Question 12 Marks
A manufacturer reckons that the value of a machine, which cost him $₹\ 15625$ will depreciate each year by $20\%.$ Find the estimated value at the end of $5$ years.
AnswerPresent value of the machine $= ₹\ 15625$
Rate of depreciation $= 20\%$
After $1$ year value of machine $= 15625 - 15625 \times \frac { 20 } { 100 } = 15625 – 3125 = ₹\ 12500$
After $2$ year value of machine $= 12500 - 12500 \times \frac { 20 } { 100 } = 12500 – 2500 = ₹\ 10000$
After $3$ year value of machine $= 10000 - 10000 \times \frac { 20 } { 100 } = 10000 – 2000 = ₹\ 8000$
$\therefore$ Sequence of values of machine after depreciation is $12500, 10000, 8000, ...$ is a G.P.
Here $a = 12500, r = \frac { 10000 } { 12500 } = \frac { 4 } { 5 }$
$\therefore a_5 = ar^4 = 12500 \times \left( \frac { 4 } { 5 } \right) ^4 = 12500 \times \frac { 256 } { 625 } = ₹5120$
Therefore, the value of machine at the end of $5$ years is $₹\ 5120$
View full question & answer→Question 22 Marks
For what values of x, the numbers $ \frac { - 2 } { 7 } , x , \frac { - 7 } { 2 }$ are in G.P.?
AnswerGiven, $ \frac { - 2 } { 7 } , x , \frac { - 7 } { 2 }$ are in G.P.
$ \therefore\frac { x } { \frac { - 2 } { 7 } } = \frac { \frac { - 7 } { 2 } } { x }$
$ \Rightarrow x ^ { 2 } = \frac { - 2 } { 7 } \times \frac { - 7 } { 2 }$
$ \Rightarrow x ^ { 2 } = 1$
$ \Rightarrow x = \pm 1$
Therefore, for x = $ \pm1$ th given numbers are in G.P.
View full question & answer→Question 32 Marks
Which term of the sequence $\frac { 1 } { 3 } , \frac { 1 } { 9 } , \frac { 1 } { 27 }$, ... is $\frac { 1 } { 19683 }$?
AnswerHere $a = \frac 13 , r = \frac { 1 } { 9 } \div \frac { 1 } { 3 } = \frac { 1 } { 3 }$ and $a_n = \frac { 1 } { 19683 }$
$\therefore a_n = ar^{n-1}$
$\Rightarrow \frac { 1 } { 19683 } = \frac { 1 } { 3 } \times \left( \frac { 1 } { 3 } \right) ^ { n - 1 }$
$\Rightarrow \left( \frac { 1 } { 3 } \right) ^ { 9 } = \left( \frac { 1 } { 3 } \right) ^ { n }$
$\Rightarrow n = 9$
Therefore, $9^{th}$ term of the given G.P. is $\frac { 1 } { 19683 }$
View full question & answer→Question 42 Marks
Which term of the sequence $2, 2\sqrt 2, 4, .... is 128?$
AnswerHere $a = 2, r = \frac { 2 \sqrt { 2 } } { 2 } = \sqrt { 2 }$ and $a_n = 128$
$\therefore a_n = ar^{n-1}$
$\Rightarrow 128 = 2 \times ( \sqrt { 2 } ) ^ { n - 1 }$
$\Rightarrow 64 = ( \sqrt { 2 } ) ^ { n - 1 }$
$\Rightarrow ( \sqrt { 2 } ) ^ { 12 } = ( \sqrt { 2 } ) ^ { n - 1 }$
$\Rightarrow n - 1 = 12$
$\Rightarrow n = 13$
Therefore, $13^{th} $ term of the given G.P. is $128$
View full question & answer→Question 52 Marks
The $4^{th}$ term of a G.P. is square of its second term and the first term $-3.$ Determine its $7^{th}$ term.
AnswerLet a be the first term and r be the common ratio of given G.P.
Here $a = -3$ and $a_4 = (a_2)^2$
Now,$a_4 = (a_2)^2$
$ \Rightarrow ar^3 = (ar)^2$
$\Rightarrow ar^3 = a^2r^2$
$\Rightarrow r = a$
$\Rightarrow r = -3$
$\therefore a _7 = ar^{7 - 1} = (-3) \times (-3)^6$
$= -3 \times 729 = -2187$
View full question & answer→Question 62 Marks
The $5^{th}, 8^{th}$ and $11^{th}$ terms of a G.P. are $p, q$ and $s$ respectively. Show that $q^2 = ps.$
AnswerLet a be the first term and r be the common ratio of given G.P.
