A SQURE STUDY MATERIAL SSC MATHEMATICS …€¦ · A SQURE STUDY MATERIAL SSC MATHEMATICS...

A SQURE STUDY MATERIAL SSC MATHEMATICS PROGRESSION

1 Prepared By: Arif Baig Cell:9703806542, Email:[email protected], web:urdu4medn.webnode.com

6.PROGRESSION EXERCISE - 6.1

1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (i) The taxi fare after each km when the fare is ` 20 for the first km and rises by `8 for each additional km. Sol: fare of first km. a=20 fare rises after first km. d=8 the fares would be 20, 28, 36, 44, 52, ……. Are in AP because common difference is same. (ii) The amount of air present in a cylinder

when a vacuum pump removes 1

4 of the air

remaining in the cylinder at a time. Sol: let the amount present in the cylinder=1024lit. Firs its removes ¼ th of the volume i.e, =1/4×1024=256 Remaing air present in the cylinder=1024-256=768 Second time it removes ¼ th of 768 i.e, =1/4×768=192 Remaing air present in the cylinder=768-192=576 Volume present in the cylinder 1024, 768,576, . . . . . . Hence difference is not same Not in AP because common difference is not same. (iii) The cost of digging a well, after every meter of digging, when it costs ` 150 for the first metre and rises by ` 50 for each subsequent metre. Sol: Cost of digging first meter=a=150 Cost of digging subsequent meters=d=50 List is 150, 200, 250,300,……… Are in AP because common difference is same. (iv) The amount of money in the account every year, when Rs 10000 is deposited a compound interest at 8 % per annum. Sol: amont deposited initially=a =10000 Rate of interet =8% So, amount of money after every year is 10000, 10800, 11664,…… .

Not in AP because common difference is not same. 2. Write first four terms of the AP, when the first term a and the common differenced are Given as follows: (i) a = 10, d = 10 Sol: Given a=10, d=10 We have a1=a=10 a2=a+d=10+10=20 a3=a+2d=10+2(10)=10+20=30 a4=a+3d=10+3(10)=10+30=40 Required terms are 10,20,30,40. (ii) a = –2, d = 0 Sol: Given a=-2, d=0 We have a1=a=-2 a2=a+d=-2+0=-2 a3=a+2d=-2+2(0)=-2+0=-2 a4=a+3d=-2+3(0)=-2+0=-2 Required terms are -2,-2,-2,-2. (iii) a = 4, d = – 3 Sol: Given a=4, d=-3 We have a1=a=4 a2=a+d=4+(-3)=4-3=1 a3=a+2d=4+2(-3)=4-6=-2 a4=a+3d=4+3(-3)=4-9=-5 Required terms are 4,1,-2,-5.

(iv) a = – 1, d =1

2

Sol: Given a=-1, d=1

2

We have a1=a=-1

a2=a+d=-1+1

2=

−2+1

2=

−1

2

a3=a+2d=-1+2(1

2)=-1+1=0

a4=a+3d=-1+3(1

2)=-1+

3

2=

−2+3

2=

1

2

Required terms are -1, −1

2,0,

1

2.

(v) a = – 1.25, d = – 0.25

Sol: Given a=-1.25, d=-0.25

We have a1=a=-1.25 a2=a+d=-1.25+(-0.25)=-1.25-0.25=-1.5 a3=a+2d=-1.25+2(-0.25)=-1.25-0.5=-1.75 a4=a+3d=-1.25+3(-0.25)=-1.25-0.75=-2 Required terms are -1.25,-1.5,-1.75,-2.25.



3. For the following APs, write the first term and the common difference: (i) 3, 1, – 1, – 3, . . . Sol: Given series of AP 3, 1, -1, -3,…….. a1, a2, a3, a4 First term a1=3 Common difference d=a2 − a1=1-3=-2 (ii) – 5, – 1, 3, 7, . . . Sol: Given series of AP -5, -1, 3, 7,…….. a1, a2, a3, a4 First term a1=-5 Common difference d=a2 − a1=-1-(-5)=-1+5=4

(iii) 1

3,

5

3,

9

3,

13

3,..........

Sol: Given series of AP

1

3,

5

3,

9

3,

13

3,..........

a1, a2, a3, a4

First term a1=1

3

Common difference

d=a2 − a1=5

3-

1

3=

5−3

3=

4

3.

(iv) 0.6, 1.7, 2.8, 3.9, . . .

Sol: Given series of AP 0.6, 1.7, 2.8, 3.9,…….. a1, a2, a3 , a4 First term a1=0.6 Common difference d=a2 − a1=1.7-0.6=1.1 4. Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. (i) 2, 4, 8, 16, . . . Sol: Given series 2, 4, 8, 16,…. a1, a2, a3, a4 We know in AP d=a2 − a1=a3 − a2 ⇒4-2=8-4 ⇒2≠4. Given series is not in AP.

(ii) 2,5

2, 3,

7

2,....

Sol: Given series

2, 5

2, 3,

7

2,....

a1, a2, a3, a4 We know in AP d=a2 − a1=a3 − a2

⇒5

2− 2 = 3 −

5

2

⇒5−4

2=

6−5

2

⇒1

2=

1

2

Given series is in AP

Here a=2, d=1

2

a5=a+4d=2+4(1

2)=2+2=4

a6=a+5d=2+5(1

2)=2+

5

2=

4+5

2=

9

2

a7=a+6d=2+6(1

2)=2+3=5

Required next three terms are 4, 9

2, 5.

(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . Sol: Given series -1.2, −3.2, 5.2, 7.2,.... a1, a2, a3 , a4 We know in AP d=a2 − a1=a3 − a2 ⇒ −3.2 − −1.2 =-5.2-(-3.2) ⇒-3.2+1.2=-5.2+3.2 ⇒-2=-2 Given series is in AP Here a=-1.2, d=-2 a5=a+4d=-1.2+4(-2)=-1.2-8=-9.2 a6=a+5d=-1.2+5(-2)=-1.2-10=-11.2 a7=a+6d=-1.2+6(-2)=-1.2-12=-13.2 Required next three terms are -9.2, -11.2, -13.2. (iv) – 10, – 6, – 2, 2, . . . Sol: Given series -10, −6, -2, 2,.... a1, a2 , a3 , a4 We know in AP d=a2 − a1=a3 − a2 ⇒ −6 − −10 =-2-(-6) ⇒-6+10=-2+6 ⇒4=4 Given series is in AP Here a=-10, d=4 a5=a+4d=-10+4(4)=-10+16=6 a6=a+5d=-10+5(4)=-10+20=10 a7=a+6d=-10+6(4)=-10+24=-14



Required next three terms are 6, 10, 14.

