1 Marks Question

Question 11 Mark

Find the 10th and nth terms of the G.P. 5, 25,125,….

Answer

We 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 = 5ⁿ
and T₁₀ = 5 (5)^10-1 = 5¹⁰

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Question 21 Mark

Insert 6 numbers between 3 and 24 such that the resulting sequence is an A.P.

Answer

Let A₁, A₂, A₃, A₄, A₅ and A₆ be six numbers between 3 and 24 such that 3, A₁, A₂, A₃, A₄, A5, A₆, 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;
A₂ = a + 2d = 3 + 2 $\times$ 3 = 9;
A₃ = a + 3d = 3 + 3 $\times$ 3 = 12;
A₄ = a + 4d = 3 + 4 $\times$ 3 = 15;
A₅ = a + 5d = 3 + 5 $\times$ 3 = 18;
A₆ = a + 6d = 3 + 6 $\times$ 3 = 21.
Therefore,six numbers between 3 and 24 are 6, 9, 12, 15, 18 and 21.

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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.

Answer

Here, 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.

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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.

Answer

Let a₁, a₂ and d₁, d₂ 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)
$\frac { 2 a _ { 1 } + 22 d _ { 1 } } { 2 a _ { 2 } + 22 d _ { 2 } } = \frac { 3 \times 23 + 8 } { 7 \times 23 + 15 }$
$\frac { a _ { 1 } + 11 d _ { 1 } } { a _ { 2 } + 11 d _ { 2 } } = \frac { 7 } { 16 }$

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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.

Answer

Let a₁, a₂, … a_n be the given A.P. Then
S_n = a₁ + a₂ + a₃ +...+ a_n-1 + a_n = $n \mathrm{P}+\frac{1}{2} n(n-1) \mathrm{Q}$
Thus,S1 = a₁ = P, S₂ = a₁ + a₂ = 2P + Q
So that a₂ = S₂ - S₁ = P + Q
Therefore,the common difference is given by d = a₂ - a₁ = (P + Q) - P = Q.

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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.

Answer

We 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.

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Question 71 Mark

Let the sequence a_n be defined as follow: a₁ = 1, a_n = a_{n - 1} + 2 for n $\ge$ 2
Find first five terms and write corresponding series.

Answer

Given,
a₁ = 1, a₂ = a₁ + 2 = 1 + 2 = 3, a₃ = a₂ + 2 = 3 + 2 = 5,
a₄ = a₃ + 2 = 5 + 2 = 7, a₅ = a₄ + 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 + ...

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Question 81 Mark

What is the 20th term of the sequence defined by a_n = (n - 1) (2 - n) (3 + n)?

Answer

Here, n = 20,
$\therefore$ a₂₀ = (20 - 1) (2 - 20) (3 + 20)
= 19 $\times$ (- 18) $\times$ (23) = - 7866.

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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)
= n² + 3n + 1
${S_n} = \sum\limits_{k = 1}^n {{k^2}} + 3\sum\limits_{k = 1}^n k + n$
$= \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } + \frac { 3 n ( n + 1 ) } { 2 } + n$
$= \frac { n ( n + 2 ) ( n + 4 ) } { 3 }$

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Question 101 Mark

If A.M. and G.M. of two positive numbers a and b are 10 and 8, respectively, find the numbers.

Answer

Given 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 = $\pm$12 .....(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

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Question 111 Mark

Insert three numbers between 1 and 256 so that the resulting sequence is a G.P.

Answer

Suppose G₁, G₂, G₃ be three numbers between 1 and 256 such that 1, G₁,G₂, G₃, 256 is a G.P.
Thus, 256 = r⁴ giving r = $\pm$4 (Taking real roots only)
For r = 4, we have G₁ = ar = 4, G₂ = ar² = 16, G₃ = ar³ = 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.

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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.

Answer

The 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
$\therefore$ S₁₀ =$\frac { 2 \left( 2 ^ { 10 } - 1 \right) } { 2 - 1 }$
= 2 (2¹⁰ - 1) = 2 (1024 - 1)
= 2 $\times$ 1023 = 2046
Hence, the numbers of ancestors preceding the person is 2046

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Question 131 Mark

Find the sum of the sequence 7, 77, 777, ........ to n.