$\therefore a_5 = p \Rightarrow ar^4 = p ...(i)$
$a_8 = q \Rightarrow ar^7 = q ...(ii)$
$a_{11} = s \Rightarrow a r^{10} = s ...(iii)$
Squaring both sides of eq. $(ii),$ we get $q^2 = (ar^7)^2$
$\Rightarrow q^2 = a^2 r^{14}$
$\Rightarrow q^2 = (ar^4) (ar^{10})$
$\Rightarrow q^2 = ps [$From eq. $(i)$ and $(iii)]$
View full question & answer→Question 72 Marks
Find the $12^{th}$ term of a G.P. whose $8^{th}$ term is $192$ and the common ratio is $2.$
AnswerLet a be the first term of given G.P. Here $r =2$ and $a_8 = 192$
$\therefore a_n = ar^{n-1}$
$\Rightarrow a _ { 8 } = a \times ( 2 ) ^ { 8 - 1 } = 192$
$\Rightarrow a \times ( 2 )^7 = 192$
$\Rightarrow a \times 128 = 192$
$\Rightarrow a = \frac { 192 } { 128 } = \frac { 3 } { 2 }$
$\therefore a_{12} = ar^{12 - 1}$
$\Rightarrow a _ { 12 } = \frac { 3 } { 2 } \times 2 ^ { 11 } = 3 \times 2 ^ { 10 }$
$= 3 \times1024 = 3072$
View full question & answer→Question 82 Marks
Find the $20^{th}$ and $n^{th}$ terms of the G.P. $\frac { 5 } { 2 } , \frac { 5 } { 4 } , \frac { 5 } { 8 }$ .......
AnswerHere, $a =\frac 52$ and r =$\frac 54 \div \frac 52 = \frac 12$
$\therefore a_n = ar^{n-1}$
$\Rightarrow a _ { 20 } = \frac { 5 } { 2 } \times \left( \frac { 1 } { 2 } \right) ^ { 20 - 1 }$
$\Rightarrow a _ { 20 } = \frac { 5 } { 2 } \times \left( \frac { 1 } { 2 } \right) ^ { 19 } = \frac { 5 } { 2 ^ { 20 } }$
and $a _ { n } = \frac { 5 } { 2 } \times \left( \frac { 1 } { 2 } \right) ^ { n - 1 } = \frac { 5 } { 2 \times 2 ^ { n - 1 } } = \frac { 5 } { 2 ^ { n } }$
View full question & answer→Question 92 Marks
Find the indicated terms of the sequence, whose nth term is $a _ { n } = ( - 1 ) ^ { n - 1 } n ^ { 3 }; a_9$
AnswerGiven: $a_n = (-1)^{n-1} n^3$
$\therefore a _ { 9 } = ( - 1 ) ^ { 9 - 1 } \times ( 9 ) ^ { 3 } = ( - 1 ) ^ { 8 } \times 729 = 729$
Therefore, $9^{th}$ term is $729.$
View full question & answer→Question 102 Marks
Find the indicated terms of the sequence, whose nth term is $a _ { n } = \frac { n ^ { 2 } } { 2 ^ { n } }; a_7$
AnswerGiven: $a _ { n } = \frac { n ^ { 2 } } { 2 ^ { n } }$
$\therefore a _ { 7 } = \frac { 7 ^ { 2 } } { 2 ^ { 7 } } = \frac { 49 } { 128 }$
Therefore, $7^{th}$ term is $\frac { 49 } { 128 }.$
View full question & answer→Question 112 Marks
Find the indicated terms of the sequence, whose nth term is $a_n = 4n - 3; a_{17}, a_{24}.$
AnswerWe have, $a_n = 4n - 3$
On putting $n = 17,$ we get
$a_{17} = 4 \times 17 - 3 = 68 - 3 = 65$
On putting $n = 24$, we get
$a_{24} = 4 \times 24 - 3 = 96 - 3 = 93$
View full question & answer→Question 122 Marks
Write the first five terms of the sequence whose $n^{th}$ term is $a _ { n } = n . \frac { n ^ { 2 } + 5 } { 4 }$ .
