Question 13 Marks
The $7^{\text {th }}$ term of an AP is -4 and its $13^{\text {th }}$ term is -16 . Find the AP.
AnswerIn the given AP, let the first term = a common difference = d
Then, $T_n = a + (n - 1)d$
$\Rightarrow T_7 = a + (7 - 1)d, and T_{13} = a + (13 - 1)d$
$\Rightarrow T_7 = -4 \Rightarrow a + 6d = -4 .....(1)$
$\Rightarrow T_{13} = -16 \Rightarrow a + 12d = -16 .....(2)$
Subtracting (1) from (2), we get
$6d = -12$
$\Rightarrow d = -2$
Putting d = -2 in (1), we get
$a + 6 \times (-2) = -4$
$\Rightarrow a - 12 = -4$
$\Rightarrow a = 8$
Thus, a = 8, d = -2
So the required AP is 8, 6, 4, 2, 0 .....
View full question & answer→Question 23 Marks
Find the sum of first n even natural numbers.
AnswerThe first n even natural numbers are 2, 4, 6, 8, 10, ..., n.
Here, a = 2 and d = (4 - 2) = 2
Sum of n terms of an AP is given by
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$=\big(\frac{\text{n}}{2}\big)\times\big[2\times2+(\text{n}-1)\times2\big]$
$=\big(\frac{\text{n}}{2}\big)\times\big[4+2\text{n}-2\big]=\big(\frac{\text{n}}{2}\big)\times(2\text{n}+2)=\text{n}(\text{n}+1)$
Hence, the required is n(n + 1).
View full question & answer→Question 33 Marks
How many term are there in the $AP 6, 10, 14, 18,....174?$
AnswerIn the given AP, we have a = 6 and d = (10 - 6) = 4
Suppose there are n term in the given AP, then
$T_n = 174$
$\Rightarrow a + (n - 1)d = 174$
$\Rightarrow 6 + (n - 1)4 = 174$
$\Rightarrow 6 + 4n - 4 = 174$
$\Rightarrow 2 + 4n = 174$
⇒ $\text{n}=\frac{172}{4}=43$
Hence there are 43 terms in the given AP
View full question & answer→Question 43 Marks
Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four termes.
AnswerThe genaral term of an AP is given by $a_n = a + (n - 1)d$
and $S_n =$ $\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big].$
Given that a = 5
Also, $\text{S}_4=\frac{1}{2}(\text{S}_8-\text{S}_4)$
$\Rightarrow\frac{4}{2}\big[2(5)+3\text{d}\big]$
$=\frac{1}{2}\Big[\frac{8}{2}[2(5)+7\text{d}]-\frac{4}{2}[2(5)+3\text{d}]\Big]$
$\Rightarrow4\big[10+3\text{d}\big]=\big[4(10+7\text{d})-2(10+3\text{d})\big]$
$\Rightarrow40+12\text{d}=40+28\text{d}-20-6\text{d}$
$\Rightarrow-10\text{d}=-20$
$\Rightarrow\text{d}=2$
Thus, the common difference is 2.
View full question & answer→Question 53 Marks
How many term are there in the AP 41, 38, 35,....8?
AnswerIn the given AP, we have $a = 41$ and $d = 38 - 41 = -3$
Suppose there are n term in the given AP, then
$T_n = 8$
$\Rightarrow a + (n - 1)d = 8$
$\Rightarrow 41 + (n - 1)(-3) = 8$
$\Rightarrow 41 - 3n + 3 = 8$
$\Rightarrow -3n = -36$
$\Rightarrow n = 12$
Hence there are 12 terms in the given AP
View full question & answer→Question 63 Marks
If the $n^{th}$ term of a progression is (4n - 10) show that it is an AP. Find its:
- First term.
- Common difference.
- $16^{th}$ term.
Answer$T_n = (4n - 10)$
$\Rightarrow T_1 = (4 \times 1 - 10) = -6$
and
$T_2 = (4 \times 2 - 10) = -2$
Thus, we have
- First term = -6
- Common difference $= (T_2 - T_1) = (-2 + 6) = 4$
- $16^{th}$ term $= a + (16 - 1)d$, where a = -6 and d = 4
$= (-6 + 15 × 4) = 54$ View full question & answer→Question 73 Marks
Find the sum of the following APs:
0.6, 1.7, 2.8, ..... to 100 terms.
AnswerHere, a = 0.6, d = 1.7 - 0.6 = 1.1 and n = 100
Now, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$\therefore\text{S}_\text{100}=\frac{\text{100}}{2}\big[2\times0.6+(100-1)(1.1)\big]$
$=50\big[1.2+108.9\big]$
$=50\times110.1$
$=5505$
View full question & answer→Question 83 Marks
Which term of the AP 121, 117, 113, .... is its first negative term?
