Question 11 Mark
Find the 10th and nth terms of the G.P. $5, 25,125,….$
AnswerWe have, $5 + 25 + 125 + ...$ is GP.
Here, $a = 5$ and $r = \frac { 25 } { 5 } = 5$
We know that, $T_n = ar^{n-1} = 5 (5)^{n-1} = 5^n$
and $T_{10} = 5 (5)^{10-1} = 5^{10}$
View full question & answer→Question 21 Mark
Insert $6$ numbers between $3$ and $24$ such that the resulting sequence is an A.P.
AnswerLet $A_1, A_2, A_3, A_4, A_5 $ and $A_6 $ be six numbers between $3$ and $24$ such that $3, A_1, A_2, A_3, A_4, A5, A_6, 24$ are in A.P. Here, $a = 3, b = 24, n = 8.$
Thus, $24 = 3 + (8 - 1) d,$ so that $d = 3.$
Therefore, $A1 = a + d = 3 + 3 = 6;$
$A_2 = a + 2d = 3 + 2 \times 3 = 9;$
$A_3 = a + 3d = 3 + 3 \times 3 = 12;$
$A_4 = a + 4d = 3 + 4 \times 3 = 15;$
$A_5 = a + 5d = 3 + 5 \times 3 = 18;$
$A_6 = a + 6d = 3 + 6 \times 3 = 21.$
Therefore,six numbers between $3$ and $24$ are $6, 9, 12, 15, 18$ and $21.$
View full question & answer→Question 31 Mark
The income of a person is ₹3,00,000, in the first year and he receives an increase of ₹10,000 to his income per year for the next 19 years. Find the total amount, he received in 20 years.
AnswerHere, we have an A.P. with a = 3,00,000, d = 10,000, and n = 20.
Using the sum formula, we obtain
$\mathrm{S}_{20}=\frac{20}{2}[600000+19 \times 10000]=10(790000)=79,00,000$
Therefore, the person received ₹79,00,000 as the total amount at the end of 20 years.
View full question & answer→Question 41 Mark
The sum of n terms of two Arithmetic Progression are in the ratio $(3n + 8) : (7n + 15).$ Find the ratio of their $12^{th}$ terms.
AnswerLet $a_1, a_2$ and $d_1, d_2$ are the first term and common difference of two A.P.'S respectively.
According to question, $\frac{{\frac{n}{2}}}{{\frac{n}{2}}}\frac{{\left[ {2{a_1} + \left( {n - 1} \right){d_1}} \right]}}{{\left[ {2{a_2} + \left( {n - 1} \right){d_2}} \right]}} = \frac{{3n + 8}}{{7n + 15}} .... (I)$
$\frac { 12 \text { th term of Ist } \mathrm { A } . \mathrm { P } } { 12 \mathrm { th } \text { term of } 2 \mathrm { ndA.P } } = \frac { a _ { 1 } + 11 d _ { 1 } } { a _ { 2 } + 11 d _ { 2 } }$
put $n = 23$ in eq $(i)$
$\frac { 2 a _ { 1 } + 22 d _ { 1 } } { 2 a _ { 2 } + 22 d _ { 2 } } = \frac { 3 \times 23 + 8 } { 7 \times 23 + 15 }$
$\frac { a _ { 1 } + 11 d _ { 1 } } { a _ { 2 } + 11 d _ { 2 } } = \frac { 7 } { 16 }$
View full question & answer→Question 51 Mark
If the sum of $n$ terms of an A.P. is $n \mathrm{P}+\frac{1}{2} n(n-1) \mathrm{Q}$, where $P$ and $Q$ are constants, find the common difference.
AnswerLet $a_1, a_2, … a_n $ be the given A.P. Then
$S_n = a_1 + a_2 + a_3 +...+ a_{n-1} + a_n = n \mathrm{P}+\frac{1}{2} n(n-1) \mathrm{Q}$
Thus,$S1 = a_1 = P, S_2 = a_1 + a_2 = 2P + Q$
So that $a_2 = S_2 - S_1 = P + Q$
Therefore,the common difference is given by $d = a_2 - a_1 = (P + Q) - P = Q.$
View full question & answer→Question 61 Mark
In an A.P. if $m^{th}$ term is n and the $n^{th}$ term is m, where $m \ne n,$ find the $p^{th}$ term.