AnswerGiven: $a_{n}=n \cdot \frac{n^{2}+5}{4}$
Putting $n = 1,2,3,4$ and $5,$ we get,
$a_{1}=1 \frac{1^{2}+5}{4}$$=1 . \frac{1+5}{4}=\frac{6}{4}=\frac{3}{2}$
$a_{2}=2 \cdot \frac{2^{2}+5}{4}$$=2 . \frac{4+5}{4}=\frac{18}{4}=\frac{9}{2}$
$a_{3}=3 . \frac{3^{2}+5}{4}$$=3 . \frac{9+5}{4}=3 \times \frac{14}{4}$$=\frac{42}{4}=\frac{21}{2}$
$a_{4}=4 \cdot \frac{4^{2}+5}{4}$$=4 . \frac{16+5}{4}=\frac{84}{4} = 21$
$a_{5}=5 \cdot \frac{5^{2}+5}{4}$$=5 . \frac{25+5}{4}=5 \times \frac{30}{4}$$=\frac{150}{4}=\frac{75}{2}$
Therefore, the first five terms are $\frac{3}{2}, \frac{9}{2}, \frac{21}{2}, 21$ and $\frac{75}{2}$
View full question & answer→Question 132 Marks
Write the first five terms of the sequence whose $n^{th}$ term is $a _ { n } = ( - 1 ) ^ { n - 1 } \cdot 5 ^ { n + 1 }$
AnswerGiven: $a _ { n } = ( - 1 ) ^ { n - 1 } \cdot 5 ^ { n + 1 }$ Putting $n = 1, 2, 3, 4$ and $5,$ we get,
$a _ { 1 } = ( - 1 ) ^ { 1 - 1 } \cdot 5 ^ { 1 + 1 } = ( - 1 ) ^ { 0 } \cdot 5 ^ { 2 } = 1 \times 25 = 25$
$a _ { 2 } = ( - 1 ) ^ { 2 - 1 } 5 ^ { 2 + 1 } = ( - 1 ) ^ { 1 } \cdot 5 ^ { 3 } = - 1 \times 125 = - 125$
$a _ { 3 } = ( - 1 ) ^ { 3 - 1 } \cdot 5 ^ { 3 + 1 } = ( - 1 ) ^ { 2 } \cdot 5 ^ { 4 } = 1 \times 625 = 625$
$a _ { 4 } = ( - 1 ) ^ { 4 - 1 } \cdot 5 ^ { 4 + 1 } = ( - 1 ) ^ { 3 } \cdot 5 ^ { 5 } = - 1 \times 3125 = - 3125$
$a _ { 5 } = ( - 1 ) ^ { 5 - 1 } \cdot 5 ^ { 5 + 1 } = ( - 1 ) ^ { 4 } \cdot 5 ^ { 6 } = 1 \times 15625 = 15625$
Therefore, the first five terms are $25, -125, 625, -3125$ and $15625.$
View full question & answer→Question 142 Marks
Write the first five terms of the sequence whose $n^{th}$ term is $a _ { n } = \frac { 2 n - 3 } { 6 }$.