(v) 3, 3+ 2,3+2 2,3+3 2,.... Sol: Given series

3, 3+ 2,3+2 2,3+3 2,... a1, a2, a3, a4 We know in AP d=a2 − a1=a3 − a2

⇒3+ 2-3=3+2 2-(3+ 2)

⇒ 2= 2 Given series is in AP

Here a=3, d= 2

a5=a+4d=3+4 2

a6=a+5d=3+5 2

a7=a+6d=3+6 2

Required next three terms are 3+4 2, 3+5 2,

3+6 2 (vi) 0.2, 0.22, 0.222, 0.2222, . . . Sol: Given series 0.2, 0.22, 0.222, 0,2222,…. a1, a2, a3, a4 We know in AP d=a2 − a1=a3 − a2 ⇒0.2-0.22=0.222-0.22 ⇒0.02≠0.002. Given series is not in AP (vii) 0, – 4, – 8, –12, . . . Sol: Given series 0, −4, -8, -12,.... a1, a2, a3 , a4 We know in AP d=a2 − a1=a3 − a2 ⇒ −4 − 0=-8-(-4) ⇒-4=-8+4 ⇒-4=-4 Given series is in AP Here a=0, d=-4 a5=a+4d=0+4(-4)=-16 a6=a+5d=0+5(-4)=-20 a7=a+6d=0+6(-4)=-24 Required next three terms are -16, -20, -24.

(viii) - 1

2, -

1

2, -

1

2, -

1

2,…….

Sol: Given series

−1

2, −

1

2, -

1

2, -

1

2

a1, a2, a3, a4 We know in AP d=a2 − a1=a3 − a2

⇒ −1

2-(−

1

2) = −

1

2− (−

1

2)

⇒ −1

2+

1

2= −

1

2+

1

2

⇒ 0 = 0 Given series is in AP

Here a=−1

2, d=0

a5=a+4d=−1

2+4(0)= −

1

2+0=−

1

2

a6=a+5d=−1

2+5(0)= −

1

2+0=−

1

2

a7=a+6d=−1

2+6(0)= −

1

2+0=−

1

2

Required next three terms are −1

2, −

1

2, −

1

2.

(ix) 1, 3, 9, 27, . . . Sol: Given series 1, 3, 9, 27,.... a1, , a3 , a4 We know in AP d=a2 − a1=a3 − a2 ⇒ 3 − 1=9-3 ⇒2≠6 Given series not in AP. (x) a, 2a, 3a, 4a, . . . Sol: Given series a, 2a, 3a, 4a,.... a1, a2, a3 , a4 We know in AP d=a2 − a1=a3 − a2 ⇒ 2a − a=3a-2a ⇒a=a Given series is in AP Here a=a, d=a a5=a+4d=a+4(a)=a+4a=5a a6=a+5d=a+5(a)=a+5a=6a a7=a+6d=a+6(a)=a+6a=7a Required next three terms are 5a, 6a, 7a. (xi) a, a2, a3, a4, . . . Sol: Given series a, a2, a3, a4,.... a1, a2, a3 , a4 We know in AP d=a2 − a1=a3 − a2 ⇒ a2 − a=a3-a2 Given series is in AP Here a=a, d=a

(xii) 2, 8, 18, 32,..... Sol: Given series

2, 8, 18, 32,... a1, a2, a3, a4 We know in AP d=a2 − a1=a3 − a2



⇒ 8- 2-= 18- 8

⇒ 2 × 2 × 2 − 2= 2 × 3 × 3 − 2 × 2 × 2

⇒ 2 2 − 2=3 2-2 2

⇒ 2 = 2 Given series is in AP

Here a= 2, d= 2

a5=a+4d= 2+4( 2)=5 2

a6=a+5d= 2+5( 2)=6 2

a7=a+6d= 2+6( 2)=7 2

Required next three terms are 5 2, 6 2, 7 2.

(xiii) 3, 6, 9, 12,..... Sol: Given series

3, 6, 9, 12,... a1, a2, a3, a4 We know in AP d=a2 − a1=a3 − a2

⇒ 6- 3-= 9- 6

⇒ 6 − 3≠3 − 6 Given series is not in AP

EXERCISE - 6.2 1. Fill in the blanks in the following table, Given that a is the first term, d the common difference and an the nth term of the AP: s.no a d n an (i) (ii) (iii) (iv) (v)

7 -18 ….. -18.9 3.5

3 …. -3 2.5 0

8 10 18 …. 105

…. 0 -5 3.6 …..

(i) sol: Given a=7, d=3, n=8, an =? we have an =a+(n-1)d a8=7+(8-1)3 =7+(7)3 =7+21 =28. (ii) sol: Given a=-18, n=10, an =0, d=? we have an =a+(n-1)d 0=-18+(10-1)d 0=-18+9d 18=9d 2=d (or) d=2. (iii) sol: Given d=-3, n=18, an =-5, a=? we have an =a+(n-1)d -5=a+(18-1)(-3) -5=a+17(-3) -5=a-51 -5+51=a

46=a (or) a=46. (iv) sol: Given a=-18.9, d=2.5, an =3.6, n=? we have an =a+(n-1)d 3.6=-18.9+(n-1)2.5 3.6=-18.9+2.5n-2.5 3.6=2.5n-21.4 3.6+21.4=2.5n 25=2.5n 10=n. (v) sol: Given a=3.5, d=0, n=105, an =? we have an =a+(n-1)d a105=3.5+(105-1)0 =3.5+(104)0 =3.5+0 =3.5 2. Find the (i) 30th term of the A.P. 10, 7, 4 ...... Sol: Given AP series 10, 7, 4, ………. a1, a2,a3 a=10, d=, a2 −, a1=7-10=-3 we have an =a+(n-1)d 30th term of AP a30 =a+29d =10+29(-3) =10-87 =-77.

(ii) 11th term of the A.P. :-3, −1

2, 2,.....

Sol: Given AP series

-3, −1

2, 2, ……….

a1, a2,a3

a=3, d=, a2 −, a1=−1

2-(-3)=

−1+6

2=

5

2

we have an =a+(n-1)d 11th term of AP a11 =a+10d

=-3+10(5

2)

=-3+5(5) =-3+25 =22. 3. Find the respective terms for the following APs. (i) a1 = 2; a3 = 26 find a2 Sol: Given a1 = 2; a3 = 26 We know



a=2, a3=a+2d=26 2+2d=26 2d=26-2 2d=24 d=12. We have a2=a+d =2+12 =14. (ii) a2 = 13; a4 = 3 find a1, a3 Sol: Given a2 = 13; a4 = 3 We know a2 =a+d= 13 ① a4 =a+3d= 3 (2) From ① and ② a+d= 13 a+3d=3 2d=-10 d=-5.

d=-5 substitute in ① a+d=13 a+(-5)=13 a=13+5 a=18. a3=a+2d =18+2(-5) =8-10 =8.

(iii)a1 = 5; a4 = 91

2 find a2 , a3

Sol: Given

a1 = 5; a4 = 91

2=

19

2

We know

a4 =a+3d=19

2

5+3d=19

2

3d=19

2-5

3d=19−10

2

3d=9

2

d=3

2.

a2=a+d=5+(3

2)=

10+3

2=

7

2=

13

2

a3=a+2d=5+2(3

2)=5+3=8.

