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

General Formula

Tn = a + (n – 1)*dwhere,

Tnis the nth term

ais the first term

d= common difference

**Question 1. Find**

**i) 10**^{th} term of AP 1, 4, 7, 10…..

^{th}term of AP 1, 4, 7, 10…..

**Solution:**

So we have a

_{1 }= a = 1d= 4-1 = 3

n= 10

So we can calculate 10

^{th}term of AP by using general formula-T

_{10}= a + (10-1)d= 1 + 9*3

= 28

Hence, the 10

^{th}term is 28.

**ii) 18**^{th} term of AP âˆš2, 3âˆš2, 5âˆš2…..

^{th}term of AP âˆš2, 3âˆš2, 5âˆš2…..

**Solution:**

So we have a

_{1}= a = âˆš2d= 3âˆš2 – âˆš2 = 2âˆš2

n= 18 (given in Question )

So we can calculate 18

^{th }term of AP by using general formula-T

_{18}= a + (18-1)d= âˆš2 + 17*(2âˆš2 )

= 35âˆš2

Hence the 18

^{th}term is 35âˆš2.

**iii) nth term of AP 13, 8, 3, -2….**

**Solution:**

So we have a1 = a = 13

d= a

_{2}– a_{1 }= 8 – 13 = -5So n

^{th}term isT

_{n}= 13 + (n-1)( – 5 )= 13 -5n + 5

= 18 – 5n

Hence the n

^{th}term is 18-5n.

**Question 2. In an A.P. show that a**_{m+n} + a_{m-n} = 2a_{m}.

_{m+n}+ a

_{m-n}= 2a

_{m}.

**Solution:**

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

a

_{m+n}= a + (m+n-1)da

_{m-n }= a + (m-n-1)dand a

_{m}= a + (m-1)dSo 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)

= 2a

_{m}= 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

T

_{n}= 248a + (n -1)d = 248

3 + (n-1)5 = 248

(n-1)5 = 245

n-1 = 49

n =50

Hence 50

^{th}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 22

^{th}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 51

^{th }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.

T

_{n}= 68a+ (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.

T

_{n}= 302a+ (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 T

_{n}< 0a + (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 T

_{n }for this sequence will beT

_{n}= 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 T_{n}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 5

^{th}term of A.P. is purely real.

(b)For T_{n}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

T

_{n}= 43Last 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 a

_{1}= a= 5common difference d= 3

T

_{n}= 80Last 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 6**^{th} and 17^{th} terms of an AP are 19 and 41 respectively. Find the 40^{th} term.

^{th}and 17

^{th}terms of an AP are 19 and 41 respectively. Find the 40

^{th}term.

**Solution:**

We are given T

_{6}=19 and T_{17}=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 T

_{40}= a + 39d= 9 + 39*2

= 9 + 78

= 87

So 40

^{th}term is 87 in this AP.

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

^{th}term of an AP is Zero, prove that its 29

^{th }term is double the 19

^{th}term.

**Solution:**

We are given 9th term of AP is 0.

a

_{9}=a+ 8d = 0 ——1We have to prove, a

_{29}= 2a_{19}We know a

_{29}= a+ 28d ——-2a

_{19}= a+ 18d ——-3From equation 1 we get

a+ 8d = 0

a= -8d

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

_{29}and a_{19}a

_{29}= -8d +28d = 20da

_{19}= -8d +18d = 10dSo its proved 2a

_{19}= a_{29}.

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

^{th}term of an AP is equal to 15 times the 15

^{th}term, show that 25

^{th}term of AP is zero.

**Solution:**

We are given 10 times the 10

^{th}term of an AP is equal to 15 times the 15^{th}term.10a

_{10}= 15a_{15}10(a + 9d) = 15( a +14d)

10a + 90d = 15a + 210d

5a= -120d

5a + 120d= 0

a + 24d = 0

a

_{25}= 0So its proved a

_{25}is equal 0.

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

^{th}and 18

^{th }term of AP are 41 and 73 respectively, find 26

^{th}term.

**Solution:**

We are given a

_{10}= a+ 9d = 41 ——-1a

_{18}= a+ 17d = 73 ——-2On 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 26

^{th}term can be calculated by,a

_{26}= a + 25d= 5 + 25*4

= 105

So the 26th term of AP is 105.

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

^{th}term is twice the 10

^{th}term. Prove that the 72

^{nd}term is twice the 34

^{th}term.

**Solution:**

We are given 24

^{th}term of an AP is twice the 10^{th}term.a

_{24}= 2a_{10}So a + 23d = 2(a + 9d)

a + 23d = 2a + 18d

a = 5d

And we know, a

_{34}= a +33d= 5d + 33d

= 38d

Similarly, a

_{72}= a +71d= 5d + 71d

= 76d

Now we can see that a

_{72}= 2a_{34}. Hence proved 72^{nd}term is twice the 34^{th}term.

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