $988 = 104 + (n - 1)13 988 = 104 + 13n - 13 988 = 91 + 13$n 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 615 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 $\left(a_n\right)$, 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 ($a_n$) = 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 + 3n - 3$
$50 = 2 + 3n$
$3n = 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 $S_n = 440$
View full question & answer→Question 625 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 635 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^{th}$ term,
$a_{13} = 3 + (13 - 1)7$
$= 3 + 12 \times 7$
$= 3 + 84$
$\Rightarrow a_{13} = 87$
Now, $n^{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 + 7n - 7$
$\Rightarrow 7n = 175$
$\Rightarrow n = 25$
Hence, $25^{th}$ term of the given A.P. is 84 more than its $13^{th}$ term.
View full question & answer→Question 645 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 × 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 655 Marks
If the $10^{\text {th }}$ term of an A.P. is 21 and the sum of its first 10 terms is 120 , find its $n ^{\text {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 665 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 675 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)^{th} term = a + (m + 1 - 1)d$
$= a + md$
and $(n + 1)^{th} term = a + (n + 1 - 1)d$
$= a + nd$
$\because$ $(m + 1)^{th} term = 2 (n + 1)^{th}$ term
$\therefore$ $a + md = 2 (a + nd)$
$\Rightarrow a + md = 2a + 2nd$
$\Rightarrow 2a - a = md - 2nd$
$a = d(m - 2n) = (m - 2n)d .....(i)$
Now $(3 m+1)^{\text {th }}$ term $=a+(3 m+1-1) d$
$= a + 3md = (m - 2n)d + 3md$
$= (m - 2n + 3m)d = (4m - 2n)d = 2(2m - n)d .....(ii)$
and $(m + n + 1)^{th} term = a + (m + n + 1 - 1)d$
$= a + (m + n)d$
$= (m - 2n)d + (m + n)d$
$= (m - 2n + m + n)d$
$= (2m - n)d .....(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 685 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 695 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(a_8) .....(i)$
and $a_{11} = 43$
We know, $a_n=a+(n-1) d$
$8^{th}$ term, $a_8 = a + (8 - 1)d$
$\Rightarrow a_8 = a + 7d$
$11^{th}$ term, $a_{11} = a + (11 - 1)d$
$\Rightarrow 43 = a + 10d$
$\Rightarrow a = 43 - 10d .....(2)$
$17^{th}$ term, $a_{17} = a + (17 - 1)d$
$\Rightarrow a_{17} = a + 16d$
$\Rightarrow a_{17} = 43 - 10d + 16d$
$\Rightarrow a_{17} = 43 + 6d$
By putting value of $a_8$ and $a_{17}$ in eq. (i)
$\Rightarrow 43 + 6d = 5 + 2(a + 7d)$
$\Rightarrow 43 + 6d = 5 + 2a + 14d$
$\Rightarrow 43 - 5 = 2a + 14d - 6d$
$\Rightarrow 38 = 2a + 8d$
$\Rightarrow 38 = 2(43 - 10d) + 8d [by eq. (ii)]$
$\Rightarrow 38 = 86 - 20d + 8d$
$\Rightarrow 38 = 86 - 12d$
$\Rightarrow 12d = 86 - 38$
$\Rightarrow\ \text{d}=\frac{48}{12}=4$
From eq. (ii) a = 43 - 10d
$= 43 - 10 \times 4$
$= 43 - 40 = 3$
We know, $n^{th}$term of A.P.
$a_n = a + (n - 1)d$
$= 3 + (n - 1)4$
$= 3 + 4n - 4$
$\Rightarrow a_n = 4n - 1$
Hence, $n^{th}$ term 4n - 1.
View full question & answer→Question 705 Marks
How many terms of the A.P. 63, 60, 57, ..... must be taken so that their sum is 693?
AnswerGiven,
A.P. 63, 60, 57, .....
and Sum of terms, $S_n = 693$
Here, First term a = 63
and Difference d = 60 - 63 = -3
We know, $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ 693=\frac{\text{n}}{2}[2(63)+(\text{n}-1)(-3)]$
$\Rightarrow 693 \times 2 = n[126 - 3n + 3]$
$\Rightarrow 1386 = n[129 - 3n]$
$\Rightarrow 1386 = 129n - 3n^2$
$\Rightarrow 3n^2 - 129n + 1386 = 0$
$\Rightarrow 3(n^2 - 43n + 462) = 0$
$\Rightarrow n^2 - 43n + 462 = 0$
$\Rightarrow n^2 - 22n - 21n + 462 = 0$
$\Rightarrow n(n - 22) - 21(n - 22) = 0$
$\Rightarrow (n - 22)(n - 21) = 0$
Now,
$n - 22 = 0$ and $n - 21 = 0$
$n = 22, n = 21$
Hence, no of terms are 21 and 22.
