Arithmetic progression

EXERCISE 19.2

QUESTION 1

Find :

(i) 10 th term of the A.P. 1,4,7,10, \ldots

Sol :

We have 

a = 1

d = 4 – 1 = 3

\left[a_{n}=a+(n-1) d\right]

a_{10}=a+(10-1) d 

=a+9 d

=1+9 \times 3

= 28

 

(ii) 18 th term of the A.P. \sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, \ldots

Sol :

We have :

a=\sqrt{2}

d=3 \sqrt{2}-\sqrt{2}=2 \sqrt{2}

\left[a_{n}=a+(n-1) d\right]

a_{18}=a+(18-1) d

=a+17 d

=\sqrt{2}+17(2 \sqrt{2})

=\sqrt{2}+34 \sqrt{2}

=35 \sqrt{2}

 

(iii) n th term of the A.P. 13,8,3,-2, \ldots

Sol :

We have :

a = 13

d = 8 – 13 = -5

a_{n}=a+(n-1) d

=13+(n-1)(-5)

=13-5 n+5

=18-5 n

 

QUESTION 2

If the sequence <a_{n}> is an A.P. show that
a_{m+n}+a_{m-n}=2 a_{m}

Sol :

Let the sequence <a_{n}> be an A.P. with the first term being A and the common difference being D .
To prove: a_{m+n}+a_{m-n}=2 a_{m}

\mathrm{LHS} : a_{m+n}+a_{m-n}

=A+(m+n-1) D+A+(m-n-1) D \left\{\because a_{n}=a+(n-1) d\right\}

=A+m D+n D-D+A+m D-n D-D

=2 A+2 m D-2 D \ldots(\mathrm{i})

 

\mathrm{RHS} : 2 a_{m}

=2[A+(m-1) D]

=2 A+2 m D-2 D \quad \ldots( ii)

From (i) and (ii), we get:

LHS = RHS

Hence proved

 

QUESTION 3

(i) Which term of the A.P. 3,8,13, \ldots is 248 ?

Sol :

a = 3

d = (8-3) = 5

Let a_{n}=248

\Rightarrow a+(n-1) d=248

\Rightarrow 3+(n-1) 5=248

\Rightarrow(n-1) 5=245

\Rightarrow n-1=49

\Rightarrow n=50

Hence, 248 is the 50 th term of the given A.P.

 

(ii) Which term of the A.P. 84,80,76, \ldots is 0 ?

Sol :

a = 84

d = (80-84) = -4

Let a_{n}=0

\Rightarrow a+(n-1) d=0

\Rightarrow 84+(n-1)(-4)=0

\Rightarrow(n-1)=21

\Rightarrow n=22

Hence , 0 is the 22nd term of given A.P

 

(iii) Which term of the A.P. 4,9,14, \ldots is 254 ?

Sol :

Here , we have 

a = 4

d = (9 – 4) = 5

Let a_{n}=254

\Rightarrow a+(n-1) d=254

\Rightarrow 4+(n-1) 5=254

\Rightarrow(n-1) 5=250

\Rightarrow(n-1)=50

\Rightarrow n=51

Hence, 254 is the 51 st term of the given A.P.

 

QUESTION 4

(i) Is 68 a term of the A.P. 7, 10,13, \ldots?

Sol :

We have 

a = 7

d= (10 – 7) = 3

Let a_{n}=68

\Rightarrow a+(n-1) d=68

\Rightarrow 7+(n-1)(3)=68

\Rightarrow(n-1)(3)=61

\Rightarrow(n-1)=\frac{61}{3}

\Rightarrow n=\frac{61}{3}+1

=\frac{64}{3}

Since n is not a natural number.So, 68 is not a term of the given A.P.

 

(ii) Is 302 a term of the A.P. 3,8,13, \ldots?

Sol :

We have 

a= 3

d = (8 – 3) = 5

Let a_{n}=302

\Rightarrow a+(n-1) d=302

\Rightarrow 3+(n-1) 5=302

\Rightarrow(n-1) 5=299

\Rightarrow(n-1)=\frac{299}{5}

\Rightarrow n=\frac{299}{5}+1

=\frac{304}{5}

Since n is not a natural number. So ,302 is not a term of the given A.P.

 

QUESTION 5

(i) Which term of the sequence 24,\dfrac{23}{4},\dfrac{22}{2}, {21}\times\dfrac{3}{4}\ldots is the first negative term?