AnswerThe given AP is 121, 117, 113, ...,
Common difference = 117 - 121 = -4
The general term of an AP is given by
$a_n = a + (n - 1)d$
$\Rightarrow a + (n - 1)d < 0$
$\Rightarrow 121 + (n - 1)(-4) <0$
$\Rightarrow 121 - 4n + 4 <0$
$\Rightarrow -4n + 125 <0$
$\Rightarrow -4n < - 125$
$\Rightarrow\text{n}>\frac{125}{4}=31.25$
So, n > 31.25
Hence, the first negative term would be the $32^{nd}$ term.
View full question & answer→Question 93 Marks
If the $10^{\text {th }}$ term of an AP is 52 and $17^{\text {th }}$ term is 20 more than its $13^{\text {th }}$ term, find the AP .
AnswerIn the given AP let first term = a
And common difference = d
Then, $T_n = a + (n - 1)d$
$T_{10} = a + (10 - 1)d, T_{17} = a + (17 - 1)d$ and $T_{13} = a + (13 - 1)d$
$T_{10} = a + 9d, T_{17} = a + 16d$ and $T_{13} = a + 12d$
Now, $T_{10}= 52$
$\Rightarrow a + 9d = 57 .....(1)$
and $T_{17} = T_{13} + 20$
$\Rightarrow a + 16d = a + 12d + 20$
$\Rightarrow 4d = 20$
$\Rightarrow d = 5$
Putting d = 5 in (1), we get
$a + 9 \times 5 = 52$
$\Rightarrow a = 52 - 45$
$\Rightarrow a = 7$
Thus, a = 7 and d = 5
So the required AP is 7, 12, 17, 22....
View full question & answer→Question 103 Marks
Find the sum of all odd numbers between 0 and 50.
AnswerOdd numbers between 0 to 50 are
1, 3, 5, 7, ....., 49
Here
First term = a = 1
last term = l = 49
There are 25 such terms
So, n = 25
We need to find sum
So, we can use formula
$\text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}+\text{l})$
Putting value in the formula
$\text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}+\text{l})$
$=\frac{25}{2}(1+49)$
$=\frac{25}{2}\times50$
$=625$
Therefore, the sum of odd number between 0 & 50 is 625.
View full question & answer→Question 113 Marks
For what value of $n$, the $n^{\text {th }}$ terms of the arithmetic progressions $63,65,67, \ldots$ and $3,10,17, \ldots$ are equal?
AnswerFirst AP is $63,65,67 \ldots$
First term $=63$, common difference $=65-63=2$
$\therefore n^{\text {th }} \text { term }=63+(n-1) 2=63+2 n-2=2 n+61$
Second AP is $3,10,17 \ldots$
First term $=3$, common difference $=10-3=7$
$n^{\text {th }} \text { term }=3+(n-1) 7=3+7 n-7=7 n-4$
The two $n ^{\text {th }}$ term are equal
$\therefore 2 n+61=7 n-4 \text { or } 5 n=61+4=65$
$\Rightarrow n=\frac{65}{4}=13$
View full question & answer→Question 123 Marks
Which term of the AP $\frac{5}{6},1,1\frac{1}{6},1\frac{1}{3},\ ...$ is 3?
Answer$\text{a}=\frac{5}{6};\text{d}=\Big(1-\frac{5}{6}\Big)=\frac{1}{6}$
In the given AP, we have
Suppose there are n terms in given AP, we have
Then,
$\text{T}_\text{n}=3$
$\Rightarrow\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}=3$
$\Rightarrow\frac{5}{6}+(\text{n}-1)\frac{1}{6}=3$
$\Rightarrow\frac{5}{6}+\frac{1}{6}\text{n}-\frac{1}{6}=3$
$\Rightarrow4+\text{n}=18\Rightarrow\text{n}=14$
$\therefore\text{n}=14$
Thus, $14^{th}$ term in the given AP is 3
View full question & answer→Question 133 Marks
Determine the $n ^{\text {th }}$ term of the AP whose $7^{\text {th }}$ term is -1 and $16^{\text {th }}$ term is 17 .
AnswerThe general term of an AP is given by
$a_n = a + (n - 1)d$
Given that $a_7= -1$
$\Rightarrow a + 6d = -1 ...(i)$
Now,
$a_{16}= 17$
$\Rightarrow a + 15d = 17...(ii)$
Subtract from (i) from (ii).