View full question & answer→Question 715 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.?
AnswerGiven, $a_4 + a_8 = 24 .....(i)$
$a_6 + a_{10}= 44 .....(ii)$
$We know, a_n = a + (n - 1)d$
Then,
$4^{th} term, a_4 = a + (4 - 1)d$
$\Rightarrow a4 = a + 3d$
$6^{th} term, a_6 = a + (6 - 1)d$
$\Rightarrow a_6 = a + 5d$
$8^{th} term, a_8 = a + (8 - 1)d$
$\Rightarrow a_8 = a + 7d$
$10^{th} term, a_{10} = a + (8 - 1)d$
$\Rightarrow a_{10} = a + 9d$
Put the value of $a_4$and $a_8$ in eq. (i)
$\Rightarrow a + 3d + a + 7d = 24$
$\Rightarrow 2a + 10d = 21 .....(iii)$
Put the value of $a_6$ and $a_{10}$ in eq (ii)
$\Rightarrow a + 5d + a + 9d = 44$
$\Rightarrow 2a + 14d = 44 .....(iv)$
By substituting eq. (iii) from (iv)
$\Rightarrow 2a + 14d - (2a + 10d) = 44 - 24$
$\Rightarrow 2a + 14d - 2a - 10d = 20$
$\Rightarrow 4d = 20$
$\Rightarrow d = 5$
By putting value of d in eq. (iii)
$\Rightarrow 2a + 10(5) = 24$
$\Rightarrow 2a + 50 = 24$
$\Rightarrow 2a = 24 - 50$
$\Rightarrow a = -13$
We know, A.P. is
$a, a + d, a + 2d, .....$
$\Rightarrow -13, -13 + 5, -13 + 2(5), .....$
$\Rightarrow -13, -8, -13 + 10, .....$
$\Rightarrow -13, -8, -3, .....$
Hence,$ A.P. is -13, -8, -3, .....$
View full question & answer→Question 725 Marks
The sum of first seven terms of an A.P. is 182 . If its $4^{\text {th }}$ and the $17^{\text {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 735 Marks
The $6^{\text {th }}$ and $17^{\text {th }}$ terms of an A.P. are 19 and 41 respectively, find the $40^{\text {th }}$ 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,
$\Rightarrow a_6 = a + (6 - 1)d$
$\Rightarrow 19 = a + 5d .....(i)$
and $17^{th}$term,
$\Rightarrow a_{17} = a + (17 - 1)d$
$\Rightarrow 41 = a + 16d .....(ii)$
by substuting eq. (i) from (ii)
$\Rightarrow 41 - 19 = (a + 16d) - (a + 5d)$
$\Rightarrow 22 = a + 16d - a - 5d$
$\Rightarrow 22 = 11d$
$\Rightarrow\ \text{d}=\frac{22}{11}$
⇒ d = 2
Now from eq. (i)
$\Rightarrow 19 = a + 5d$
$\Rightarrow 19 = a + 5 \times 2$
$\Rightarrow a = 19 - 10$
$\Rightarrow a = 9$
We have to find $40^{th}$ term then putting n = 40
$a_{40} = a + (40 - 1)d$
$= 9 + 39 \times 2$
$\Rightarrow a_{40} = 87$
Hence, $40^{th}$ term is 87.
View full question & answer→Question 745 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 755 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.
$4+8+12+16+\ldots . . .+52$ [for left]
and $4+8+12+16+\ldots . . .+52$ [for right].
Total distance covered by Ruchi for placing these flags
= 2 × (4 + 8 + 12 + ..... +52).
$=2\times\Big[\frac{13}{2}\{2\times4+(13-1)\times(8-4)\}\Big]$
$\begin{Bmatrix}\because\frac{\text{Sumpf n terms of an A.P.}}{\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]} \end{Bmatrix}$
$=2\times\Big[\frac{13}{2}(8+12\times4)\Big]$
$[\because$ 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 $14^{\text {th }}$ flag in her left position or $27^{\text {th }}$ flag in her right position
$=(2+2+2+\ldots . .+13 \text { times })$
$=2 \times 13=26 m$
Hence, the required maximum distance she travelled carrying a flag is 26 m .
View full question & answer→Question 765 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 775 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 $11^{\text {th }}$ term $(n=11)$,
$a_{11}=a+(11-1) d$
$=a+10 d$
For the $20^{\text {th }}$ term $( n =20)$,
$a_{20}=a+(20-1) d$
$=a+19 d$
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.
View full question & answer→