(Each question 2 marks)

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.

Answer

Present 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

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Question 22 Marks

Find the sum to n terms in each of the series $3 \times 8 + 6 \times 11 + 9 \times 14 + \ldots \ldots$

Answer

$\text { Given: } 3 \times 8+6 \times 11+9 \times 14+\ldots . . . . \text { to } n \text { terms. }$
$\therefore a_n=\left[n^{\text {th }} \text { term of } 3,6,9, \ldots . . . . . . .\right]\left[n^{\text {th }} \text { term of } 8,11,14, \ldots . . . . . . .\right]$
$=[3+(n-1) \times 3][8+(n-1) \times 3]$
$=3 n(3 n+5)=9 n^2+15 n$
$\therefore S_n=\sum_{k=1}^n a_k \sum_{=k=1}^n\left(9 k^2+15 k\right)$
$=\left[9.1^2+15.1\right]+\left[9.2^2+15.2\right]+\left[9.3^2+15.3\right] \ldots \ldots . . .+\left[9 n^2+15 n\right]$
$=9\left(1^2+2^2+3^2+\ldots \ldots \ldots . . n^2\right)+15(1+2+3+\ldots \ldots . . n)$
$=9 \frac{n(n+1)(2 n+1)}{6}+15 \frac{n(n+1)}{2}$
$=\frac{n(n+1)}{2}(6 n+3+15)=$
$=\frac{n(n+1)}{2}(6 n+18)$
$=3 n(n+1)(n+3)$

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Question 32 Marks

Find the sum to $n$ terms in each of the series $5^2+6^2+7^2+\ldots \ldots+20^2$

Answer

Given: $5^2+6^2+7^2+\ldots \ldots . .+20^2$
$=\left(1^2+2^2+3^2+\ldots \ldots+20^2\right)-\left(1^2+2^2+3^2+4^2\right)$
$\sum\limits_{n = 1}^{20} {{n^2}} - \sum\limits_{n = 1}^4 {{n^2}} $
$= \frac { 20 ( 20 + 1 ) ( 40 + 1 ) } { 6 } - \frac { 4 ( 4 + 1 ) ( 8 + 1 ) } { 6 }$
$=\frac { 20 \times 21 \times 41 } { 6 } - \frac { 20 \times 9 } { 6 }$
$= \frac { 20 } { 6 } ( 861 - 9 )=2840$

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Question 42 Marks

Find the sum to n terms of the series $3 \times 1 ^ { 2 } + 5 \times 2 ^ { 2 } + 7 \times 3 ^ { 2 } + \ldots$

Answer

Given: $3 \times 1^2+5 \times 2^2+7 \times 3^2+\ldots \text { to } \mathrm{n} \text { terms }$
$\therefore a_n=\left[\mathrm{n}^{\text {th }} \text { term of } 3,5,7, \ldots \ldots \ldots \ldots\right]\left[n^{\text {th }} \text { term of } 1,2,3,4, \ldots \ldots \ldots . .\right]^2$
$=(2 \mathrm{n}+1)(\mathrm{n})^2=2 \mathrm{n}^3+\mathrm{n}^2=2 n^3+n^2$
$\therefore \mathrm{~S}_n=\sum_{k=1}^n a_k=\sum_{k=1}^n\left(2 k^3+k^2\right)$
$=\left[2.1^3+1^2\right]+\left[2.2^3+2^2\right]+\left[2.3^3+3^2\right]+\ldots \ldots+\left[2 n^3+n^2\right]$
$=2\left(1^3+2^3+3^3+\ldots \ldots . . n^3\right)+\left(1^2+2^2+3^2+\ldots \ldots n^2\right)$
$= 2 \left[ \frac { n ( n + 1 ) } { 2 } \right] ^ { 2 } + \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 }$
$= 2{{{n^2}{{(n + 1)}^2}} \over 4} + {{n(n + 1)(2n + 1)} \over 6}$
$= \frac { n ( n + 1 ) } { 2 } \left[ n ( n + 1 ) + \frac { 2 n + 1 } { 3 } \right]$
$= \frac { n ( n + 1 ) } { 2 } \left[ \frac { 3 n ^ { 2 } + 3 n + 2 n + 1 } { 3 } \right]$
$= \frac { n ( n + 1 ) \left( 3 n ^ { 2 } + 3 n + 2 n + 1 \right) } { 6 }$
= $\frac { n ( n + 1 ) \left( 3 n ^ { 2 } + 5 n + 1 \right) } { 6 }$

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Question 52 Marks

For what values of x, the numbers $ \frac { - 2 } { 7 } , x , \frac { - 7 } { 2 }$ are in G.P.?

Answer

Given, $ \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.

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Question 62 Marks

Which term of the sequence $\frac { 1 } { 3 } , \frac { 1 } { 9 } , \frac { 1 } { 27 }$, ... is $\frac { 1 } { 19683 }$?

Answer

Here 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 $\frac { 1 } { 19683 }$

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Question 72 Marks

Which term of the sequence 2, 2$\sqrt 2$, 4, .... is 128?