$9d = 18$
$\Rightarrow d = 2$
Substituting in (i), we get $a = -13.$
Thus, the $n^{th}$ term will be,
$a_n = a + (n - 1)d$
$\Rightarrow a_n = -13 + (n - 1)(2)$
$\Rightarrow a_n = -13 + 2n - 2$
$\Rightarrow a_n = 2n - 15.$
View full question & answer→Question 143 Marks
The sum of the $2^{\text {nd }}$ and the $7^{\text {th }}$ terms of an $A P$ is 30 . If its $15^{\text {th }}$ term is 1 less than twice its $8^{\text {th }}$ term, find the $A P$.
AnswerThe genaral term of an AP is given by $a_n = a + (n - 1)d$
Given that $a_2 + a_7= 30$
$\Rightarrow a + d +a + 6d = 30$
$\Rightarrow 2a + 7d = 30 .....(i)$
Next, $a_{15} = 2a_8 - 1$
$\Rightarrow a + 14d = 2(a + 7d) - 1$
$\Rightarrow a + 14d = 2a + 14d - 1$
$\Rightarrow a = 1$
Substituting in (i), we get d = 4.
So, the AP is a, a + d, a + 2d, a + 3d.
that is,$ 1, 5, 9, 13,.....$
View full question & answer→Question 153 Marks
Find the sum of the following arithmetic series:$(-5) + (-8) + (-11) + .... + (-230)$
Answerwe have
$a = -5,$
$d = -8 - (-5) = -8 + 5 = -3$
Let the total number of terms be n.
Then $T_n = -230$
$⇒ \text{a} + (\text{n} - 1)\text{d} = -230$
$\Rightarrow-5+(\text{n}-1)(-3)=-230$
$\Rightarrow(\text{n}-1)(-3)=-225$
$\Rightarrow\text{n}-1=\frac{-225}{-3}$
$\Rightarrow\text{n}-1=75$
$\Rightarrow\text{n}=76$
Now, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[\text{a}+\text{l}\big]$
$\therefore\text{S}_\text{76}=\frac{\text{76}}{2}\big[-\text{5}+(-230)\big]$
$=38\times(-235)=-8930$
View full question & answer→Question 163 Marks
The $16^{\text {th }}$ term of an AP is 5 times its $3^{\text {rd }}$ term. If its $10^{\text {th }}$ term is 41 , find the sum of its first 15 terms.
AnswerThe general term of an AP is given by $a_n = a + (n - 1)d$
Given that $a_{16} = 5a_3$
$\Rightarrow a + 15d = 5(a + 2d)$
$\Rightarrow a + 15d = 5a + 10d$
$\Rightarrow 4a = 5d ....(i)$
Now,$a_{10} = 41$
$\Rightarrow a + 9d = 41$
$\Rightarrow 4a + 36d = 164$
$\Rightarrow 5d + 36d = 164 ....(from (i))$
$\Rightarrow 41d = 164$
$\Rightarrow d = 4$ substituting in (i),
we gwt a = 5. We know that, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
Sum of the first 15 term $=\text{S}_{15}$
$=\frac{15}{2}\big[2(5)+14(4)\big]$
$=5[10+56]$ $=495$
View full question & answer→Question 173 Marks
The $4^{\text {th }}$ term of an AP is zero. Prove that its $25^{\text {th }}$ term is triple its $11^{\text {th }}$ term.
AnswerIn the first term of given AP = a and common difference = d
Then, $T_n = a + (n - 1)d$
$\Rightarrow T_4 = a + (4 - 1)d, T_{25}= a + (25 - 1)d$, and $T_{11} = a + (11 - 1)d$
$\Rightarrow T_4 = a + 3d, T_{25} = a + 24d$, and $T_{11}= a + 10d$
Now, $T_4 = 0$
$\Rightarrow a + 3d = 0$
$\Rightarrow a = -3d$
$\therefore$ $T_{25} = a + 24d = (-3d + 24d) = 21d$
And $T_{11} = a + 10d = -3d + 10d = 7d$
$\therefore$ $T_{25} = 21d = 3 \times (7d) = 3 \times T_{11}$
Then, $T_n = a + (n - 1)d$
$\Rightarrow T_4 = a + (4 - 1)d, T_{25} = a + (25 - 1)d$, and $T_{11} = a + (11 - 1)d$
$\Rightarrow T_4 = a + 3d, T_{25} = a + 24d$, and $T_{11} = a + 10d$
Now, $T_4 = 0$
$\Rightarrow a + 3d = 0$
$\Rightarrow a = -3d$
$\therefore$ $T_{25} = a + 24d = (-3d + 24d) = 21d$
And $T_{11} = a + 10d = -3d + 10d = 7d$
$\therefore$ $T_{25} = 21d = 3 \times (7d) = 3 \times T_{11}$
Hence $25^{\text {th }}$ term is triple its $11^{\text {th }}$ term
View full question & answer→Question 183 Marks
The $13^{\text {th }}$ term of an AP is 4 times its $3^{\text {rd }}$ term. If its $5^{\text {th }}$ term is 16 , find the sum of its first 10 terms.