AnswerWe have $a_m = a + (m - 1) d = n, ......(i)$
and $a_n = a + (n - 1) d = m ......(ii)$
Solving $(i)$ and $(ii),$ we obtain
$(m - n) d = n - m,$ or $d = - 1, .....(iii)$
and $a = n + m - 1 ......(iv)$
Thus, $a_p = a + (p - 1)d$
$= n + m - 1 + ( p - 1)(-1) = n + m - p$
Thus, the $p^{th} $ term is $n + m - p.$
View full question & answer→Question 71 Mark
Let the sequence $a_n$ be defined as follow: $a_1 = 1, a_n = a_{n - 1} + 2 $for $n \ge 2$
Find first five terms and write corresponding series.
AnswerGiven,
$a_1 = 1, a_2 = a_1 + 2 = 1 + 2 = 3, a_3 = a_2 + 2 = 3 + 2 = 5,$
$a_4 = a_3 + 2 = 5 + 2 = 7, a_5 = a_4 + 2 = 7 + 2 = 9$
Thus,the first five terms of the sequence are $1, 3, 5, 7$ and $9$
The corresponding series is $1 + 3 + 5 + 7 + 9 + ...$
View full question & answer→Question 81 Mark
What is the 20th term of the sequence defined by $a_n = (n - 1) (2 - n) (3 + n)?$
AnswerHere, $n = 20,$
$\therefore a_{20} = (20 - 1) (2 - 20) (3 + 20)$
$= 19 \times (- 18) \times (23) = - 7866.$
View full question & answer→Question 91 Mark
Find the sum to n terms of the series $5 + 11 + 19 + 29 + 41 + ........$
Answer$S_n = 5 + 11 + 19 + 29 + --- + a_{n-1}+ a_n$
$S_n = 5 + 11 + 19 + ----- + a_{n-1}+ a_{n-2}+ a_n$
On subtracting
$0 = 5 + [6+8+10+12+---+(n-1)$ terms$] – a_n$
$a _ { n } = 5 + \frac { ( n - 1 ) [ 12 + ( n - 2 ) \times 2 ] } { 2 }$
$= 5 + (n - 1) (n + 4)$
$= n^2 + 3n + 1$
${S_n} = \sum\limits_{k = 1}^n {{k^2}} + 3\sum\limits_{k = 1}^n k + n$
$= \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } + \frac { 3 n ( n + 1 ) } { 2 } + n$
$= \frac { n ( n + 2 ) ( n + 4 ) } { 3 }$
View full question & answer→Question 101 Mark
If A.M. and G.M. of two positive numbers a and b are 10 and 8, respectively, find the numbers.
AnswerGiven that A.M. $= \frac{a+b}{2}=10 .....(i)$
and G.M. $= \sqrt{a b}=8 ......(ii)$
From (i) and (ii), we obtain
$a + b = 20 ......(iii)$
$ab = 64 ....(iv)$
Substituting the value of a and b from (3), (4) in the identity $(a - b)^2 = (a + b)^2 - 4ab,$
we obtain,
$(a - b)^2 = 400 - 256 = 144$
or $a - b = \pm12 .....(v)$
Solving (iii) and (v), we
$a = 4, b = 16$ or $a = 16, b = 4$
Thus, the numbers a and b are $4, 16$ or $16, 4$ respectively
View full question & answer→Question 111 Mark
Insert three numbers between $1$ and $256$ so that the resulting sequence is a G.P.
AnswerSuppose $G_1, G_2, G_3 $ be three numbers between 1 and 256 such that $1, G_1,G_2, G_3, 256$ is a G.P.
Thus, $256 = r^4 $ giving $r = \pm4 ($Taking real roots only$)$
For $r = 4$, we have $G_1 = ar = 4, G_2 = ar^2 = 16, G_3 = ar^3 = 64$
Similarly, for $r = - 4,$ numbers are $- 4,16$ and $- 64$
Therefore,we can insert $4, 16, 64$ between $1$ and $256$ so that the resulting sequences are in G.P.