Answer

Here $\mathrm{a}=2, \mathrm{r}=\frac{2 \sqrt{2}}{2}=\sqrt{2}$ and $\mathrm{a}_{\mathrm{n}}=128$
$\therefore a_n=a r^{n-1}$
$\Rightarrow 128=2 \times(\sqrt{2})^{n-1}$
$\Rightarrow 64=(\sqrt{2})^{n-1}$
$\Rightarrow(\sqrt{2})^{12}=(\sqrt{2})^{n-1}$
$\Rightarrow n-1=12$
$\Rightarrow n=13$
Therefore, $13^{\text {th }}$ term of the given G.P. is 128

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Question 82 Marks

The $4^{\text {th }}$ term of a G.P. is square of its second term and the first term -3 . Determine its $7^{\text {th }}$ term.

Answer

Let $a$ be the first term and r be the common ratio of given G.P.
Here $a=-3$ and $a_4=\left(a_2\right)^2$
Now, $a_4=\left(a_2\right)^2$
$\Rightarrow a r^3=(a r)^2$
$\Rightarrow a r^3=a^2 r^2$
$\Rightarrow \mathrm{r}=\mathrm{a}$
$\Rightarrow r=-3$
$\therefore a_7=a r^{7-1}=(-3) \times(-3)^6$
$=-3 \times 729=-2187$

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Question 92 Marks

The $5^{\text {th }}, 8^{\text {th }}$ and $11^{\text {th }}$ terms of a G.P. are $p, q$ and $s$ respectively. Show that $q^2=p s$.

Answer

Let $a$ be the first term and r be the common ratio of given G.P.
$\therefore a_5=p \Rightarrow a r^4=p \ldots \text {..(i) }$
$a_8=q \Rightarrow a r^7=q \ldots \text { (ii) }$
$a_{11}=s \Rightarrow a r^{10}=s \ldots \text { (iii) }$
Squaring both sides of eq. (ii), we getq ${ }^2=\left(a r^7\right)^2$
$\Rightarrow q^2=a^2 r^{14}$
$\Rightarrow q^2=\left(a r^4\right)\left(a r^{10}\right)$
$\Rightarrow q^2=p s[\text { From eq. (i) and (iii)] }$

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Question 102 Marks

Find the $12^{\text {th }}$ term of a G.P. whose $8^{\text {th }}$ term is 192 and the common ratio is 2.

Answer

Let a be the first term of given G.P. Here $\mathrm{r}=2$ and $\mathrm{a}_8=192$
$\therefore a_n=a r^{n-1}$
$\Rightarrow a_8=a \times(2)^{8-1}=192$
$\Rightarrow a \times(2)^7=192$
$\Rightarrow a \times 128=192$
$\Rightarrow a=\frac{192}{128}=\frac{3}{2}$
$\therefore a_{12}=a r^{12}-1$
$\Rightarrow a_{12}=\frac{3}{2} \times 2^{11}=3 \times 2^{10}$
$=3 \times 1024=3072$

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Question 112 Marks

Find the $20^{th}$ and $n^{th}$ terms of the G.P. $\frac { 5 } { 2 } , \frac { 5 } { 4 } , \frac { 5 } { 8 }$ .......

Answer

Here, 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 $a _ { n } = \frac { 5 } { 2 } \times \left( \frac { 1 } { 2 } \right) ^ { n - 1 } = \frac { 5 } { 2 \times 2 ^ { n - 1 } } = \frac { 5 } { 2 ^ { n } }$

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Question 122 Marks

If $\frac { a ^ { n } + b ^ { n } } { a ^ { n - 1 } + b ^ { n - 1 } }$ is the A. M. between a and b. Then find the value of n.

Answer

$\frac { a ^ { n } + b ^ { n } } { a ^ { n - 1 } + b ^ { n - 1 } } = \frac { a + b } { 2 }$
$2 a^n+2 b^n=(a+b)\left(a^{n-1}+b^{n-1}\right)$
$2 a^n+2 b^n=a^n+a b^{n-1}+b \cdot a^{n-1}+b^n$
$a^n+b^n=a \cdot b^{n-1}+b \cdot d^{n-1}$
$a^n-b \cdot a^{n-1}=a \cdot b^{n-1}-b^n$
$a^{n-1}(a-b)=b^{n-1}(a-b)$
$\left( \frac { a } { b } \right) ^ { n - 1 } = \frac { a - b } { a - b }$$\left( \frac { a } { b } \right) ^ { n - 1 } = \left( \frac { a } { b } \right) ^ { 0 } \left( \because \left( \frac { a } { b } \right) ^ { 0 } = 1 \right)$
$n - 1 = 0$
$n = 1$

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Question 132 Marks

Insert five numbers between 8 and 26 so that the resulting sequence is an A.P.

Answer

Let $A_1, A_2, A_3, A_4$ and $A_5$ be five numbers between 8 and 26 such that $8, A_1, A_2, A_3, A_4, A_5, 26$ are in $A_1 P$ Here, $\mathrm{a}=8$ and $\mathrm{a}_7=26$ and let d be the common difference.
$\therefore a_7=a+(7-1) d=26$
$\Rightarrow 8+6 d=26$
$\Rightarrow 6 d=18$
$\Rightarrow d=3$
Now, $A_1=a+d=8+3=11$
$A_2=a+2 d=8+2 \times 3=8+6=14$
$A_3=a+3 d=8+3 \times 3=8+9=17$
$A_4=a+4 d=8+4 \times 3=8+12=20$
$A_5=a+5 d=8+5 \times 3=8+15=23$
Hence the five numbers are $11,14,17,20$ and 23

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