AnswerThe general term of an AP is given by
$a_n=a+(n-1) d$
Given that $a_{13} = 4a_3$
$\Rightarrow a + 12d = 4(a + 2d)$
$\Rightarrow a + 12d = 4a + 8d$
$\Rightarrow 3a = 4d ....(i)$
Now,
$a_5 = 16$
$\Rightarrow a + 4d = 16$
$\Rightarrow 3a + 12d = 48$
$\Rightarrow 4d + 12d = 48$ ....(from (i))
$\Rightarrow 16d = 48$
$\Rightarrow d = 3$
substituting in (i), we gwt a = 4.
We know that, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
Sum of the first 10 term
$=\text{S}_{10}$
$=\frac{10}{2}\big[2(4)+9(3)\big]$
$=5[8+27]$
$=175$
View full question & answer→Question 193 Marks
Find the sum of first n terms of an AP whose nth term is (5 - 6n). Hence, find the sum of its first 20 terms.
AnswerThe $n ^{\text {th }}$ term of the given A.P. is given by, $T_n=5-6 n$
Let a be the first term and $d$ be the common difference of this A.P..
Then,
$a = T_1= 5 - 6(1) = 5 - 6 = -1$
$T_2 = 5 - 6(2) = 5 - 12 = -7$
$\Rightarrow d = T_2 - T_1 = -7 = -7 - (-1) + 1 = -6$
$\therefore$ Sum of first n terms $=\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$=\frac{\text{n}}{2}\big[2(-1)+(\text{n}-1)(-6)\big]$
$=\frac{\text{n}}{2}\big[-2-6\text{n}+6)\big]$
$=\frac{\text{n}}{2}\big[4-6\text{n}\big]$
$=\text{n}\big[2-3\text{n}\big]$
$\therefore$ Sum of first 20 terms $= S_{20} = 20[2 - 3 \times 20] = 20 \times (-58) = -1160$
View full question & answer→Question 203 Marks
The $19^{\text {th }}$ term of an AP is equal to 3 times its $6^{\text {th }}$ term. If its $9^{\text {th }}$ term is 19 , find the $A P$.
AnswerThe genaral term of an AP is given by $a_n=a+(n-1) d$ Given that $a_{19}=3 a_6$
$\Rightarrow a + 18d = 3(a + 5d)$
$\Rightarrow a + 18d = 3a + 15d$
$\Rightarrow 2a = 3d ....(i)$
Next, $a_9= 19$
$\Rightarrow a + 8d = 19$
$\Rightarrow 2a + 16d = 38$
$\Rightarrow 3d + 16d = 38$
$\Rightarrow 19d = 38$
$\Rightarrow d = 2$
Substituting in (i), we get a = 3
Thus, the AP is 3, 5, 7, 9, ....
View full question & answer→Question 213 Marks
Which term of the AP 72, 68, 64, 60, ..... is 0?
AnswerIn the given AP, we have a =72 and d = 68 - 72 = -4
Suppose there are n terms in given AP, we have
$T_n = 0$
$\Rightarrow a + (n - 1)d = 0$
$\Rightarrow 72 + (n - 1)(-4) = 0$
$\Rightarrow 72 - 4n + 4 = 0$
$\Rightarrow 4n = 76$
$\Rightarrow n = 19$
Hence, the $19^{th}$ term in the given AP is 0.
View full question & answer→Question 223 Marks
How many terms of the$\text{AP}\ 20, 19\frac{1}{3},18\frac{2}{3},\ ...$must be taken so that their sum is 300? Explain the double answer.
AnswerHere$\text{a}=20,\ \text{d}=19\frac{1}{3}-20=\frac{58}{3}-20$
$=\frac{58-60}{3}=\frac{-2}{3}$
Let the required number of terms be n.