Sol :

24,\dfrac{23}{4},\dfrac{22}{2}, {21}\times\dfrac{3}{4}\ldots

This is an A.P

Here , we have

a = 24

d=\left(23 \frac{1}{4}-24\right)

= -\dfrac{3}{4}

Let the first negative term be a_{n}

Then , we have :

a_{n}<0

\Rightarrow a+(n-1) d<0

\Rightarrow 24+(n-1)(-3 / 4)<0

\Rightarrow 24-\frac{3 n}{4}+\frac{3}{4}<0

\Rightarrow 24+\frac{3}{4}<\frac{3 n}{4}

\Rightarrow \frac{99}{4}<\frac{3 n}{4}

\Rightarrow \frac{99}{4}<\frac{3 n}{4}

\Rightarrow 99<3 n

\Rightarrow n>33

Thus, the 34 th term is the first negative term of the given A.P.

 

(ii) Which term of the sequence 12+8 i, 11+6 i, 10+4 i, \ldots is (a) purely real (b) purely imaginary?

Sol :

12+8 i, 11+6 i, 10+4 i . .

This is an A.P

Here , we have 

a=12+8 i

d=(11+6 i-12-8 i)

=(-1-2 i)

Let the real term be a_{n}=a+(n-1) d

a_{n}=(12+8 i)+(n-1)(-1-2 i)

=(12+8 i)+(-n+1-2 i n+2 i)

=12+8 i-n+1-2 i n+2 i

=(13-n)+(8-2 n+2) i

=(13-n)+(10-2 n) i

Let nth term be purely real 

\therefore(10-2 n)=0 

n = 5

So, 5 th term is purely real.

Let n th term be purely imaginary. Then, 13-n=0

\therefore n=13

So, 13 th term is purely imaginary.

REASON :complex number is a number which can be represented in the form of ‘z=a+ib’ , where a & b are real number and i is an imaginary number, (which is a square root of -1). If a=0 then z is purely imaginary number but if b=0 then z is purely real number.

 

QUESTION 6

(i) How many terms are there in the A.P. 7,10,13, \ldots 43 ?

Sol :

Here , we have

a = 7

d = (10 – 7) =3

a_{n}=43

Let there be n terms in the given A.P.

a_{n}=a+(n-1) d

\Rightarrow 43=7+(n-1) 3

\Rightarrow 36=(n-1) 3

\Rightarrow 12=(n-1)

\Rightarrow 13=n

Thus, there are 13 terms in the given A.P.

 

(ii) How many terms are there in the A.P. -1,-\frac{5}{6},-\frac{2}{3},-\frac{1}{2}, \ldots, \frac{10}{3} ?

Sol :

Here , we have

a = -1

d=\left(\frac{-5}{6}-(-1)\right)

=\left(1-\frac{5}{6}\right)=\frac{1}{6}

a_{n}=\frac{10}{3}

Let there be n terms in the given A.P.

a_{n}=a+(n-1) d

\Rightarrow \frac{10}{3}=-1+(n-1) \frac{1}{6}

\Rightarrow \frac{13}{3}=(n-1) \frac{1}{6}

\Rightarrow \frac{13}{3}=(n-1) \frac{1}{6}

\Rightarrow 26=(n-1)

\Rightarrow 27=n

Thus, there are 27 terms in the given A.P.

 

QUESTION 7

The first term of an A.P. is 5 , the common difference is 3 and the last term is 80 ; find the number of
terms.

Sol :

a=5 

d=3

a_{n}=80 Let the number of terms be <em>n</em> Then, we have: a_{n}=a+(n-1) d \Rightarrow 80=5+(n-1) 3 \Rightarrow 75=(n-1) 3 \Rightarrow 25=(n-1) \Rightarrow 26=n

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Thus , there are 26 terms in the given A.P
 
<strong><span style="background-color: #ffff00;">QUESTION 8</span></strong>
The 6 th and 17 th terms of an A.P. are 19 and 41 respectively, find the 40 th term.
Sol :
Given :


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a_{6}=19 \Rightarrow a+(6-1) d=19\left[a_{n}=a+(n-1) d\right] \Rightarrow a+5 d=19 \quad \ldots(1) AND a_{17}=41 \Rightarrow a+(17-1) d=41 \quad\left[a_{n}=a+(n-1) d\right] \Rightarrow a+16 d=41 \quad \ldots(2) Solving the two equations, we get, 16 d-5 d=41-19 \Rightarrow 11 d=22 \Rightarrow d=2 Puttingd=2in the equation (1) , we get : a+5 \times 2=19 \Rightarrow a=19-10 \Rightarrow a=9

