Class 11 RD Sharma Solutions- Chapter 19 Arithmetic Progressions- Exercise 19.2 | Set 1

General Formula

Tn = a + (n – 1)*d

where,

Tn is the nth term

a is the first term

d = common difference

i) 10th term of AP 1, 4, 7, 10…..

Solution:

So we have a1 = a = 1

d= 4-1 = 3

n= 10

So we can calculate 10th term of AP by using general formula-

T10= a + (10-1)d

= 1 + 9*3

= 28

Hence, the 10th term is 28.

ii) 18th term of AP âˆš2, 3âˆš2, 5âˆš2…..

Solution:

So we have a1 = a = âˆš2

d= 3âˆš2 – âˆš2 = 2âˆš2

n= 18 (given in Question )

So we can calculate 18th term of AP by using general formula-

T18= a + (18-1)d

= âˆš2 + 17*(2âˆš2 )

= 35âˆš2

Hence the 18th term is 35âˆš2.

iii) nth term of AP 13, 8, 3, -2….

Solution:

So we have a1 = a = 13

d= a2 – a1 = 8 – 13 = -5

So nth term is

Tn= 13 + (n-1)( – 5 )

= 13 -5n + 5

= 18 – 5n

Hence the nth term is 18-5n.

Question 2. In an A.P. show that am+n + am-n = 2am.

Solution:

We can prove this with the help of general formula. Let’s solve LHS.

am+n= a + (m+n-1)d

am-n = a + (m-n-1)d

and am = a + (m-1)d

So now,

LHS = a + (m+n-1)d + a + (m-n-1)d

= 2a + (m+n-1+m-n-1)d

= 2a+ (2m-2)d

= 2(a+(m-1)d)

= 2am= RHS

Question 3. i) Which term of the A.P. is 3, 8, 13,…… is 248.

Solution:

So we are given a = 3

d= 8 – 3 = 5

Tn = 248

a + (n -1)d = 248

3 + (n-1)5 = 248

(n-1)5 = 245

n-1 = 49

n =50

Hence 50th term of this AP is 248.

ii) Which term of A.P. is 84, 80, 76…… is 0.

Solution:

So we are given a = 84

d= 80 – 84 = -4

Tn = 0

a + (n -1)d = 0

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

(n-1)(-4) = -84

n-1 = 21

n =22

Hence 22th term of this AP is 248.

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

Solution:

So we are given a = 4

d= 9 – 4 = 5

Tn = 254

a + (n -1)d = 254

4 + (n-1)5 = 254

(n-1)5 = 250

n-1 = 50

n =51

Hence 51th term of this AP is 248.

Question 4. i) Is 68 a term of the A.P. 7, 10, 13,……?

Solution:

We are given a = 7

d= 10 – 7 = 3

We can find whether 68 is a term of this AP by finding the valid n for 68 by using general formula. If there is no valid integer n value for 68 then it will not be a term of this AP.

Tn = 68

a+ (n-1)d = 68

7 + (n-1)3 = 68

(n-1)3= 61

n-1 = 61/3

n= 64/3

Hence we can see there is no valid integer value of n for 68 so 68 is not the term of the AP.

ii) Is 302 is a term of A.P. 3, 8, 13, ….?

Solution:

We are given a = 3

d= 8 – 3 = 5

We can find whether 302 is a term of this AP by finding the valid n for 302 by using general formula. If there is no valid integer n value for 302 then it will not be a term of this AP.

Tn = 302

a+ (n-1)d = 302

3 + (n-1)5 = 302

(n-1)5= 299

n-1 = 299/5

n= 304/5

Hence we can see there is no valid integer value of n for 302 so 302 is not the term of the AP.

Question 5.i) Which term of the sequence 24, 23Â¼, 22Â½, 22Â¾, …… is the first negative number?

Solution:

We are given a = 24

d= 23Â¼ – 24 = 93/4- 24 = -3/4

So to find first negative number we can find n value from Tn < 0

a + (n- 1)d < 0

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

24 -3n/4 +3/4 < 0

99/4 – 3n/4 <0

3n/4> 99/4

By solving this inequality we found

n>33

So we can say 34th term of AP will be first negative number.