Then, $\text{S}_\text{n}=300$
$\Rightarrow\frac{\text{}\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]=300$
$\Rightarrow\frac{\text{n}}{2}\Big[2(20)+(\text{n}-1)\Big(\frac{-2}{3}\Big)\Big]=300$
$\Rightarrow\text{n}\big[40-\frac{2}{3}\text{n}+\frac{2}{3}\big]=600$
$\Rightarrow\text{n}\big[\frac{122}{3}-\frac{2}{3}\text{n}\big]=600$
$\Rightarrow\frac{122}{3}\text{n}-\frac{2}{3}\text{n}^2=600$
$\Rightarrow122\text{n}-2\text{n}^2=1800$
$\Rightarrow61\text{n}-\text{n}^2=900$
$\Rightarrow\text{n}^2-61\text{n}+900=0$
$\Rightarrow\text{n}^2-25\text{n}-36\text{n}+900=0$
$\Rightarrow\text{n}(\text{n}-25)-36(\text{n}-25)=0$
$\Rightarrow(\text{n}-25)(\text{n}-36)=0$
$\Rightarrow\text{n}-25=0$ or $\text{n}-36=0$
$\Rightarrow\text{n}=25$ or $\text{n}=36$
$\therefore$ Sum of first 25
terms = Sum of first 36 terms = 300
This means that the sum of all terms from $26^{nd}$ to $36^{th}$ term is zero.
View full question & answer→Question 233 Marks
Find the $6^{th}$ term from the end of the AP 17, 14, 11, ...., (-40).
AnswerHere $a = 17, d = (14 - 17) = -3, l = -40$
And $n = 6$
Now, $n^{th}$ term from the end = [l - (n - 1)d]
$= [-40 - (6 - 1)(-3)]$
$= [-40 + 5 \times 3]$
$= [-40 + 15] = -25$
Hence. the $6^{th}$ term from the end is -25.
View full question & answer→Question 243 Marks
Find the middel term of the AP 10, 7, 4, ....., (-62).
AnswerThe given AP is $10, 7, 4, .... ,(-62)$
$a = 10$ and$ d = 7 - 10 = -3$
$a_n = a + (n - 1)d$
$\Rightarrow -62 = 10 + (n - 1)(-3)$
$\Rightarrow -72 = (n - 1)(-3)$
$\Rightarrow -72 = -3n + 3$
$\Rightarrow 3n = 75$
$\Rightarrow n = 25$
So, there are 25 term in the AP.
The middel term $=\Big(\frac{\text{n}+1}{2}\Big)=\Big(\frac{26}{2}\Big)=13$
So, the $13^{th}$ term is given by
$a_{13} = a + 12d = 10 + 12(-3) = -26.$
View full question & answer→Question 253 Marks
How many two-digit numbers are duvisible by 3?
AnswerThe two-digit numbers divisible by 3 start from
12, 15, 18, 21, ..., 99
Here,
$a = 12$
$d = 3$
$a_n = a + (n - 1)d$
$\Rightarrow 99 = 12 + (n - 1)(3)$
$\Rightarrow 99 = 12 + 3n - 3$
$\Rightarrow 90 = 3n$
$\Rightarrow n = 30$
This, 30 two-digit number are divisible by 3.
View full question & answer→Question 263 Marks
Find:
The $15^{th}$ term of the AP -40, -15, 10, 35, ....
AnswerThe given AP is $-40, -15, 10, 35, ....$
$a = -40$ and $d = -15 - (-40)$
$= -15 + 40 = 25$
$a_n = a + (n - 1)d$
$\Rightarrow a_{15} = -40 + 14(25)$
$\Rightarrow a_{15} = 310$
So, the $15^{th}$ term is 310.
View full question & answer→Question 273 Marks
Find the sum of the following arithmetic series:
$5 + (-41) + 9 + (-39) + 13 + (-37) + 17 + ..... + (-5) + 81 + (-3).$
Answerlet there be two series in the given series
first 5 + 9 + 13 .... 81
and the second be -41 + (-39) + (-37) .... -3
number of terms in first series
$a_n= a + (n - 1)d$
$81 = 5 + (n - 1)d$
$76 = (n - 1)4$
$19 = n - 1$
$n = 20$
similarly in second series there are 20 terms
now, sum of the two series
sum of given series = sum of first series + sum of second series
$=\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}+\frac{\text{n}}{2}\{2\text{a}+(\text{n}-1)\text{d}\}$
$=\frac{\text{n}}{2}\big[\{2\times5+19\times4\}+\{\text{2}\times-41+19\times2)\}\big]$
$=10\big[\{10+76\}+\{-82+32\}\big]$
$=10\times\big[86-44\big]$
$=10\times42$
$=420$
View full question & answer→Question 283 Marks
Which term of the AP 5, 15, 25, .... will be 130 more than its $31^{st}$ term?