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We know:


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a_{40}=a+(40-1) d \quad\left[a_{n}=a+(n-1) d\right] =a+39 d =9+39 \times 2

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=9+78 =87
 
<strong><span style="background-color: #ffff00;">QUESTION 9</span></strong>
If 9 th term of an A.P. is zero, prove that its 29 th term is double the 19 th term.
Sol :
To prove:

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a_{29}=2 a_{19}

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Given : 


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a_{9}=0 \Rightarrow a+(9-1) d=0 \quad\left[a_{n}=a+(n-1) d\right] \Rightarrow a+8 d=0 \Rightarrow a=-8 d \ldots(\mathrm{i})

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Proof:
LHS:

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a_{19}=a+(19-1) d =a+18 d = -8d+18d[from (i)] = 10d \dots (ii)

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RHS :

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a_{29}=a+(29-1) d =a+28 d =-8 d+28 d [from (i)] = 20 d \dots (iii)

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Hence proved
 
<strong><span style="background-color: #ffff00;">QUESTION 10</span></strong>
If 10 times the 10 th term of an A.P. is equal to 15 times the 15 th term, show that 25 th term of the A.P. is zero.
Sol :
Given :


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10 a_{10}=15 a_{15} \Rightarrow 10[a+(10-1) d]=15[a+(15-1) d] \Rightarrow 10(a+9 d)=15(a+14 d) \Rightarrow 10 a+90 d=15 a+210 d \Rightarrow 0=5 a+120 d \Rightarrow 0=a+24 d \Rightarrow a=-24 d \ldots(\mathrm{i})

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To show:

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a_{25}=0

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Proof :


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\Rightarrow \mathrm{LHS} : a_{25}=a+(25-1) d =a+24 d =-24 d+24 d[from (i)] =0=\mathrm{RHS}

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Hence , proved
 
<strong><span style="background-color: #ffff00;">QUESTION 11</span></strong>
The 10 th and 18 th terms of an A.P. are 41 and 73 respectively. Find 26 th term.
Sol :
Given :


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a_{10}=41 \Rightarrow a+(10-1) d=41 \left[a_{n}=a+(n-1) d\right] \Rightarrow a+9 d=41 AND a_{18}=73 \Rightarrow a+(18-1) d=73\left[a_{n}=a+(n-1) d\right] \Rightarrow a+17 d=73 Solving the two equations, we get: \Rightarrow 17 d-9 d=73-41 \Rightarrow 8 d=32 \Rightarrow d=4 \quad \ldots(i) Putting the value in first equation, we get : a+9 \times 4=41 \Rightarrow a+36=41 \Rightarrow a=5 \quad \ldots(ii)

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a_{26}=a+(26-1) d \quad\left[a_{n}=a+(n-1) d\right] \Rightarrow a_{26}=a+25 d \Rightarrow a_{26}=5+25 \times 4from~(i)~and~(ii) \Rightarrow a_{26}=5+100=105

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<strong><span style="background-color: #ffff00;">QUESTION 12</span></strong>
In a certain A.P. the 24 th term is twice the 10 th term. Prove that the 72 nd term is twice the 34 th
term.
Sol :
Given :


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a_{24}=2 a_{10} \Rightarrow a+(24-1) d=2[a+(10-1) d] \Rightarrow a+23 d=2(a+9 d) \Rightarrow a+23 d=2 a+18 d \Rightarrow 5 d=a \quad \ldots(\mathrm{i})

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To prove:

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a_{72}=2 a_{34}

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Proof :
LHS

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: a_{72}=a+(72-1) d \Rightarrow a_{72}=a+71 d \Rightarrow a_{72}=5 d+71 d[from (i)] \Rightarrow a_{72}=76 d

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\mathrm{RHS} : 2 a_{34}=2[a+(34-1) d] \Rightarrow 2 a_{34}=2(a+33 d) \Rightarrow 2 a_{34}=2(5 d+33 d)[from (i)] \Rightarrow 2 a_{34}=2(38 d) \Rightarrow 2 a_{34}=76 d \therefore \mathrm{RHS}=\mathrm{LHS}

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Hence, proved.
 
<strong><span style="background-color: #ffff00;">QUESTION 13</span></strong>
If

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(m+1)th term of an A.P. is twice the(n+1)th term, prove that(3 m+1)th term is twice the(m+n+1)$ th term.

Sol :

Given :

 

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