Selina Concise Class 10 Math Chapter 11 Exercise 11B Geometric Progression Solutions

Selina Concise Class 10 Math Chapter 11 Exercise 11B Geometric Progression Solutions

 

Geometric Progression

Exercise: 11B

 

Question: 1

Which term of the geometric Progression:

-10, 5/√3, -5/6, ——, is -5/72  ?

= Solution:- Given geometric progression is -10, 5/√3, -5/6, ——

Here, a1 = -10, a2 = 5/√3, a3 = -5/6, ——-

Now, We have to find common ratio,

a2/a1 = 5√3/-10 = 5/√3 x 1/-10 = 1/-2√3

a3/a2 = (-5/6)/(5/√3) = -5/6 x √3/5 = -√3/6

= -√3/2 x 3

= -√3/ 2 x √3 x √3

= -1/ 2√3

Therefore, a2/a1 = a3/a2 = —— = -1/2√3

Now, We have to find ‘n’ –

We know that,

tn = arn-1

Given that,

tn = -5/72

-5/72 = (-10) (-12√3)n-1

-5/72 x 10 = (-1)/(2√3)n-1

5/720 = (-1/2√3) n-1

1/144 = (-1/2√3) n-1

1/12 x 12 = (-1/2√3)n-1

1/3 x 4 x 3 x 4 = (-1/2√3)n-1

1/4 x 4 x 3 x 3 = (-1/2√3)n-1

1/2 x 2 x 2 x 2 x 3= (-1/2√3)n-1

1/2 x 2 x 2 x 2 x √3 x √3 x √3 x √3 = (-1/2√3)n+1

(-1/2√3)4 = (-1/2√3)n-1

Base is same so the exponents are equal to each other

4 = n – 1

4+1 = n

N = 5

Therefore, The 5th term of the given Geometric Progression is -5/72

 

Question 2:

The fifth term of a geometric progression is 81 and its second term is 24. Find the geometric progression.

=. Solution:-

Given that

The fifth term of a geometric progression is 81.

T5 = 81

And second term is 24

t2 = 24

We know that,

tn = arn-1

t5 = ar5-1

81 = ar4 —- (1)

Also, tn = arn-1

t2 = ar2-1

24 = ar —- (2)

Divide equation (1) by (2)

We get

81/24 = ar4/ar

81/24 = r4/r

81/24 = r x r x r x r/r

81/24 = r x r x r

81/24 = r3

27/8 = r3

Taking cube root on both side

3√27/8 = 3√r3

3/2 = r

Put r = 3/2 in equation

We get

81 = ar4

81 = a (3/2)4

81 = a x 3/2 x 3/2 x 3/2 x 3/2

2 x 81/3 = a x 27/8

54 = a x 27/8

54 x 8/27 = a

2 x 8 = a

A = 16

Therefore, The geometric progression is a, ar, ar2, ar3——

a = 16

ar = 16 x 3/2 = 8 x 3 = 24

ar2 = 16 x 3/2 x 3/2

= 8 x 3 x 3/2

= 4 x 3 x 3

= 36

Ar3 = 16 x 3/2 x 3/2 x 3/2

= 8 x 3 x 3/2 x 3/2

= 4 x 3 x 3 x 3/2

= 2 x 3 x 3 x 3

Ar3 = 54

Therefore, The geometric progression is 16, 24, 36, 54, ——–

 

Question 9:

Fourth and seventh terms of a geometric progression are 1/18 and -1/486 respectively

Find the geometric progression.

= Solution:

Given that, The fourth term of geometric progression

t4 = 1/18

And the seventh term of geometric progression

t7 = -1/486

We know that,

tn = ar n-1

tn = arn-1

t4 = ar4-1

t4 = ar3

1/18 = ar3 —- (i)

 

And,

t7 = ar7-1

-1/486 = ar6 —- (ii)

 

Divided equation (2) by (1)

-1/486/1/18 = ar6/ar3

-1/486 x 18/1 = r6/r3

-1/27 = r3

Taking cube root on both side

√-1/27 = √r3

-1/3 = r

Therefore, r = -1/3

Put r = -1/3 in equation (i)

We get,

1/18 = a(-1/3)3

1/18 = a x (-1/3) x (-1/3) x (-1/3)

1/18 x 3/-1 = a x (-1/3) x (-1/3)

1/18 x 3/-1 = a x (-1/3) x (-1/3)

1/-6 = a x (-1/3) x (-1/3)

3/(-6) x (-1) = a x (-1/3)

1/2 = a x (-1/3)

3/2 x (-1) = a

-3/2 = a

a = -3/2

Therefore, The geometric progression is a, ar, ar2, ar3, —-

a= -3/2

ar = (-3/2) x (-1/3)

ar = 1/2

ar2 = (-3/2) x (-1/3) x (-1/3)

= 1/2 x (-1/3)

ar2 = -1/6

Therefore, The geometric progression is -3/2, 1/2, -1/6—–.

 

Question 4:

If the first and the third terms of a geometric progression are 2 and 8 respectively, find its second term.

= Solution:-

Given that, the first term of a geometric progression is

t1 -= 2

And the third of a geometric progression is 8

t3 = 8

We know that the general form of geometric progression.

tn = arn-1

t1 = ar1-1

t1 = a

2 = a —– (1)

And tn = arn-1

t3 = ar3-1

t3 = ar2

8 = ar2 —– (2)

Dividing equation (2) by (1) ,

we get ar2/a = 8/2

r2 = 4

Taking square root on both sides

√r2 = √ 4

r = + – 2

Now, We have to find 2nd term of the geometric progression

t2 = ar2-1

= ar

= 2(2) (Because, a = 2, r = 2)

Or, if a = 2 and r = -2 then

t2 = ar

= 2 x (-2)

t2 = -4

 

Question: 5

The product of 3rd and 8th terms of a G.P is 243. If it’s 4th term is 3, find it’d 7th term

= Solution:

The product of 3rd and 8th terms of a geometric progression is 243

We know that,

tn = arn-1

t3 = ar3-1

t3 = ar2 —- (1)

and,

t8 = ar8-1

t8 = ar7 —- (2)

Multiplying equation (i) and (2)

t3 x t8 = ar2 x ar7

Therefore, ar2 x ar7 = 243 (Because, Given)

a2 r9 = 243

a x a x r3 x r6 = 243

ar3 x ar6 = 243 — (A)

Also given that,

T4 = ar4-1

3 = ar3

Put ar3 = 3 in equation (A)

We get,

(3) ar6 = 243

ar6 = 243/3

ar6 = 81 —–(B)

Now, We have to find 7th term,

t7 = ar7-1

t7 = ar6

From equation (B)

t7 = 87 (Because, ar6 = 81)

Therefore, The 7th term of the geometric progression is 81.

 

 

Here is your solution of Selina Concise Class 10 Math Chapter 11 Exercise 11B Geometric Progression

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