AnswerThe given AP is $5, 15, 25 ....$
$\therefore$ $a = 5, d = 15 - 5 = 10$
We have, $T_n = 130 + T_{31}$
$\Rightarrow a + (n - 1)d = 130 + 5 + (31 - 1) \times 10$
$\Rightarrow 5 + (n - 1) \times 10 = 130 + 5 + (31 - 1) \times 10$
$\Rightarrow 5 + 10n - 10 = 135 + 300$
$\Rightarrow 10n - 5 = 435$ or 10n = 453 + 5
$\therefore\text{n}=\frac{440}{10}=44$
Thus, the required term is $44^{th}$
View full question & answer→Question 293 Marks
If the $p^{\text {th }}$ term of an $A P$ is $q$ and its $q^{\text {th }}$ term is $p$ then show that its $(p+q)^{\text {th }}$ term is zero.
AnswerLet a be the term and d be the common difference
$p^{th} term = a + (p - 1)d = q(given) .....(1)$
$q^{th} term = a + (q - 1)d = p(given) .....(2)$
subtracting (2) from (1)
$(p - q)d = q - p$
$(p - q)d = -(p - q)$
$\therefore$ $d = -1$
Putting $d = -1 in (1)$
$a - (p - 1) = q$
$\therefore$ $a = p + q - 1$
$\therefore$ $(p + q)^{th} $term $= a + (p + q - 1)d$
$= (p + q - 1) = (p + q - 1) = 0$
View full question & answer→Question 303 Marks
A sum of ₹ 2800 is to be used to award four prizes. If each prize after the first is ₹ 200 less than the preceding prize, find the value of each of the prizes.
AnswerLet the first price be ₹ a.
Since each prize after the first is ₹ 200
Less than the preceding prize,
So, the prizes from the first to the last are
₹ a, ₹(a - 200), ₹(a - 400), ₹(a - 600)
The common difference = (a - 200) - a = -200
$\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow2800=\frac{4}{2}\big[2\text{a}+3(-200)\big]$
$\Rightarrow1400=2\text{a}-600$
$\Rightarrow2\text{a}=2000$
$\Rightarrow\text{a}=1000$
So, the prizes are ₹ 1000, ₹ 800, ₹ 600 and ₹ 400.
View full question & answer→Question 313 Marks
Which term of the AP 21, 18, 15, .... is -81?
AnswerThe given AP is 21, 18, 15, ...
a = 21 and $d = 18 - 21 = - 3$
$a_n = a + (n - 1)d$
$\Rightarrow -18 = 21 + (n - 1)(-3)$
$\Rightarrow -18 = 21 + (n -1)(-3)$
$\Rightarrow -18 = 21 - 3n + 3$
$\Rightarrow 3n = 105$
$\Rightarrow n = 35$
So, -18 is the $35^{th}$ term.
View full question & answer→Question 323 Marks
How many terms of the AP 63, 60, 57, 54, ..... must be taken so that their sum is 693? Explain the double answer.
AnswerHere $a = 63, d = (60 - 63) = -3$
Let the required number of terms be n.
Then, $\text{S}_\text{n}=693$
$\Rightarrow\frac{\text{}\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]=693$
$\Rightarrow\frac{\text{n}}{2}\big[2(63)+(\text{n}-1)(-3)\big]=693$
$\Rightarrow\text{n}\big[126-3\text{n}+3\big]=1386$
$\Rightarrow\text{n}\big[129-3\text{n}\big]=1386$
$\Rightarrow129\text{n}-3\text{n}^2=1386$
$\Rightarrow3\text{n}^2-129\text{n}+1386=0$
$\Rightarrow\text{n}^2-43\text{n}+462=0$
$\Rightarrow\text{n}^2-21\text{n}-22\text{n}+462=0$
$\Rightarrow\text{n}(\text{n}-21)-22(\text{n}-21)=0$
$\Rightarrow(\text{n}-21)(\text{n}-22)=0$
$\Rightarrow\text{n}=21=0$ or $\text{n}-22=0$
$\Rightarrow\text{n}=21$ or $\text{n}=22$
$\therefore$ Sum of first 21 terms = Sum of first 22 terms = 693
This means that the $22^{nd}$ term is zero.
View full question & answer→Question 333 Marks
The sum of the first n terms of an AP is given by $Sn = (3n^2 - n)$. Find its:
- $n^{th}$ term
- First term
- Common difference.