(iv) a1 = -4; a6 = 6 find a2 , a3 , a4 , a5 Sol: Given a1 = -4; a6 =6

We know a6 =a+5d=6 -4+5d=6

5d=6+4 5d=10 d=2. a2=a+d=-4+2=-2 a3=a+2d=-4+2(2)=-4+4=0 a4=a+3d=-4+3(2)=-4+6=2 a5=a+4d=-4+4(2)=-4+8=4. (v) a2 = 38; a6 = -22 find a1, a3 , a4 , a5 Sol: Given a2 =38; a6 =-22 We know a2 =a+d=38 ① a6 =a+5d=-22 ② From ① and ② a+d= 38 a+5d=-22 4d=-60 d=-15. d=-15 substitute in ① a+d=38 a+(-15)=38 a=38+15 a=53. a1=a=53 a3=a+2d=53+2(-15)=53-30=23 a4=a+3d=53+3(-15)=53-45=8 a5=a+4d=53+4(-15)=53-60=-7. 4. Which term of the AP : 3, 8, 13, 18, . . . ,is 78? Sol: Given series in AP 3, 8, 13, 18, . . . a1, a2 a3 , a4 Here a=3, d=a2 − a1=8-3=5, an =78 Let the nth term is 78. We have an =a+(n-1)d 78=3+(n-1)5 78=3+5n-5 78=5n-2 78+2=5n 80=5n 16=n. 16th term is 78. 5. Find the number of terms in each of the following APs : (i) 7, 13, 19, . . . , 205 Sol: Given series in AP 7, 13, 19, . . . a1, a2 a3 , Here a=7, d=a2 − a1=13-7=6, an =205 Let the n terms in AP We have an =a+(n-1)d 205=7+(n-1)6 205=7+6n-6



78=6n+1 205-1=6n 204=6n 34=n. 34 terms are there in AP

(ii) 18, 151

2, 13, …….-47

Sol: Given series in AP

18, 151

2, 13, . . .

a1, a2 a3 ,

Here a=18, d=a2 − a1=151

2− 18=

31

2− 18 =

31−36

2=

−5

2, an =-47

Let the n terms in AP We have an =a+(n-1)d

-47=18+(n-1)( −5

2)

-47=18-5n

2+

5

2

-47-18-5

2=−

5n

2

-65-5

2=−

5n

2

−130−5

2=−

5n

2

-135=-5n 27=n 27 terms are there in AP. 6. Check whether, –150 is a term of the AP : 11, 8, 5, 2 . . . sol: Given AP series 11, 8, 5, 2, . . . . a1, a2, a3 , a4 Here a=11, d=a2 − a1=8-11=-3 Let -150 is a nth term of AP We have, an =a+(n-1)d -150=11+(n-1)(-3) -150=11-3n+3 -150=-3n+14 -150-14=-3n -164=-3n

164

3=n

But n should be a positive integer So, -150 is not a term of the Given AP series. 7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73. Sol: Given a11 = 38, a16 = 73 We have a11 = a + 10d = 38 (1) a16 = a + 15d = 73 (2) From (1) and (2) a + 10d = 38 a + 15d = 73 5d=35

d=7 d =7 subsituting in (1) a + 10d = 38 a+10(7)=38 a+70=38 a=38-70 a=-32. 31st term of AP a31 = a + 30d =-32+30(7) =-32+210 =178. 8. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero? Sol: Given a3 = 4, a9 = −8 We have a3 = a + 2d = 4 (1) a9 = a + 8d = −8 (2) From (1) and (2) a + 2d = 4 a + 8d = −8 6d=-12 d=-2 d =-2 subsituting in (1) a + 2d = 4 a+2(-2)=4 a-4=4 a=4+4=8 let the nth termof AP is zero. an =a+(n-1)d 0=8+(n-1)(-2) 0=8-2n+2 0=10-2n 2n=10 n=5 5th term of AP is zero 9. The 17th term of an AP exceeds its 10th term by 7. Find the common difference. Sol: let 17th term of AP a17=a+16d 10th term of AP a10=a+9d According to the problem a10+7=a17 a+9d+7=a+16d -16d+9d=a-a-7 -7d=-7 d=1. 10. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms? Sol: let first AP series a1, a2 , a3 ,…….



Here a100=a+99d Second AP series b1, b2, b3,……. Here b100=b+99d According to the problem a100-b100=100 a+99d-(b+99d)=100 a+99d-b-99d=100 a-b=100. We have a1000=a+999d , b1000=b+999d = a1000 − b1000 =a+999d-(b+999d) =a+999d-b-999d =a-b we know a-b=100 =100. 11. How many three-digit numbers are divisible by 7? Sol: we know that The list of three-digit numbers divisible by 7 is : 105, 112, 119, . . . , 994 are in AP Here, a = 105, d = a2 − a1 = 112 − 105 = 7, an = 994. Let the n terms of three-digit numbers divisible by 7 is As an = a + (n – 1) d, we have 994 = 105 + (n – 1) 7 i.e., 994 =105+ 7n – 7 i.e., 994=7n+98 i.e., 994-98=7n 896=7n 128=n So, there are 128 three-digit numbers divisible by 7. 12. How many multiples of 4 lie between 10 and 250? Sol: we know that The list of multiples of 4 lies between 10 and 250 : 12, 16, 20, . . . , 248 are in AP Here, a = 12, d = a2 − a1 = 16 − 12 = 4, an = 248. Let the n terms multiples of 4 lies between 10 and 250 As an = a + (n – 1) d, we have 248 = 12+ (n – 1) 4 i.e., 248 =12+ 4n – 4 i.e., 248=4n+8 i.e., 248-8=4n 240=4n 60=n So, there are 60 multiples of 4 lies between 10 and 250.

13. For what value of n, are the nth terms of two APs: 63, 65, 67, . .. and 3, 10, 17, . . .equal? Sol: For 63, 65, 67, ….. Here a=63, d = a2 − a1 = 65 − 63 = 2 nth term of AP As an = a + (n – 1) d an =63+(n-1)2 =63+2n-2 =2n+61. For 3, 10, 13, ….. Here a=3, d = a2 − a1 = 10 − 3 = 7 nth term of AP As an = a + (n – 1) d an =3+(n-1)7 =3+7n-7 =7n-4 Let for n value nth terms are equal 2n+61=7n-4 2n-7n=-4-61 -5n=-65 n=13. 14. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12. Sol: Given 3rd term of AP is 16 a3 = a + 2d=16 (1) We have a7 = a + 6d , a5 = a + 4d According to the problem a7=a5+12 a+6d=a+4d+12 6d-4d=a-a+12 2d=12 d=6. d=6 substituting in (1) a + 2(6)=16 a+12=16 a=16-12 a=4. a1=a=4 a2=a+d=4+6=10 a3=a+2d=4+2(6)=4+12=16. Required AP series is 4, 10, 16,… 15. Find the 20th term from the end of the AP : 3, 8, 13, . . ., 253. Sol: Given AP series 3, 8, 13, …….., 253. Here a =3, d = a2 − a1 = 8 − 3 = 5 , an =253 Let the n terms in this series As an =a+(n-1)d 253=3+(n-1)(5) 253 =3+5n-5 253=5n-2 253+2=5n 255=5n