View full question & answer→Question 121 Mark
A person has $2$ parents, $4$ grandparents, $8$ great grandparents and so on. Find the numbers of his ancestors during the ten generations preceding his own.
AnswerThe number of ancestors is: $2, 4, 8,16, ....$
It is in GP since;
$\frac 42 = 2$
$\frac 84 = 2$
$\frac {16}8 = 2 ..$.
Therefore, common ratio, $r = 2$
$a = 2$
and $n = 10$
$\because S_n = as r > 1$
$\therefore S_{10} =\frac { 2 \left( 2 ^ { 10 } - 1 \right) } { 2 - 1 }$
$= 2 (2^{10} - 1) = 2 (1024 - 1)$
$= 2 \times 1023 = 2046$
Hence, the numbers of ancestors preceding the person is 2046
View full question & answer→Question 131 Mark
Find the sum of the sequence $7, 77, 777, ........$ to $n.$
Answer$S_n = 7 + 77 + 777 + ---- + n$ terms
$= \frac { 7 } { 9 } [ 9 + 99 + 999 + - - - - + n$ terms $]$
$= \frac { 7 } { 9 }[(10 - 1 ) + (10^2 - 1) + (10^3 - 1) + --- + n$ terms$]$
$= \frac { 7 } { 9 }[(10^1 + 10^2 + ---- +n$ terms$) - (1 + 1 + 1 ---- + n$ terms$)]$
$= \frac { 7 } { 9 } \left[ \frac { 10 \left( 10 ^ { n } - 1 \right) } { 10 - 1 } - n \right]$
$= \frac { 7 } { 9 } \left[ \frac { 10 \left( 10 ^ { n } - 1 \right) } { 9 } - n \right]$
View full question & answer→Question 141 Mark
The sum of first three terms of a GP is $\frac { 13 } { 12 }$ and their product is $- 1.$ Find the terms.
AnswerLet the three numbers be $\frac { a } { r },$ a and ar.
Their $sum = \frac { a } { r } + a +ar = \frac { 13 } { 12 }$
$\Rightarrow a \left[ \frac { 1 } { r } + 1 + r \right] = \frac { 13 } { 12 }$
$\Rightarrow a \left[ \frac { 1 + r + r ^ { 2 } } { r } \right] = \frac { 13 } { 12 }$
$\Rightarrow a (1 + r + r^2) = \frac { 13 } { 12 } r ...(i)$
Their product $= \frac { a } { r } \times a\times ar = - 1 \Rightarrow a^3 = - 1$
$\Rightarrow a = - 1 [$taking cube root on both sides$] ...(ii)$
On putting the value of a in Eq. $(i),$ we get
$(- 1) [1 + r + r^2] = \frac { 13 } { 12 } r$
$\Rightarrow - 1 - r - r^2 = \frac { 13 } { 12 } r$
$\Rightarrow - 12 - 12r - 12r^2 = 13r$
$\Rightarrow 12r^2 + 25r + 12 = 0$
$\Rightarrow 12r^2+ 16r + 9r + 12 = 0$
$\Rightarrow 4r(3r + 4) + 3(3r + 4) = 0$
$\Rightarrow (4r + 3) (3r + 4) = 0$
$\Rightarrow$ Either $3r + 4 = 0$ or $4r + 3 = 0$
$\Rightarrow r = -\frac { 4 } { 3 }$ or $r = \frac { - 3 } { 4 }$
When $a = - 1$ and $r = - \frac { 4 } { 3 },$ then the numbers are
$\frac { - 1 } { - 4 / 3 } , - 1 , - 1 \times \frac { - 4 } { 3 } i . e . , \frac { 3 } { 4 } , - 1 , \frac { 4 } { 3 }$
And when $a = - 1$ and $r = - \frac { 3 } { 4 }$, then the numbers are
$\frac { - 1 } { - 3 / 4 } , - 1 , - 1 \times \frac { - 3 } { 4 } \text { i.e., } \frac { 4 } { 3 } , - 1 , \frac { 3 } { 4 }$
View full question & answer→Question 151 Mark
How many terms of GP 3, $\frac { 3 } { 2 } , \frac { 3 } { 4 }$, ... are needed to give the sum $\frac { 3069 } { 512 }$?