Answer

S_n = 7 + 77 + 777 + ---- + n terms
$= \frac { 7 } { 9 } [ 9 + 99 + 999 + - - - - + n \text { terms } ]$
$= \frac { 7 } { 9 }$[(10 - 1 ) + (10² - 1) + (10³ - 1) + --- + n terms]
$= \frac { 7 } { 9 }$[(10¹ + 10² + ---- +n terms) - (1 + 1 + 1 ---- + n terms)]
$= \frac { 7 } { 9 } \left[ \frac { 10 \left( 10 ^ { n } - 1 \right) } { 10 - 1 } - n \right]$
$= \frac { 7 } { 9 } \left[ \frac { 10 \left( 10 ^ { n } - 1 \right) } { 9 } - n \right]$

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Question 141 Mark

The sum of first three terms of a GP is $\frac { 13 } { 12 }$ and their product is - 1. Find the terms.

Answer

Let 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 + r²) = $\frac { 13 } { 12 } r$ ...(i)
Their product = $\frac { a } { r }$ $\times$ a$\times$ ar = - 1 $\Rightarrow$ a³ = - 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 + r²] = $\frac { 13 } { 12 } r$
$\Rightarrow$ - 1 - r - r² = $\frac { 13 } { 12 } r$
$\Rightarrow$ - 12 - 12r - 12r² = 13r
$\Rightarrow$ 12r² + 25r + 12 = 0
$\Rightarrow$ 12r²+ 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
$\frac { - 1 } { - 3 / 4 } , - 1 , - 1 \times \frac { - 3 } { 4 } \text { i.e., } \frac { 4 } { 3 } , - 1 , \frac { 3 } { 4 }$

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Question 151 Mark

How many terms of GP 3, $\frac { 3 } { 2 } , \frac { 3 } { 4 }$, ... are needed to give the sum $\frac { 3069 } { 512 }$?

Answer

Given 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 - $\frac { 3069 } { 3072 } = \frac { 3072 - 3069 } { 3072 }$
$\Rightarrow$ $\frac { 1 } { 2 ^ { n } } = \frac { 3 } { 3072 } = \frac { 1 } { 1024 }$
$\Rightarrow$ 2ⁿ = 1024 $\Rightarrow$ 2ⁿ = 2¹⁰
On comparing the powers, we get
n = 10
Hence, 10 terms are needed to give the sum $\frac { 3069 } { 512 }$.

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Question 161 Mark

Find the sum of first n terms and the sum of first 5 terms of the geometric series $1 + \frac { 2 } { 3 } + \frac { 4 } { 9 } + - --$

Answer

a = 1, r = $\frac{2}{3}$
$S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$
$=\frac{1\left[1-\left(\frac{2}{3}\right)^{n}\right]}{1-\frac{2}{3}}$
$=3\left[1-\left(\frac{2}{3}\right)^{n}\right]$
$S_{5}=3\left[1-\left(\frac{2}{3}\right)^{5}\right]=\frac{211}{81}$

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Question 171 Mark

Write the first three terms of $a_{n}=\frac{n-3}{4}$

Answer

Here, $a_{n}=\frac{n-3}{4}$. 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.

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Question 181 Mark

In a G.P., the 3^rd term is 24 and the 6^th term is 192. Find the 10^th term.

Answer

Given, a₃ = ar² = 24 ...(i)
and a₆ = ar⁵ = 192 ...(ii)
Dividing (ii) by (i), we get r = 2.
Putting, r = 2 in (i), we get a = 6.
Hence a₁₀ = 6 (2)⁹ = 3072

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Question 191 Mark

Write the first three terms of a_n = 2n + 5

Answer

Here a_n = 2n + 5
Substituting n = 1, 2, 3, we obtain
a₁ = 2(1) + 5 = 7, a₂ = 9, a₃ = 11
Thus, the required terms are 7, 9 and 11.

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Question 201 Mark

Which term of the G.P., 2,8,32, ... up to n terms is 131072?

Answer

Let 131072 be the n^th term of the given G.P. Here a = 2 and r = 4.
Thus,131072 = a_n = 2(4)^n-1 or 65536 = 4^n-1
This gives 4⁸ = 4^n-1.
So that n - 1 = 8, i.e., n = 9. Therefore, 131072 is the 9^th term of the G.P.

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