AnswerGiven: $a _ { n } = \frac { 2 n - 3 } { 6 }$ Putting $n = 1, 2, 3, 4$ and $5,$ we get,
$a _ { 1 } = \frac { 2 \times 1 - 3 } { 6 } = \frac { 2 - 3 } { 6 } = \frac { - 1 } { 6 }$
$a _ { 2 } = \frac { 2 \times 2 - 3 } { 6 } = \frac { 4 - 3 } { 6 } = \frac { 1 } { 6 }$
$a _ { 3 } = \frac { 2 \times 3 - 3 } { 6 } = \frac { 6 - 3 } { 6 } = \frac { 3 } { 6 } = \frac { 1 } { 2 }$
$a _ { 4 } = \frac { 2 \times 4 - 3 } { 6 } = \frac { 8 - 3 } { 6 } = \frac { 5 } { 6 }$
$a _ { 5 } = \frac { 2 \times 5 - 3 } { 6 } = \frac { 10 - 3 } { 6 } = \frac { 7 } { 6 }$
Therefore, the first five terms are $\frac { - 1 } { 6 } , \frac { 1 } { 6 } , \frac { 1 } { 2 } , \frac { 5 } { 6 }$ and $\frac { 7 } { 6 }.$
View full question & answer→Question 152 Marks
Write the first five terms of the sequence whose $n^{th}$ term is $a_n = 2^n$
AnswerGiven: $a_n = 2^n$ Putting $n = 1, 2, 3, 4$ and $5,$ we get,
$a_1 = 2^1 = 2$
$a_2 = 2^2 = 4$
$a_3 = 2^3 = 8$
$a_4 = 2^4 = 16$
$a_5 = 2^5 = 32$
Therefore, the first five terms are $2, 4, 8, 16$ and $32.$
View full question & answer→Question 162 Marks
Write the first five terms of the sequence whose $n^{th}$ term is $a _ { n } = \frac { n } { n + 1 }$
AnswerGiven: $a _ { n } = \frac { n } { n + 1 }$ Putting $n = 1, 2, 3, 4$ and $5,$ we get,
$a _ { 1 } = \frac { 1 } { 1 + 1 } = \frac { 1 } { 2 }$
$a _ { 2 } = \frac { 2 } { 2 + 1 } = \frac { 2 } { 3 }$
$a _ { 3 } = \frac { 3 } { 3 + 1 } = \frac { 3 } { 4 }$
$a _ { 4 } = \frac { 4 } { 4 + 1 } = \frac { 4 } { 5 }$
$a _ { 5 } = \frac { 5 } { 5 + 1 } = \frac { 5 } { 6 }$
Therefore, the first five terms are $\frac { 1 } { 2 } , \frac { 2 } { 3 } , \frac { 3 } { 4 } , \frac { 4 } { 5 }$ and $\frac { 5 } { 6 }.$
View full question & answer→Question 172 Marks
The Fibonacci sequence is defined by $1 = a_1 = a_2 $ and $a_n = a_{n-1}+ a_{n-2}, n > 2.$
Find $\frac { a _ { n + 1 } } { a _ { n } }$, for $n = 1, 2, 3, 4, 5.$
AnswerGiven, $1 = a_1 = a_2$_
and $a_n = a_{n-1} + a_{n-2}, n > 2$
On putting $n = 3, 4, 5, 6$ respectively, we get
For $n = 3, a_3 = a_{3-1} + a_{3-2}= a_2 + a_1 = 1 + 1 = 2$
For $n = 4, a_4= a_{4-1} + a_{4-2}= a_3 + a_2= 2 + 1 = 3$
For $n = 5, a_5 = a_{5-1} + a_{5-2}= a_4 + a_3 = 3 + 2 = 5$
For $n = 6, a_6 = a_{6-1}+ a_{6-2} = a_5+ a_4 = 5 + 3 = 8$
Now, $\frac { a _ { n + 1 } } { a _ { n } }$, for $n = 1, 2, 3, 4, 5.$
For $n = 1, \frac { a _ { 2 } } { a _ { 1 } } = \frac { 1 } { 1 } = 1$
For $n = 2, \frac { a _ { 3 } } { a _ { 2 } } = \frac { 2 } { 1 } = 2$
For $n = 3, \frac { a _ { 4 } } { a _ { 3 } } = \frac { 3 } { 2 }$
For $n = 4, \frac { a _ { 5 } } { a _ { 4 } } = \frac { 5 } { 3 }$
For $n = 5, \frac { a _ { 6 } } { a _ { 5 } } = \frac { 8 } { 5 }$
Hence, the terms are $1, 2, \frac { 3 } { 2 } , \frac { 5 } { 3 }$ and $\frac { 8 } { 5 }$
View full question & answer→Question 182 Marks
Write the first five terms of the sequence and obtain the corresponding series $a_1 = a_2 = 2, a_n = a_{n-1} - 1, n > 2.$