Answer$S_n = (3n^2- n)$ (given)
$\therefore$ $S_{n - 1} = [3(n - 1)^2 - (n - 1)]$
$= [3n^2 + 3 - 6n - n + 1]$
$= [3n^2 - 7n + 4]$
The $n^{th}$ term is given by
$T_n = (S_n - S_{n - 1})$
$= [(3n^2 - n) - (3n^2- 7n + 4)]$
$= 6n - 4$
$\therefore$ $n^{th}$ term = 6n - 4
Putting n = 1 in (1), we get
$T_1 = (6 \times 1 - 4) = 2$
$\therefore$ First term = 2
Putting n = 1 in (1), we get $T_2 = (6 \times 2 - 4) = 8$
$\therefore$ $d = (T_1 - T_2) = 8 - 2 = 6$
View full question & answer→Question 343 Marks
Is 184 a term of the AP 3, 7, 11, 15, ....?
AnswerThe given AP is 3, 7, 11, 15
Common difference = 7 - 3 = 4
So, every term of the AP will be an odd number,
Since even + odd = odd
184 is an even number and hence cannot be a term of the given AP.
View full question & answer→Question 353 Marks
The sum of the first n terms of an AP is $\Big(\frac{5\text{n}^2}{2}+\frac{3\text{n}}{2}\Big).$ Find the $n^{th}$ term and the $20^{th}$ term of this AP.
Answer$\text{S}_\text{n}=\Big(\frac{5\text{n}^2}{2}+\frac{3\text{n}}{2}\Big)\dots(1)$
It is given that
Now, $20^{th}$ term
= (sum of first 20 term) - (sum of first 19 terms)
Putting - 20 in (1) we get
$\text{S}_\text{20}=\Big[\frac{5\times(20)^2}{2}+\frac{3\times20}{2}\Big]=[1000+30]$
$=1030$
Putting n = 19 in (1), we get
$\text{S}_\text{19}=\Big[\frac{5\times(19)^2}{2}+\frac{3\times19}{2}\Big]=\Big[\frac{5\times19\times19}{2}+\frac{57}{2}\Big]$
$\Big[\frac{1805}{2}+\frac{57}{2}\Big]=\Big[\frac{1862}{2}\Big]=931$
$\therefore$ $T_{20} = (S_{20} - S_{19}) = (1030 - 931)= 99$
Hence, the $20^{th}$ term is 99
View full question & answer→Question 363 Marks
The sum of the first 7 term of an AP is 49 and the sum of its first 17 terms is 289. Find the sum of its first n terms.
AnswerLet a be the first term and d be the common difference of the given A.P.
Then, we have
$\text{S}_7=49$
$\Rightarrow\frac{7}{2}[2\text{a}+6\text{d}]=49$
$\Rightarrow\frac{7\times2}{2}[\text{a}+3\text{d}]=49$
$\Rightarrow\text{a}+3\text{d}=7\dots\text{(i)}$
Also, $\text{S}_{17}=289$
$\Rightarrow\frac{17}{2}[2\text{a}+16\text{d}]=289$
$\Rightarrow\frac{17\times2}{2}[\text{a}+8\text{d}]=289$
$\Rightarrow\text{a}+8\text{d}=17\dots\text{(ii)}$
Subtracting (i) from (ii), we get
$5\text{d}=10$
$\Rightarrow\text{d}=2$
$\Rightarrow\text{a}=7-3(2)=7-6=1$
$\therefore\text{S}_\text{n}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
$=\frac{\text{n}}{2}\big[2\text{(1)}+(\text{n}-1)\text{2}\big]$
$=\frac{\text{n}}{2}\big[2+2\text{n}-2\big]$
$=\frac{\text{n}}{2}\times2\text{n}$
$=\text{n}^2$
Thus, the first term is 1 and the common difference is 2.
View full question & answer→Question 373 Marks
Find the sum of two middel most term of the $\text{AP}-\frac{4}{3},-1,\frac{-2}{3},\ ....4\frac{1}{3}.$
AnswerThe given AP is $\frac{4}{3},-1,\frac{-2}{3},\ ....4\frac{1}{3}.$
$\text{a}=-\frac{4}{3}$ and $\text{d}=-1-\Big(-\frac{4}{3}\Big)=-1+\frac{4}{3}=\frac{1}{3}$
$\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow4\frac{1}{3}=-\frac{4}{3}+(\text{n} - 1)\Big(\frac{1}{3}\Big)$
$\Rightarrow\frac{13}{3}=-\frac{4}{3}+\frac{\text{n}}{3}-\frac{1}{3}$
$\Rightarrow\frac{13}{3}+\frac{4}{3}+\frac{1}{3}=\frac{\text{n}}{3}$
$\Rightarrow\text{n}=18$
So, there are 18 term in the AP.
The two middel most term will be $\Big(\frac{\text{n}}{2}\Big)^\text{th}$ and $\Big(\frac{\text{n}}{2}+1\Big)^\text{th}$ term
So, the terms are the $9^{th}$ and $10^{th}$ term.
