2 Marks Questions - MATHS STD 11 Science Questions - Vidyadip

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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 = $\frac { 10000 } { 12500 } = \frac { 4 } { 5 }$
$\therefore$ a₅ = ar⁴ = 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

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 32 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 42 Marks

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

Answer

Here 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

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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.

Answer

Let a be the first term and r be the common ratio of given G.P.
Here a = -3 and a₄ = (a₂)²
Now,a₄ = (a₂)²
$ \Rightarrow$ ar³ = (ar)²
$\Rightarrow$ ar³ = a²r²
$\Rightarrow$ r = a
$\Rightarrow$ r = -3
$\therefore$ a ₇ = ar^{7 - 1} = (-3) $\times$ (-3)⁶
= -3 $\times$ 729 = -2187

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Question 62 Marks

The 5^th, 8^thand 11^th terms of a G.P. are p, q and s respectively. Show that q² = ps.

Answer

Let a be the first term and r be the common ratio of given G.P.
$\therefore $ a₅ = p $\Rightarrow$ ar⁴ = p ...(i)
a₈ = q $\Rightarrow$ ar⁷ = q ...(ii)
a₁₁ = s $\Rightarrow$ a r¹⁰ = s ...(iii)
Squaring both sides of eq. (ii), we getq² = (ar⁷)²
$\Rightarrow$ q² = a² r¹⁴
$\Rightarrow$ q² = (ar⁴) (ar¹⁰)
$\Rightarrow $ q² = ps [From eq. (i) and (iii)]

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Question 72 Marks

Find the 12^th term of a G.P. whose 8^th term is 192 and the common ratio is 2.

Answer

Let a be the first term of given G.P. Here r =2 and a₈ = 192
$\therefore$a_n = arⁿ^-1
$\Rightarrow a _ { 8 } = a \times ( 2 ) ^ { 8 - 1 } = 192$
$\Rightarrow a \times ( 2 )^7 = 192$
$\Rightarrow a \times 128 = 192$
$\Rightarrow a = \frac { 192 } { 128 } = \frac { 3 } { 2 }$
$\therefore$a₁₂ = ar^{12 - 1}
$\Rightarrow a _ { 12 } = \frac { 3 } { 2 } \times 2 ^ { 11 } = 3 \times 2 ^ { 10 }$
= 3 $\times$1024 = 3072

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Question 82 Marks

Find the 20^th and n^thterms 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 92 Marks

Find the indicated terms of the sequence, whose nth term is $a _ { n } = ( - 1 ) ^ { n - 1 } n ^ { 3 }$; a₉

Answer

Given: a_n = (-1)^n-1 n³

$\therefore a _ { 9 } = ( - 1 ) ^ { 9 - 1 } \times ( 9 ) ^ { 3 } = ( - 1 ) ^ { 8 } \times 729 = 729$

Therefore, 9^th term is 729.

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Question 102 Marks

Find the indicated terms of the sequence, whose nth term is $a _ { n } = \frac { n ^ { 2 } } { 2 ^ { n } }$; a₇

Answer

Given: $a _ { n } = \frac { n ^ { 2 } } { 2 ^ { n } }$

$\therefore a _ { 7 } = \frac { 7 ^ { 2 } } { 2 ^ { 7 } } = \frac { 49 } { 128 }$

Therefore, 7^th term is $\frac { 49 } { 128 }.$

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Question 112 Marks

Find the indicated terms of the sequence, whose nthterm is a_n = 4n - 3; a₁₇, a₂₄.

Answer

We have, a_n = 4n - 3
On putting n = 17, we get
a₁₇ = 4 $\times$ 17 - 3 = 68 - 3 = 65
On putting n = 24, we get
a₂₄ = 4 $\times$ 24 - 3 = 96 - 3 = 93

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Question 122 Marks

Write the first five terms of the sequence whose n^th term is $a _ { n } = n . \frac { n ^ { 2 } + 5 } { 4 }$ .

Answer

Given: $a_{n}=n \cdot \frac{n^{2}+5}{4}$
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 $\frac{75}{2}$

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Question 132 Marks

Write the first five terms of the sequence whose n^th term is $a _ { n } = ( - 1 ) ^ { n - 1 } \cdot 5 ^ { n + 1 }$

Answer

Given: $a _ { n } = ( - 1 ) ^ { n - 1 } \cdot 5 ^ { n + 1 }$

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.