51=n 20th term from the end of AP is a31 a31=a+30d =3+30(5) =3+150 =158. 16. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP. Sol: let 4th term of AP a4=a+3d 8th term of AP a8=a+7d Given a4 + a8=24 a+3d+a+7d=24 2a+10d=24 2(a+5d)=24 a+5d=12. ① let 6th term of AP a6=a+5d 10th term of AP a10=a+9d Given a6 + a10=44 a+5d+a+9d=44 2a+14d=44 2(a+7d)=44 a+7d=22. ② from ① and ② a+5d=12 a+7d=22 2d=10 d=5. d=5 substituting in ① a+5d=12 a+5(5)=12 a+25=12 a=12-25 a=-13. a1=a=-13 a2=a+d=-13+5=-8 a3=a+2d=-13+2(5)=-13+10=-3 Required AP series is -13, -8, -3,….. 17. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000? Sol: Given in 1995 subba rao salary is 5000 and annual increment is 200. 5000, 5200,5400,……..7000. subba rao salary in AP Here a=5000, d= a2 − a1 = 5200 − 5000 =200 , an =7000 Let after n years subba rao salary is 7000. As an =a+(n-1)d 7000=5000+(n-1)200 7000=5000+200n -200

7000=200n+4800 7000-4800=200n 2200=200n 11=n. So, 1995+11=2006 In 2006 subba rao income reach to 7000. 18. Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs 1.75. If in the nth week, her weekly savings

become Rs 20.75, find n.

Sol: Given

Ramakali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs 1.75 5, 6.75, 8.50,…….20.75 ramakali savings are in AP

Here a=5, d = a2 − a1 = 6.75 − 5 = 1.75 , an =20.75

Let after n years her weekly savings become 20.75

As an =a+(n-1)d 20.75=5+(n-1)(1.75) 20.75=5+1.75n-1.75 20.75=1.75n+3.25 20.75-3.25=1.75n 17.5=1.75n 10=n. After 10 years her weekly savings become 20.75.

EXERCISE - 6.3 1. Find the sum of the following APs: (i) 2, 7, 12, . . ., to 10 terms. Sol: Given AP series 2, 7, 12, . . . . Here a=2, d = a2 − a1 = 7 − 2 = 5 , n=10

As Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

Sum to 10 terms

S10=10

2 2(2) + 10 − 1 5

=5(4+(9)5) =5(4+45) =5(49) =245. (ii) –37, –33, –29, . . ., to 12 terms. Sol: Given AP series -37, -33, -29, . . . .



Here a=-37, d = a2 − a1 = −33 − −37 =−33 + 37 = 4 , n=12

As Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

Sum to 12 terms

S12=12

2 2(−37) + 12 − 1 4

=6(-74+(11)4) =6(-74+44) =6(-30) =-180. (iii) 0.6, 1.7, 2.8, . . ., to 100 terms. Sol: Given AP series 0.6, 1.7, 2.8, . . . . Here a=0.6, d = a2 − a1 = 1.7 − 0.6 = 1.1 , n=100

As Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

Sum to 100 terms

S100=100

2 2(0.6) + 100 − 1 (1.1)

=50(1.2+(99)(1.1)) =50(1.2+108.9) =50(110.1) =5505.

(iv) 1

15,

1

12,

1

10, .....,to 11terms.


1

15,

1

12,

1

10, . . . .

Here a=1

15, d = a2 − a1 =

1

12−

1

15=

5−4

60 =

1

60,

n=11

As Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

Sum to 11 terms

Sn=11

2 2(

1

15) + 11 − 1

1

60

=11

2 (2(

1

15)+(10)

1

60)

=11

2 (

2

15+

1

6)

=11

2 (

8+10

60)

=11

2 (

18

60)

=11

2 (

3

10)

=33

20.

2. Find the sums Given below :

(i)7 +101

2+ 14, ....+ 84


7+101

2+14 . . . .+84

Here a=7, d = a2 − a1 = 101

2− 7 =

21

2− 7 =

21−14

2=

7

2 , 𝑡𝑛=84,

We need to find ‘n’ As an =a+(n-1)d

84=7+(n-1) 7

2

84=7+7𝑛

2−

7

2

84=14+7𝑛−7

2

168=7n+7 168-7=7n 161=7n 23=n

As Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

Sum to 23 terms

=23

2 2(7) + 23 − 1

7

2

=23

2 (14+(22)

7

2)

=23

2 (14+(11)7)

=23

2 (14+77)

=23

2(91)

=2093

2.

(ii) 34 + 32 + 30 + . . . + 10 Sol: Given AP series 34+32+30 +. . . .+10 Here a=34, d = a2 − a1 = 32 − 34 = −2 , 𝑡𝑛=10, We need to find ‘n’ As an =a+(n-1)d 10=34+(n-1) (−2) 10=34-2n+2 10=-2n+36 2n=36-10 2n=26 n=13

As Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

Sum to 13 terms

=13

2 2(34) + 13 − 1 (−2)

=13

2 (68+ 12 (−2))

=13

2 (68-24)

=13

2 (44)

= 13(22) =286. (iii) –5 + (–8) + (–11) + . . . + (–230) Sol: Given AP series –5 + (–8) + (–11) + . . . + (–230) Here a=-5, d = a2 − a1 = −8 − −5 = −8 +5 = −3 , 𝑡𝑛=-230, We need to find ‘n’ As an =a+(n-1)d -230=-5+(n-1) (−3) -230=-5-3n+3 -230=-3n-2 3n=230-2



3n=228 n=76

As Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

Sum to 76 terms

=76

2 2(−5) + 76 − 1 (−3)

=38 (-10+ 75 (−3)) =38 (-10-225) =38 (-235) = -8930. 3. In an AP: (i) Given a = 5, d = 3, an = 50, find n and sn . Sol: Given a = 5, d = 3, an = 50 As an =a+(n-1)d 50=5+(n-1)3 50=5+3n-3 50=3n+2 50-2=3n 48=3n 16=n,

We have Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

Sum to 16 terms

=16

2 2(5) + 16 − 1 (3)

=8 (10+ 15 (3)) =8 (10+45) =8 (55) = 440. (ii) Given a = 7, a13= 35, find d and s13. Sol: Given a = 7, a13= 35 As an =a+(n-1)d 35=7+(13-1)d 35=7+12d 35-7=12d 28=12n

28

12=3n

7

3=n,

We have Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

Sum to 13 terms

=13

2 2(7) + 13 − 1 (

7

3)

=13

2 (14+ 12 (

7

3))

= 13

2 (14+4(7))

= 13

2 (14+28)

= 13

2(42)