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

Solution:

We are given a = 12 + 8i

d= (11 + 6i) -(12 + 8i) = 11 +6i – 12 – 8i =-1 – 2i

So Tn for this sequence will be

Tn = a + (n-1)d

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

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

= 13 – n + 10i -2in

= 13 – n + (10 – 2n)i

(a) For Tn to be purely real, imaginary part should be equal to 0.

So we know that 10 – 2n must be 0

10 – 2n =0

so n= 5

Hence 5th term of A.P. is purely real.

(b) For Tn to be purely imaginary, real part should be equal to 0.

So we know that 13 – n must be 0

13 – n =0

so n= 13

Hence 13th term of A.P. is purely imaginary.

Question 6. i) How many terms are in AP 7, 10, 13, ….., 43?

Solution:

We are given a= 7

d= 10 – 7 = 3

Tn = 43

Last term of AP is 43. So if we calculate the position of 43 then we will get the terms in this AP.

a + (n-1)d =43

7 + (n-1)3 = 43

(n-1)3 = 36

n-1 = 12

n = 13

So there are total 13 terms in this AP.

ii) How many terms are there in AP -1, -5/6, -2/3, -1/2……, 10/3?

Solution:

We are given a= -1

d= -5/6 – (-1) = 1-5/6 = 1/6

Tn = 10/3

Last term of AP is 10/3. So if we calculate the position of 10/3 then we will get the terms in this AP.

a + (n-1)d =10/3

-1 + (n-1)(1/6) = 10/3

(n-1)(1/6) = 13/3

n-1 = 13*6/3

n-1= 26

n = 27

So there are total 27 terms in this AP.

Question 7. The first term of AP is 5. The common difference is 3 and last term is 80. Find the number of terms.

Solution:

We are given

first term a1= a= 5

common difference d= 3

Tn = 80

Last term of AP is 80. So if we calculate the position of 80 then we will get the terms in this AP.

a + (n-1)d =80

5 + (n-1)3 = 80

(n-1)3 = 75

n-1 = 75/3

n = 25+1

n = 26

So there are total 26 terms in this AP.

Question 8. The 6th and 17th terms of an AP are 19 and 41 respectively. Find the 40th term.

Solution:

We are given T6=19 and T17=41.

a + 5d = 19 —– 1

a+ 16d = 41 —– 2

On solving equation 1 and 2

a + 5d – a – 16d = 19 – 41

-11d = – 22

d = 2

and a =19 – 10

a= 9

So T40 = a + 39d

= 9 + 39*2

= 9 + 78

= 87

So 40th term is 87 in this AP.

Question 9. If 9th term of an AP is Zero, prove that its 29th term is double the 19th term.

Solution:

We are given 9th term of AP is 0.

a9=a+ 8d = 0 ——1

We have to prove, a29 = 2a19

We know a29= a+ 28d ——-2

a19= a+ 18d ——-3

From equation 1 we get

a+ 8d = 0

a= -8d

So putting a = -8d in eq 2 & 3 will give us the value of a29 and a19

a29 = -8d +28d = 20d

a19 = -8d +18d = 10d

So its proved 2a19 = a29.

Question 10. If 10 times the 10th term of an AP is equal to 15 times the 15th term, show that 25th term of AP is zero.

Solution:

We are given 10 times the 10th term of an AP is equal to 15 times the 15th term.

10a10 = 15a15

10(a + 9d) = 15( a +14d)

10a + 90d = 15a + 210d

5a= -120d

5a + 120d= 0

a + 24d = 0

a25 = 0

So its proved a25 is equal 0.

Question 11. The 10th and 18th term of AP are 41 and 73 respectively, find 26th term.

Solution:

We are given a10 = a+ 9d = 41 ——-1

a18 = a+ 17d = 73 ——-2

On solving 1 & 2

a + 9d – a -17d = 41 – 73

-8d = -32

d= 4

By substituting value of d in equation 1 we get

a= 41-9*4

a= 5

So value of 26th term can be calculated by,

a26 = a + 25d

= 5 + 25*4

= 105

So the 26th term of AP is 105.

Question 12. In certain AP the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.

Solution:

We are given 24th term of an AP is twice the 10th term.

a24 = 2a10

So a + 23d = 2(a + 9d)

a + 23d = 2a + 18d

a = 5d

And we know, a34= a +33d

= 5d + 33d

= 38d

Similarly, a72= a +71d

= 5d + 71d

= 76d

Now we can see that a72 = 2a34 . Hence proved 72nd term is twice the 34th term.

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