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

**Geometric Progression**

__Exercise: 11B__

__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, a_{1} = -10, a_{2} = 5/√3, a_{3} = -5/6, ——-

Now, We have to find common ratio,

a_{2}/a_{1} = 5√3/-10 = 5/√3 x 1/-10 = 1/-2√3

a_{3}/a_{2 }= (-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, a_{2}/a_{1} = a_{3}/a_{2} = —— = -1/2√3

Now, We have to find ‘n’ –

We know that,

**t _{n} = ar^{n-1}**

Given that,

t_{n} = -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 5^{th} 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.

T_{5} = 81

And second term is 24

t_{2} = 24

We know that,

**t _{n} = ar^{n-1}**

t_{5} = a^{r5-1}

81 = ar^{4} —- (1)

Also, t_{n} = ar^{n-1}

t_{2} = ar^{2}-1

24 = ar —- (2)

Divide equation (1) by (2)

We get

81/24 = ar^{4}/ar

81/24 = r^{4}/r

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

81/24 = r x r x r

81/24 = r^{3}

27/8 = r^{3}

Taking cube root on both side

3√27/8 = 3√r^{3}

3/2 = r

Put r = 3/2 in equation

We get

81 = ar^{4}

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, ar^{2}, ar^{3}——

a = 16

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

ar^{2} = 16 x 3/2 x 3/2

= 8 x 3 x 3/2

= 4 x 3 x 3

= 36

Ar^{3} = 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

Ar^{3} = 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

t_{4} = 1/18

And the seventh term of geometric progression

t_{7} = -1/486

We know that,

**t _{n} = ar ^{n-1}**

t_{n} = ar^{n-1}

t_{4} = ar^{4-1}

t_{4 }= ar^{3}

1/18 = ar^{3} —- (i)

And,

t_{7} = ar^{7-1}

-1/486 = ar^{6} —- (ii)

Divided equation (2) by (1)

-1/486/1/18 = ar^{6}/ar^{3}

-1/486 x 18/1 = r^{6}/r^{3}

-1/27 = r^{3}

Taking cube root on both side

√-1/27 = √r^{3}

-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, ar^{2}, ar^{3}, —-

a= -3/2

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

ar = 1/2

ar^{2} = (-3/2) x (-1/3) x (-1/3)

= 1/2 x (-1/3)

ar^{2} = -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

t_{1} -= 2

And the third of a geometric progression is 8

t^{3} = 8

We know that the general form of geometric progression.

**t _{n} = ar^{n-1}**

t_{1} = ar^{1-1}

t_{1} = a

2 = a —– (1)

And t_{n} = ar^{n-1}

t_{3} = ar^{3-1}

t_{3} = ar^{2}

8 = ar^{2} —– (2)

Dividing equation (2) by (1) ,

we get ar^{2}/a = 8/2

r^{2} = 4

Taking square root on both sides

√r^{2} = √ 4

r = + – 2

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

t_{2} = ar^{2-1}

= ar

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

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

t_{2} = ar

= 2 x (-2)

t_{2} = -4

** **

**Question: 5**

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

= Solution:

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

We know that,

**t _{n }= ar^{n-1}**

t_{3} = ar^{3-1}

t_{3} = ar^{2} —- (1)

and,

t8 = ar^{8-1}

t_{8} = ar^{7} —- (2)

Multiplying equation (i) and (2)

t_{3} x t_{8 }= ar^{2} x ar^{7}

Therefore, ar^{2} x ar^{7} = 243 (Because, Given)

a^{2} r^{9} = 243

a x a x r^{3} x r^{6} = 243

ar^{3} x ar^{6} = 243 — (A)

Also given that,

T_{4} = ar^{4-1}

3 = ar^{3}

Put ar^{3} = 3 in equation (A)

We get,

(3) ar^{6} = 243

ar^{6} = 243/3

ar^{6} = 81 —–(B)

Now, We have to find 7^{th} term,

t_{7} = ar^{7-1}

t_{7} = ar^{6}

From equation (B)

t_{7} = 87 (Because, ar^{6} = 81)

Therefore, The 7^{th} 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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