$\text{a}_9=\text{a}+8\text{d}=-\frac{4}{3}+8\Big(\frac{1}{3}\Big)=-\frac{4}{3}+\frac{8}{3}=\frac{4}{3}$
$\text{a}_{10}=\text{a}+9\text{d}=-\frac{4}{3}+9\Big(\frac{1}{3}\Big)=-\frac{4}{3}+\frac{9}{3}=\frac{5}{3}$
So, the sum $=\frac{4}{3}+\frac{5}{3}=3.$
View full question & answer→Question 383 Marks
If 4 times the $4^{\text {th }}$ term of an AP is equal to 18 times its $18^{\text {th }}$ term then find its $22^{\text {nd }}$ term.
AnswerThe general term of an AP is given by
$a_n = a + (n - 1)d$
Given that $4a_4= 18a_{18}$
$\Rightarrow 4(a + 3d) = 18(a + 17d)$
$\Rightarrow 2(a + 3d) = 9(a + 17d)$
$\Rightarrow 2a + 6d = 9a + 153d$
$\Rightarrow 7a = -147d$
$\Rightarrow a = -21a ....(i)$
Now,
$a_{22}= a + 21d$
$\Rightarrow a_{22} = -21d + 21d$
$\Rightarrow a_{22} = 0$
Thus, the $22^{nd}$ term is 0.
View full question & answer→Question 393 Marks
Find the sum of the following arithmetic series:
$7+10\frac{1}{2}+14+...+84$
Answerwe have
$\text{a}=7,\text{d}=10\frac{1}{2}-7=\frac{21}{2}-7=\frac{21-14}{2}=\frac{7}{2}$
Let the total number of terms be n.
Then $T_n = 84$
$⇒ \text{a} + (\text{n} - 1)\text{d} = 84$
$\Rightarrow7+(\text{n}-1)\frac{7}{2}=84$
$\Rightarrow(\text{n}-1)\frac{7}{2}=77$
$\Rightarrow(\text{n}-1)=\frac{77\times2}{7}$
$\Rightarrow\text{n}-1=22$
$\Rightarrow\text{n}=23$
Now, $\text{S}_\text{n}=\frac{\text{n}}{2}\big[\text{a}+\text{l}\big]$
$\therefore\text{S}_\text{23}=\frac{\text{23}}{2}\big[\text{7}+84\big]$
$=\frac{23}{2}\times91=\frac{2093}{2}=1046\frac{1}{2}$
View full question & answer→Question 403 Marks
The $9^{\text {th }}$ term of an AP is -32 and The sum of its $11^{\text {th }}$ and $13^{\text {th }}$ terms is -94 . Find the common difference of the AP.
AnswerThe general term of an AP is given by
$a_n = a + (n - 1)d$
Given that $a_9= -32$
$\Rightarrow a + 8d = -32 ...(i)$
Now,
$a_{11} = a_{13} = -94$
$\Rightarrow a + 10d + a + 12d = -94$
$\Rightarrow 2a + 22d = -94$
$\Rightarrow a + 11d = -47 ...(ii)$
Subtract from (i) from (ii)
$3d = -15$
$\Rightarrow d = -5$
So, the common difference is -5.
View full question & answer→Question 413 Marks
Find the middel term of the AP 6, 13, 20, ....., 216.
AnswerThe given AP is 6, 13, 20, .... ,216
a = 6 and $d = 13 - 6 = 7$
$a_n = a + (n - 1)d$
$\Rightarrow 216 = 6 + (n - 1)(7)$
$\Rightarrow 210 = (n - 1)(7)$
$\Rightarrow 210 = 7n - 7$
$\Rightarrow 7n = 217$
$\Rightarrow n = 31$
So, there are 31 term in the AP.
The middel term $=\Big(\frac{\text{n}+1}{2}\Big)=\Big(\frac{32}{2}\Big)=16$
So, the $16^{th}$ term is given by
$a_{16} = a + 15d = 6 + 15(7) = 111.$
View full question & answer→Question 423 Marks
Which term of the AP 3, 8, 13, 18, .....is 88?
AnswerIn the given AP, we have a = 3 and d = 8 - 3 = 5
Suppose there are n terms in given AP, then
$T_n = a + (n - 1)d$
$\Rightarrow 3 + (n - 1)5 = 88$
$\Rightarrow 3 + 5n - 5 = 88$
$\Rightarrow 5n = 90$
⇒ $\text{n}=\frac{90}{5}=18$
Hence, the $18^{th}$ term of given AP is 88.
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