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Question 142 Marks

Write the first five terms of the sequence whose n^th term is $a _ { n } = \frac { 2 n - 3 } { 6 }$.

Answer

Given: $a _ { n } = \frac { 2 n - 3 } { 6 }$

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 $\frac { 7 } { 6 }.$

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Question 152 Marks

Write the first five terms of the sequence whose n^th term is a_n = 2ⁿ

Answer

Given: a_n = 2ⁿ

Putting n = 1, 2, 3, 4 and 5, we get,

a₁ = 2¹ = 2

a₂ = 2² = 4

a₃ = 2³ = 8

a₄ = 2⁴ = 16

a₅ = 2⁵ = 32

Therefore, the first five terms are 2, 4, 8, 16 and 32.

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Question 162 Marks

Write the first five terms of the sequence whose n^th term is $a _ { n } = \frac { n } { n + 1 }$

Answer

Given: $a _ { n } = \frac { n } { n + 1 }$

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 $\frac { 5 } { 6 }.$

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Question 172 Marks

The Fibonacci sequence is defined by 1 = a₁ = a₂ 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.

Answer

Given, 1 = a₁ = a₂
and a_n = a_n-1 + a_n-2, n > 2
On putting n = 3, 4, 5, 6 respectively, we get
For n = 3, a₃ = a_3-1 + a_3-2= a₂ + a₁ = 1 + 1 = 2
For n = 4, a₄= a_4-1 + a_4-2= a₃ + a₂= 2 + 1 = 3
For n = 5, a₅ = a_5-1 + a_5-2= a₄ + a₃ = 3 + 2 = 5
For n = 6, a₆ = a_6-1+ a_6-2 = a₅+ a₄ = 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 $\frac { 8 } { 5 }$

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Question 182 Marks

Write the first five terms of the sequence and obtain the corresponding series a₁ = a₂ = 2, a_n = a_n-1 - 1, n > 2.

Answer

Given: a₁ = a₂ = 2, a_n = a_n-1 - 1, n > 2
Putting n = 3, 4 and 5, we get
a₃ = a_3-1 - 1 = a₂ _{- 1} = 2 - 1 = 1
a₄ = a_4-1 - 1 = a₃ _{- 1} = 1 - 1 = 0
a₅ = a_5-1 - 1 = a₄ _{- 1} = 0 - 1 = -1
Hence the first five terms are 2, 2, 1, 0, -1.
Therefore, corresponding series is 2 + 2 + 1 + 0 + (-1) + .......

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Question 192 Marks

Write the first five terms of the sequence and obtain the corresponding series a₁ = -1, $a _ { n } = \frac { a _ { n - 1 } } { n } , n \geq 2$

Answer

Given: a₁ = -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 $- 1 + \left( {{{ - 1} \over 2}} \right) + \left( {{{ - 1} \over 6}} \right) + \left( {{{ - 1} \over {24}}} \right) + \left( {{{ - 1} \over {120}}} \right).........$

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Question 202 Marks

Write the first five terms of the sequence and obtain the corresponding series a₁ = 3, a_n = 3a_n-1 + 2 for all n > 1

Answer

Given: a₁ = 3, a_n-1 + 2 for all n > 1

Putting n = 2, 3, 4 and 5, we get

a₂ = 3a_2-1 + 2 = 3a_2-1 + 2 = 3a₁ + 2 = 3 $\times$ 3 + 2 = 9 + 2 = 11

a₃ = 3a_3-1 + 2 = 3a₂ + 2 = 3 $\times$ 11 + 2 = 33 + 2 = 35

a₄ = 3a_4-1 + 2 = 3a₃ + 2 = 3 $\times$ 35 + 2 = 105 + 2 = 107

a₅ 3a_5-1 + 2 = 3a₄ + 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 + ……….

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Question 212 Marks

Find the indicated terms of the sequence, whose nth term is $a _ { n } = \frac { n ( n - 2 ) } { n + 3 }$; a₂₀

Answer

Given: $a_{n}=\frac{n(n-2)}{n+3}$

$\therefore a_{20}=\frac{20(20-2)}{20+3}$$=\frac{20 \times 18}{23}=\frac{360}{23}$

Therefore, 20^th term is $\frac{360}{23}$.

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Question 222 Marks

Write the first five terms of the sequence whose n^th term is a_n = n (n + 2)

Answer

Given: 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.

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2 Marks Questions - MATHS STD 11 Science Questions - Vidyadip