=13(21) =273. (iii) Given a12 = 37, d = 3, find a and s12. Sol: Given a12 = 37, d = 3

As an =a+(n-1)d 37=a+(12-1)3 37=a+(11)3 37=a+33 37-33=a 4=a,

We have Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

Sum to 12 terms

=12

2 2(4) + 12 − 1 (3)

=6 (8+ 11 (3)) =6 (8+33) =6 (41) = 246 . (iv) Given a3 = 15, s10 = 125, find d and a10 . Sol: Given a3 = 15, s10 = 125 We have a3=a+2d=15 (1)

s10 =10

2 2𝑎 + 10 − 1 𝑑 =125

=5(2a+9d)=125 =2a+9d=25 (2) From (1) and (2) 2×(1) 2a+4d=30 (2) 2a+9d=25 5d=-5 d=-1 d=-1 substituting in (1) a+2(-1)=15 a-2=15 a=15+2 a=17, As an =a+(n-1)d a10=17+(10-1)(-1) =17+9(-1) =17-9 =8. (v) Given a = 2, d = 8, sn = 90, find n and an . Sol: Given a = 2, d = 8, sn = 90

We have Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

90=𝑛

2 2(2) + 𝑛 − 1 8

180=n(4+8n-8) 180=n(8n-4) 180=8𝑛2-4n 8𝑛2-4n-180=0 4(2𝑛2-n-45)=0 2𝑛2-n-45=0 2𝑛2+9n-10n-45=0 n(2n+9)-5(2n+9)=0 (n-5)(2n+9)=0 n-5=0, 2n+9=o n=5, 2n=-9



n=−9/2,(discarded) an= 𝑎+ 𝑛−1 𝑑 =2+(5-1)8=2+(4)8=2+32=34. (vi) Given an = 4, d = 2, sn = –14, find n and a. Sol: Given that an=l=4, d=2,Sn=-14

an= 𝑎+ 𝑛−1 𝑑 4=a+(n-1)2 4=a+2n-2 4+2=a+2n 6=a+2n a=6-2n ∴ Sn=𝑛/2(𝑎+𝑙) -14=𝑛/2(6−2𝑛+4) -28=n(-2n+10)

-28=-2n2+10n

2n2-10n-28=0

2(n2-5n-14)=0

n2-5n-14=0

n2+2n-7n-14=0 n(n+2)-7(n+2)=0 (n+2)(n-7)=0 n+2=0, n-7=0 n=-2(discarded), n=7

then

a=6-2n=6-2(7)=6-14=-8. (vii) Given l= 28, sn= 144, and there are total 9 terms. Find a. Sol: Given l= 28, sn= 144, n=9

As sn=𝑛

2(a+l)

144=9

2(a+28)

288=9(a+28) 32=a+28 32-28=a 4=a. 4. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum? Sol: Given first term a1=17, last term l=a+(n-1)d=350, d=9 a and d values substituting in ‘l’ a+(n-1)d=350 17+(n-1)9=350 17+9n-9=350 8+9n=350 9n=350-8 9n=342 n=38. Sum to n terms

As Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

=38

2 2(17) + 38 − 1 9

=19 34 + 37 9 =19(34+333) =19(367) =6973. 5. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively. Sol: Given AP 2nd and 3rd terms are 14 and 18 a2=a+d=14 (1) a3=a+2d=18 (2) We have d= a3 − a2=18-14=4 d=4 substituting in (1) a+d=14 a+4=14 a=14-4 a=10. Sum to first 51 terms

as Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

S15=51

2 2(10) + 51 − 1 4

S15=51

2 20 + (50)4

S15=51

2 20 + 200

S15=51

2 220

S15=51(110) =5610. 6. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms. Sol: Given sum to 7 terms of an AP is 49 and sum to 17 terms of an AP is 289

as Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

S7=7

2 2𝑎 + 7 − 1 𝑑 =49 , S17=

17

2 2𝑎 +

17−1𝑑=289 7(2a+6d)=98, 17(2a+16d)=578 2a+6d=14, 2a+16d=34 2(a+3d)=14, 2(a+8d)=34 a+3d=7, (1) a+8d=17 (2) from (1) and (2) a+3d=7 a+8d=17 5d=10 d=2 d=2 substitutuing in (1) a+3d=7 a+3(2)=7 a+6=7



a=7-6 a=1. Sum to n terms

As Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

=𝑛

2 2(1) + 𝑛 − 1 2

=𝑛

2 2 + 2𝑛 − 2

=𝑛

2(2n)

=𝑛2 . 7. Show that a1, a2 , . . ., an , . . . form an AP where an is defined as below : (i) an = 3 + 4n (ii) an = 9 – 5n Also find the sum of the first 15 terms in each case. i) sol: Given an = 3 + 4n we have a1 = 3 + 4(1)=3+4=7 a2 = 3 + 4(2)=3+8=11 a3 = 3 + 4(3)=3+12=15 In AP we know d=a2 − a1=a3 − a2 11-7=15-11 4=4. Hence series in AP Here a=7, d=4, Sum to 15 terms

As Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

=15

2 2 7 + 15 − 1 4

=15

2 14 + 14 4

=15

2 14 + 56

=15

2[70]

=15[35] =525. i) sol: Given an = 9-5n we have a1 = 9-5(1)=9-5=4 a2 = 9-5(2)=9-10=-1 a3 = 9-5(3)=9-15=-6 In AP we know d=a2 − a1=a3 − a2 -1-4=-6-(-1) -5=-5. Hence series in AP Here a=4, d=-5, Sum to 15 terms

As Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

=15

2 2 4 + 15 − 1 (−5)

=15

2 8 + 14 (−5)

=15

2 8 − 70

=15

2[-62]

=15[-31] =-465. 8. If the sum of the first n terms of an AP is 4n – n2 , what is the first term (remember the first term is s1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms. Sol: Given sum of the first n terms of an AP is 4n – n2 Sn=4n – n2 S1=4(1) – 12=4-1=3 S2=4(2) – 22=8-4=4 We have a1= S1=3 a2= S2 − S1=4-3=1 d = a2 − a1 = 1 − 3 = −2 3rd term a3=a+2d=3+2(-2)=3-4=-1 10th term a10=a+9d=3+9(-2)=3-18=-15 nth term an=a+(n-1)d =3+(n-1)(-2) =3-2n+2 =5-2n. 9. Find the sum of the first 40 positive integers divisible by 6. Sol: let first positive integers divisible by 6 6,12,18,. . . . . . Here a=6 d = a2 − a1 = 12 − 6 = 6 ,

We have Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

Sum to first 40 positive integers divisible by 6

S40=40

2 2(6) + 40 − 1 6

=20(12+(39)6) =20(12+234) =20(246) =4920. 10. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their Over all academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes sol: let sum of Rs 700 is Sn =700,

seven cash prizes to students so n=7,

each prize is Rs 20 less than its preceding prize so d=20,

As we have Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑



700=7

2 2𝑎 + 7 − 1 20

1400=7 2𝑎 + 6 20 200=2a+120 200-120=2a 80=2a 40=a 1st prize a7=a+6d=40+6(20)=40+120=160 2nd priz a6=a+5d=40+5(20)=40+100=140 3rd prize a5=a+4d=40+4(20)=40+80=120 4th prize a4=a+3d=40+3(20)=40+60=100 5th prize a3=a+2d=40+2(20)=40+40=80 6th prize a2=a+d=40+20=60 7th prize 𝑎1=a=40. The prizes are 160,140,120,100,80,60,40.

11. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students? Sol: Given that Planting of trees by each class=3section×class Planting of trees by first, second, third, . . . . classes

3×1, 3×2,3×3. . . . . . . . 3×12

Series in GP

3, 6, 9,. . . . . . . . . . .36

a1 a2 a3 a12

here

a=3, d=6-3=3 l=36

Total plants planted by first to twele classes

∴ Sn=𝑛

2 𝑎 + 𝑙

=12

2 3 + 36

=6[39]

=234.

Hence, total 234 plants planted.

12. A spiral is made up of successive semicircles, with

centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . . as shown in Fig. 5.4. What is the total length of such a spiral made up of thirteen

consecutive semicircles? (Take π =22

7 )

[Hint : Length of successive semicircles is l1, l2, l3 , l4 . . . with centres at A, B, A, B, . . ., respectively.] sol: Given that Radius of successive semicircles l1, l2, l3, l4, . . . . l13. Is respectively

0.5, 1.0, 1.5, 2.0,.. . . . . . . . .

Length of spiral 13=l1+l2+l3+. . . . . . l13

∴ cicumference of semicircles=𝜋𝑟

Total length of spiral=𝜋×0.5+𝜋×1.0+𝜋×1.5+. . . . . . . .

=𝜋[0.5+1.0+1.5+. . . . . . . .are in GP]

a1 a2 a3

here

a=0.5, d=1-0.5=0.5 n=13

∴ Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

=𝜋[13

2 2(0.5) + 13 − 1 (0.5) ]

=𝜋[13

2 1 + 12 0.5 ]

=𝜋[13

2 1 + 6 ] ∴ 𝜋 =

22

7

=22

7[

13

2 7 ]

=11[13]

=143cm

13. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how may rows are the 200 logs placed and how many logs are in the top row?

Sol: Total logs=200

Sn=200

Logs in 1st ,2nd , 3rd ,. . . . . rows

20, 19, 18,. . . . . . . . ..are in GP



a1 a2 a3

here

a=20, d=19-20=-1

∴ Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

200=𝑛

2 2(20) + 𝑛 − 1 (−1)

400=n[40-n+1]

400=n[-n+41]

400=-n2+41n

n2-41n+400=0

400n2

-16n -25n

n2-16n-25n+400=0

n(n-16)-25(n-16)=0

(n-16)(n-25)=0

n-16=0, n-25=0

n=16, n=25( 25 rows does not exist because of only 20 logs in bottom row)

hence total rows are 16.

Total logs in top row

a16=a+15d=20+15(-1)=20-15=5

14. In a bucket and ball race, a bucket is placed at the starting point, which is 5 m from the first ball, and the other balls are placed 3 m apart in a straight line. There are ten balls in the line. A competitor starts from the bucket, picks up the nearest ball, runs back with it, drops it in the bucket, runs back to pick up the next ball, runs to the bucket to drop it in, and she continues in the same way until all the balls are in the bucket. What is the total distance the competitor has to run? [Hint : To pick up the first ball and the second ball, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)] sol: Given that Distance of first ,second, third,. . . . ball from bucket 5, 5+3, 5+3+3,.. . . . . . . . . . .

5, 8, 11, . . . . . . . . . . .

Distance rum by competitor to drop the first, second, third, . . . tenth ball in bucket

2×5, 2×8, 2×11, . . . . . . . . . .

10, 16, 22, . . . . . . . . . . are in GP

a1 a2 a3

here, a=10, d=16-10=6

total distance run by competitor to drop all 10 balls in bucket.

∴ Sn=𝑛

2 2𝑎 + 𝑛 − 1 𝑑

=10

2 2(10) + 10 − 1 6

=5[20+(9)6]

=5[20+54]

=5[74]

=370m.

Hence the total distance competitor has to run

is 370m.

EXERCISE - 6.4

1. In which of the following situations, does the list of numbers involved make a G.P.? (i) Salary of Hema, when her salary is 5,00,000/- for the first year and expected to receive yearly increase of 10% . sol: given that Hema’s annual salary is 5,00,000 and its increase yearly 10% So, salary of first year a=5,00,000

salary of second year

A=P 1 +𝑅

100 𝑛

= 5,00,000 1 +10

100

1

=5,00,000(11

10)=5,50,000

Salary of third year

= 5,50,000 1 +10

100

1

=5,50,000(11

10)=6,05,000

Let the first, second , third ,. . . . year salary in GP.



500000, 550000,605000, . . . …. . . .

a1 a2 a3

for GP

r=𝑎2

𝑎1=

𝑎3

𝑎2

=550000

500000=

605000

550000

=11

10=

11

10

(ii) Number of bricks needed to make each step, if the stair case has total 30 steps Bottom step needs 100 bricks and each successive step needs 2 brick less than the previous step. sol: Given that No. of bricks needed for bottom step=100 each successive step needs 2 brick less than the previous step hence bricks needed for second step from bottom =100-2=98

bricks needed for third step from bottom =98-2=96

bricks needed for fourth step from bottom =96-2=94

So, series is

100, 98, 96, 94, . . . . . . . .

Hence the situation form AP not GP.

(iii) Perimeter of the each triangle, when the mid points of sides of an equilateral triangle whose side is 24 cm are joined to form another triangle, whose mid points in turn are joined to form still another triangle and the process continues indefinitely. 24 24 24 sol: Given that side of a equilateral triangle=24

Perimeter of equilateral triangle=24+24+24=72

side of a each inside equilateral triangle=24/2=12

Perimeter of each inside equilateral triangle=12+12+12=36

Perimeters of equilateral triangles

72, 36, 18, . . . . . . . . .

a1 a2 a3

for GP

r=𝑎2

𝑎1=

𝑎3

𝑎2

=36

72=

18

36

=1

2=

1

2

Hence the situation form GP.

2. Write three terms of the G.P. when the first term ‘a’ and the common ratio ‘r’ are Given? (i) a = 4 r = 3 Sol: Given a=4, r=3 Let three terms in GP are 𝑎1, 𝑎2 and 𝑎3 We have 𝑎1=a=4 𝑎2 = 𝑎𝑟1=4(3)=12 𝑎3 = 𝑎𝑟2= 4(3)2=4(9)=36 Required three terms in GP are 4, 12, 36.

(ii) a= 5 , r=1

5

Sol: Given a= 5, r=1

5

Let three terms in GP are 𝑎1, 𝑎2 and 𝑎3 We have

𝑎1=a= 5

𝑎2 = 𝑎𝑟1= 5 (1

5)=

1

5

𝑎3 = 𝑎𝑟2= 5(1

5)2= 5 (

1

25)= 5(

1

5.