AnswerGiven: $a_1 = a_2 = 2, a_n = a_{n-1} - 1, n > 2$
Putting $n = 3, 4$ and $5,$ we get
$a_3 = a_{3-1} - 1 = a_2{- 1} = 2 - 1 = 1$
$a_4 = a_{4-1} - 1 = a_3{- 1} = 1 - 1 = 0$
$a_5 = a_{5-1} - 1 = a_4{- 1} = 0 - 1 = -1$
Hence the first five terms are $2, 2, 1, 0, -1.$
Therefore, corresponding series is $2 + 2 + 1 + 0 + (-1) + .......$
View full question & answer→Question 192 Marks
Write the first five terms of the sequence and obtain the corresponding series $a_1 = -1, a _ { n } = \frac { a _ { n - 1 } } { n } , n \geq 2$
AnswerGiven: $a_1 = -1, a _ { n } = \frac { a _ { n - 1 } } { n } , n \geq 2$ Putting $n = 2, 3, 4$ and $5,$ we get
$a _ { 2 } = \frac { a _ { 2 - 1 } } { 2 } = \frac { a _ { 1 } } { 2 } = \frac { - 1 } { 2 }$
$a _ { 3 } = \frac { a _ { 3 - 1 } } { 3 } = \frac { a _ { 2 } } { 3 } = \frac { - 1 / 2 } { 3 } = \frac { - 1 } { 6 }$
$a _ { 4 } = \frac { a _ { 4 - 1 } } { 4 } = \frac { a _ { 3 } } { 4 } = \frac { - 1 / 6 } { 4 } = \frac { - 1 } { 24 }$
$a _ { 5 } = \frac { a _ { 5 - 1 } } { 5 } = \frac { a _ { 4 } } { 5 } = \frac { - 1 / 24 } { 5 } = \frac { - 1 } { 120 }$
Hence the first five terms are $- 1,{{ - 1} \over 2},{{ - 1} \over 6},{{ - 1} \over {24}},{{ - 1} \over {120}}$
$\therefore$ Corresponding series is $- 1 + \left( {{{ - 1} \over 2}} \right) + \left( {{{ - 1} \over 6}} \right) + \left( {{{ - 1} \over {24}}} \right) + \left( {{{ - 1} \over {120}}} \right).........$
View full question & answer→Question 202 Marks
Write the first five terms of the sequence and obtain the corresponding series $a_1 = 3, a_n = 3a_{n-1} + 2$ for all $n > 1$
AnswerGiven: $a_1 = 3, a_{n-1} + 2$ for all $n > 1$Putting $n = 2, 3, 4$ and $5,$ we get
$a_2 = 3a_{2-1} + 2 = 3a_{2-1} + 2 = 3a_1 + 2 = 3 \times 3 + 2 = 9 + 2 = 11$
$a_3 = 3a_{3-1} + 2 = 3a_2 + 2 = 3 \times 11 + 2 = 33 + 2 = 35$
$a_4 = 3a_{4-1} + 2 = 3a_3 + 2 = 3 \times 35 + 2 = 105 + 2 = 107$
$a_5 3a_{5-1} + 2 = 3a_4 + 2 = 3 \times 107 + 2 = 321 + 2 = 323$
Hence the first five terms are $3,11,35,107,323.$
Therefore, corresponding series is $3 + 11 + 35 + 107 + 323 + ……….$
View full question & answer→Question 212 Marks
Find the indicated terms of the sequence, whose nth term is $a _ { n } = \frac { n ( n - 2 ) } { n + 3 }; a_{20}$
AnswerGiven: $a_{n}=\frac{n(n-2)}{n+3}$
$\therefore a_{20}=\frac{20(20-2)}{20+3}$
$=\frac{20 \times 18}{23}=\frac{360}{23}$
Therefore, $20^{th}$ term is $\frac{360}{23}$.
View full question & answer→Question 222 Marks
Write the first five terms of the sequence whose $n^{th} $ term is $a_n = n (n + 2)$
AnswerGiven: $a_n = n(n + 2)$
Putting $n = 1, 2, 3, 4$ and $5,$ we get,
$a _ { 1 } = 1 ( 1 + 2 ) = 1 \times 3 = 3$
$a _ { 2 } = 2 ( 2 + 2 ) = 2 \times 4 = 8$
$a _ { 3 } = 3 ( 3 + 2 ) = 3 \times 5 = 15$
$a _ { 4 } = 4 ( 4 + 2 ) = 4 \times 6 = 24$
$a _ { 5 } = 5 ( 5 + 2 ) = 5 \times 7 = 35$
Therefore, the first five terms are $3, 8, 15, 24$ and $35.$
View full question & answer→