AnswerGiven GP is $3, \frac { 3 } { 2 } , \frac { 3 } { 4 }, ...$
Here, $a = 3, r = \frac { 3 } { 2 } \div 3 = \frac { 1 } { 2 }$
Let $n$ be the number of terms needed.
Then, $S_n = \frac { 3069 } { 512 }$
$\Rightarrow \frac { a \left( 1 - r ^ { n } \right) } { 1 - r } = \frac { 3069 } { 512 } [\because r < 1]$
$\Rightarrow \frac { 3 \left\{ 1 - \frac { 1 } { 2 ^ { n } } \right\} } { 1 - \frac { 1 } { 2 } } = \frac { 3069 } { 512 }$
$\Rightarrow 6 \left( 1 - \frac { 1 } { 2 ^ { n } } \right) = \frac { 3069 } { 512 }$
$\Rightarrow 1 - \frac { 1 } { 2 ^ { n } } = \frac { 3069 } { 3072 }$
$\Rightarrow \frac { 1 } { 2 ^ { n } } = 1 - \frac { 3069 } { 3072 } = \frac { 3072 - 3069 } { 3072 }$
$\Rightarrow \frac { 1 } { 2 ^ { n } } = \frac { 3 } { 3072 } = \frac { 1 } { 1024 }$
$\Rightarrow 2^n = 1024 \Rightarrow 2^n = 2^{10}$^
On comparing the powers, we get
$n = 10$
Hence,$10$ terms are needed to give the $\sum \frac { 3069 } { 512 }.$
View full question & answer→Question 161 Mark
Find the sum of first n terms and the sum of first 5 terms of the geometric series $1 + \frac { 2 } { 3 } + \frac { 4 } { 9 } + - --$
Answera = 1, r = $\frac{2}{3}$
$S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$
$=\frac{1\left[1-\left(\frac{2}{3}\right)^{n}\right]}{1-\frac{2}{3}}$
$=3\left[1-\left(\frac{2}{3}\right)^{n}\right]$
$S_{5}=3\left[1-\left(\frac{2}{3}\right)^{5}\right]=\frac{211}{81}$
View full question & answer→Question 171 Mark
Write the first three terms of $a_{n}=\frac{n-3}{4}$
AnswerHere, $a_{n}=\frac{n-3}{4}$. Thus, $a_{1}=\frac{1-3}{4}=-\frac{1}{2}, a_{2}=-\frac{1}{4}, a_{3}=0$
Therefore the first three terms are $-\frac{1}{2},-\frac{1}{4}$ and 0.
View full question & answer→Question 181 Mark
In a G.P., the $3^{\text {rd }}$ term is $24$ and the $6^{\text {th }}$ term is $192 $. Find the $10^{\text {th }}$ term.
AnswerGiven, $a_3=a r^2=24 \ldots..(i)$
and $a_6=a r^5=192 \ldots (ii)$
Dividing (ii) by (i), we get $r=2$.
Putting, $r=2$ in (i), we get $a=6$.
Hence $a_{10}=6(2)^9=3072$
View full question & answer→Question 191 Mark
Write the first three terms of $a_n = 2n + 5$
AnswerHere $a_n=2 n+5$
Substituting $\mathrm{n}=1,2,3$, we obtain
$a_1=2(1)+5=7, a_2=9, a_3=11$
Thus, the required terms are $7,9$ and $11.$
View full question & answer→Question 201 Mark
Which term of the G.P., $2,8,32, ... $ up to n terms is $131072?$
AnswerLet $131072$ be the $\mathrm{n}^{\text {th }}$ term of the given G.P. Here $\mathrm{a}=2$ and $\mathrm{r}=4$.
Thus, $131072=a_n=2(4)^{n-1}$ or $65536=4^{n-1}$
This gives $4^8=4^{n-1}$.
So that $\mathrm{n}-1=8$, i.e., $\mathrm{n}=9$.
Therefore, $131072$ is the $9^{\text {th }}$ term of the G.P.
View full question & answer→