1

5)=

1

5 5

Required three terms in GP are 5, 1

5,

1

5 5.

(iii) a = 81 r =−1

3

Sol: Given a=81, r=−1

3

Let three terms in GP are 𝑎1, 𝑎2 and 𝑎3 We have 𝑎1=a=81

𝑎2 = 𝑎𝑟1=81 (−1

3)=27(-1)=-27

𝑎3 = 𝑎𝑟2= 81(−1

3)2=81 (

1

9)=9(1)=9

Required three terms in GP are 81, -27, 9.



(iv) a=1

64, r=2

Sol: Given a=1

64, r=2

Let three terms in GP are 𝑎1, 𝑎2 and 𝑎3 We have

𝑎1=a=1

64

𝑎2 = 𝑎𝑟1=1

64 (2)=

1

32

𝑎3 = 𝑎𝑟2= 1

64(2)2=

1

64 (4)=

1

16

Required three terms in GP are 1

64,

1

32,

1

16.

3. Which of the following are G.P. ? If they are G.P. Write three more terms? (i) 4, 8, 16 ..... Sol: 4, 8, 16, . . . . 𝑎1, 𝑎2 , 𝑎3 In GP we know

r= 𝑎2

𝑎1=

𝑎3

𝑎2

8

4=

16

8

2=2. Given series in GP Here a=4, r=2 𝑎4 = 𝑎𝑟3=4(2)3=4(8)=32 𝑎5 = 𝑎𝑟4=4(2)4=4(16)=64 𝑎6 = 𝑎𝑟5=4(2)5=4(32)=128 Required three terms in GP are 32, 64, 128……

(ii) 1

3 ,

−1

6,

1

12 .....

Sol: 1

3 ,

−1

6,

1

12 , . . . .

𝑎1, 𝑎2 , 𝑎3 In GP we know

r= 𝑎2

𝑎1=

𝑎3

𝑎2

−1

61

3

=1

12−1

6

−1

2=

−1

2.

Given series in GP

Here a=1

3, r=

−1

2

𝑎4 = 𝑎𝑟3=1

3(−1

2)3=

1

3 (

−1

8)=

−1

24

𝑎5 = 𝑎𝑟4=1

3(−1

2)4=

1

3 (

1

16)=

1

48

𝑎6 = 𝑎𝑟5=1

3(−1

2)5=

1

3 (

−1

32)=

−1

96

Required three terms in GP are −1

24,

1

48,

−1

96,…….

(iii) 5, 55, 555, ... Sol: 5, 55, 555, . . . . 𝑎1, 𝑎2 , 𝑎3 In GP we know

r= 𝑎2

𝑎1=

𝑎3

𝑎2

55

5=

555

55

11≠111

11.

Given series is not in GP. (iv) -2, -6, -18 ..... Sol: -2, -6, -18, . . . . 𝑎1, 𝑎2 , 𝑎3 In GP we know

r= 𝑎2

𝑎1=

𝑎3

𝑎2

−6

−2=

−18

−6

3=3. Given series in GP Here a=-2, r=3 𝑎4 = 𝑎𝑟3=-2(3)3=-2(9)=-18 𝑎5 = 𝑎𝑟4=-2(3)4=-2(27)=-54 𝑎6 = 𝑎𝑟5=-2(3)5=-2(81)=-168 Required three terms in GP are -18, -54, -168……

(v) 1

2 ,

1

4,

1

6 .....

Sol: 1

2 ,

1

4,

1

6 , . . . .


r= 𝑎2

𝑎1=

𝑎3

𝑎2

1

41

2

=1

61

4

1

2≠

3

2.

Given series is not in GP (vi) 3, −32,33 , ..... Sol: 3, −32,33, . . . . 𝑎1, 𝑎2 , 𝑎3 In GP we know

r= 𝑎2

𝑎1=

𝑎3

𝑎2

−32

3=

33

−32

−9

3=

27

−9

-3=-3. Given series in GP Here a=3, r=-3 𝑎4 = 𝑎𝑟3=3(−3)3=3(-27)=-81 𝑎5 = 𝑎𝑟4=3(−3)4=3(81)=243 𝑎6 = 𝑎𝑟5=3(−3)5=3(-243)=-729 Required three terms in GP are-81, 243, -729, ……

(vii) x, 1, 1

x ,.....

Sol: x, 1, 1

𝑥, . . . .




r= 𝑎2

𝑎1=

𝑎3

𝑎2

1

𝑥=

1

𝑥

1

1

𝑥=

1

𝑥

Given series in GP

Here a=x, r=1

𝑥

𝑎4 = 𝑎𝑟3=x(1

𝑥)3=x(

1

𝑥3)= 1

𝑥2

𝑎5 = 𝑎𝑟4=x(1

𝑥)4=x(

1

𝑥4)= 1

𝑥3

𝑎6 = 𝑎𝑟5=x(1

𝑥)5=x(

1

𝑥5)= 1

𝑥4

Required three terms in GP are 1

𝑥2, 1

𝑥3, 1

𝑥4……

(viii)1

2 , -2,

8

2 ,.....

Sol: 1

2 , -2,

8

2 , . . . .


r= 𝑎2

𝑎1=

𝑎3

𝑎2

−2

1

2

=

8

2

−2

−2

1. 2

1=

8

2.

1

−2

- 2 2=-2 2. Given series in GP

Here a=1

2, r= -2 2

𝑎4 = 𝑎𝑟3=1

2(−2 2)3=

1

2 (-8. 2. 2. 2)=-

8.2=-16

𝑎5 = 𝑎𝑟4=1

2(−2 2)4=

1

2

(16 2. 2. 2. 2)=16.2. 2=32 2

𝑎6 = 𝑎𝑟5=1

2(−2 2)5=

1

2 (-

32 2. 2. 2. 2. 2)=-32.2.2=-128

Required three terms in GP are -16, 32 2, -128…… (ix) 0.4, 0.04, 0.004, ..... Sol: 0.4, 0.04, 0.004, . . . . 𝑎1, 𝑎2 , 𝑎3 In GP we know

r= 𝑎2

𝑎1=

𝑎3

𝑎2

0.04

0.4=

0.004

0.04

0.1=0.1. Given series in GP Here a=0.4, r=0.4 𝑎4 = 𝑎𝑟3=0.4(0.1)3=0.4(0.001)=0.0004 𝑎5 = 𝑎𝑟4=0.4(0.1)4=0.4 (0.0001)=0.00004 𝑎6 = 𝑎𝑟5=0.4(0.1)5=0.4 (0.00001)=0.000004

Required three terms in GP are 0.0004, 0.00004, 0.000004…… 4. Find x so that x, x + 2, x + b are consecutive terms of a geometric progression. Sol: Given three consecutive terms of GP x, x+2, x+b 𝑎1 𝑎2 , 𝑎3 In GP we know

r= 𝑎2

𝑎1=

𝑎3

𝑎2

𝑥+2

𝑥=

𝑥+5

𝑥+2

(x+2)(x+2)=x(x+5) 𝑥2+2x+2x+4=𝑥2+5x 𝑥2+4x-𝑥2-5x=4 -x=4 x=-5.

EXERCISE-6.5 1. For each geometric progression find the common ratio ‘r’, and then find 𝑎𝑛

(i)3, 3

2,

3

4,

3

8 .........

Sol: Given series of GP

3, 3

2,

3

4,

3

8 .........

𝑎1 𝑎2 𝑎3

Here a=3, r= 𝑎2

𝑎1=

3

2

3=

1

2

nth term of GP

𝑎𝑛 = 𝑎𝑟𝑛−1=3(1

2)𝑛−1

(ii) 2, -6, 18, -54 Sol: Given series of GP 2, -6, -18, -54......... 𝑎1 𝑎2 𝑎3

Here a=2, r= 𝑎2

𝑎1=

−6

2=-3

nth term of GP 𝑎𝑛 = 𝑎𝑟𝑛−1=2(−3)𝑛−1 (iii) -1, -3, -9, -18 .... Sol: Given series of GP -1, -3, -9, -18......... 𝑎1 𝑎2 𝑎3

Here a=-1, r= 𝑎2

𝑎1=

−3

−1=3

nth term of GP 𝑎𝑛 = 𝑎𝑟𝑛−1=-1(3)𝑛−1

(iv) 5, 2, 4

5,

8

25 .........

Sol: Given series of GP

5, 2, 4

5 ,

8

25 .........

𝑎1 𝑎2 𝑎3



Here a=5, r= 𝑎2

𝑎1=

2

5

nth term of GP

𝑎𝑛 = 𝑎𝑟𝑛−1=5(2

5)𝑛−1

2. Find the 10th and nth term of G.P. : 5, 25, 125, ..... Sol: Given series of GP 5, 25, 125, ......... 𝑎1 𝑎2 𝑎3

Here a=5, r= 𝑎2

𝑎1=

25

5=5

We have 10th term of GP 𝑎10 = 𝑎𝑟9=5(5)9=(5)10 nth term of GP 𝑎𝑛 = 𝑎𝑟𝑛−1=5(5)𝑛−1=(5)1+𝑛−1=(5)𝑛 3. Find the indicated term of each geometric Progression

(i) a1 = 9; r=1

3 ; find a7

Sol: Given a1 = 9; r=1

3

We have 7th term of GP

𝑎7 = 𝑎𝑟6=9(1

3)6=9.

1

729=

1

81

(ii) a1 = -12; r=1

3 ;find a6

Sol: Given a1 = -12; r=1

3

We have 7th term of GP

𝑎6 = 𝑎𝑟5=-12(1

3)5=-12.

1

243=

−4

81.

4. Which term of the G.P. (i) 2, 8, 32, ..... is 512 ? Sol: Given GP series 2, 8, 32,………. 𝑎1 𝑎2 𝑎3

Here a=2, r= 𝑎2

𝑎1=

8

2=4

Let nth term of GP is 512. We have 𝑎𝑛 = 𝑎𝑟𝑛−1 512=2(4)𝑛−1 256=(4)𝑛−1 44=(4)𝑛−1 4=n-1 4+1=n 5=n. So, 5th term of GP is 512.

(ii) 3, 3, 3 3 ...... is 729 ? Sol: Given GP series

3, 3, 3 3,………. 𝑎1 𝑎2 𝑎3

Here a= 3, r= 𝑎2

𝑎1=

3

3= 3

Let nth term of GP is 729. We have 𝑎𝑛 = 𝑎𝑟𝑛−1

729= 3 ( 3)𝑛−1

729=( 3)1+𝑛−1

36=( 3)𝑛

36=(31

2 )𝑛

36=3𝑛

2

6=𝑛

2

12=n. So, 12th term of GP is 729.

(iii) 1

3,

1

9,

1

27,…… is

1

2187 ?

Sol: Given GP series 1

3,

1

9,

1

27,……….

𝑎1 𝑎2 𝑎3

Here a=1

3, r=

𝑎2

𝑎1=

1

91

3

=1

3

Let nth term of GP is 1

2187.

We have 𝑎𝑛 = 𝑎𝑟𝑛−1

1

2187=

1

3 (

1

3)𝑛−1

1

2187=(

1

3)1+𝑛−1

(1

3)7=(

1

3)𝑛

7=n.

So, 7th term of GP is 1

2187.

5. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2. Sol: Given common ratio r=2. 8th term of GP 𝑎8 = 𝑎𝑟7=192 (∴r=2) 𝑎(2)7=192 a(128)=192

a=192

128=

3

2

12th term of GP

𝑎12 = 𝑎𝑟11=3

2(2)11=3(2)10=3(1024)=3072.

6. The 4th term of a geometric progression is 2

3

and the seventh term is 16

81 . Find the geometric

series. Sol: Given 4th term of GP



𝑎4 = 𝑎𝑟3=2

3 …… 1

7th term of GP

𝑎7 = 𝑎𝑟6=16

81 …… 2

From (1) and (2)

𝑎7

𝑎4=

𝑎𝑟6

𝑎𝑟3=

168123

𝑟3 =16

81×

3

2

𝑟3 =8

27

𝑟3=(2

3)3

r=2

3

r=2

3 substituting in (1)

𝑎𝑟3=2

3

𝑎(2

3)3=

2

3

a(8

27)=

2

3

a=2

3×

27

8

a=9

4.

𝑎1=a=9

4

𝑎2 = 𝑎𝑟1= 9

4(

2

3)1=

3

2

𝑎3 = 𝑎𝑟2= 9

4(

2

3)2=

9

4×

4

9=1

Required GP series is 9

4,

3

2, 1

7. If the geometric progressions 162, 54, 18 .....

and 2

81,

2

27,

2

9 have their nth term equal. Find the

value of n. Sol: Given series 162, 54, 18 ..... 𝑎1 𝑎2 𝑎3

Here a=162, r=54

162=

1

3

nth term of GP

𝑎𝑛 = 𝑎𝑟𝑛−1=162(1

3)𝑛−1.

For 2

81,

2

27,

2

9 ,……

𝑎1 𝑎2 𝑎3

Here a=2

81 , r=

2

272

81

=3

nth term of GP

𝑎𝑛 = 𝑎𝑟𝑛−1=2

81 (3)𝑛−1

Given both nth terms are equal

162(1

3)𝑛−1=

2

81 (3)𝑛−1

81×81(1

3)𝑛−1=(3)𝑛−1

34×34 (3−1)𝑛−1=(3)𝑛−1 38×3−𝑛+1=(3)𝑛−1 38−𝑛+1=(3)𝑛−1 39−𝑛=(3)𝑛−1 9-n=n-1 -n-n=-1-9 -2n=-10 n=5. The 5th term og two GP are equal.

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