## Rajasthan Board RBSE Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.3

Question 1.

(i) Find the 7^{th} term of series 1 + 3 + 9 + 27 + …

(ii) Find the 10th term of

Solution:

(i) Given series = 1 + 3 + 9 + 27 + ….

Hedre, a = 1, r = 3/1 = 3

7^{th} term = T_{1} (From T_{n} – ar^{n-1})

= 1 × (3)^{7-1}

= (3)^{6} = 729

Question 2.

(i) Which term of series 64 + 32 + 16 + 8 + … is 1/64 ?

(ii) Which term of series 6 + 3 + 3/2 + 3/4 + … is 3/256 ?

Solution:

(i) Given series =64 + 32 + 16 + 8 +….

Question 3.

Find the common ratio and 12^{th} term of G.P. 5 + 10 + 20 + 40 + …

Solution:

Given

series = 5 + 10 + 20 + 40 + ….

n^{th} term = T_{n} = ar^{n – 1
}= 5 × (2)^{n – 1} = 5.2^{n – 1}

Question 4.

Find the fifth term from the last term of G.P. 2, 6, 18, 54, … 118098.

Solution:

Given

series = 2, 6, 18, 54,…..,118098

Common ratio r = 6/2 = 3

Last term = 118098

⇒ ar^{n – 1} = 118098

⇒ 2 × (3)^{n – 1} = 118098

⇒ (3)^{n – 1} = 59049 = (3)^{10
}On comparing, n – 1 = 10

⇒ n = 11

5^{th} term from last =ar^{n – 5
}= 2 × (3)^{11 – 5 }=2 × (3)^{6
}= 2 × 72 = 1458

Question 5.

Third term of a G.P. is 32 and 7th term is 8192, then find 10th term of GP.

Solution:

Given, T_{3}= ar^{2} = 32

T_{7} = ar^{6} = 8192

Divide equation (ii) by (i), we get

⇒ r^{4 }= 256

⇒ r^{4 }= 4^{4
}On comparing r = 4

Then, from equation (i)

ar^{2} = 32

⇒ a × 4^{2} = 32

⇒ a = 2

10th term = T_{10 }= ar^{9
}= 2 × (4)^{9} = 524288

Question 6.

Find G.P., whose 3^{rd} term is 1 and 7^{th} term is 16.

Solution:

Let a is first term of G.P. and r is common ratio, then

Third term T_{3 }= 1

⇒ ar^{2} = 1

Seventh term T_{7} = 16

ar^{6} = 16

Divide equation (ii) from (i), we get

⇒ r^{4 }= 16 = 2^{4
}Hence, r = 2

ar^{2 }= 1

a × (2)^{2} = 1

Question 7.

(i) Find 3 G.M. between 3 and 48.

(ii) Find 6 G.M. between 2 and 256.

Solution:

(i) Let G_{1}, G_{2}, G_{3} be three GM. between 3 and 48, then

3, G_{1}, G_{2}, G_{3}, 48 will be in G.P.

Here a = 3,

Last term (5th term) = 48 and n = 5

3.r^{4} = 48 [∵ l = ar^{n – 1}]

⇒ r^{4} = 16 = 2^{4
}⇒ r = 2

G_{1} = 3 × 2 = 6

G_{2} = 3 × 2^{2} = 12

G_{3} = 3 × 2^{3} = 24

Hence, three G.M. are 6, 12, 24.

(ii) Let G_{1}, G_{2}, G_{3}, G_{4}, G_{5}, G_{6} be six G.M. between 2 and 256, then

2, G_{1}, G_{2}, G_{3}, G_{4}, G_{5}, G_{6}, 256 will be in G.P.

Here a = 6, last term (8th term) = 256 and n = 8

Hence, 2.r^{7} = 256 [∵ l = ar^{n – 1}]

⇒ r^{7}= 128 = 2^{7
}⇒ r = 2

G_{1} = 2 × 2 = 4

G_{2} = 4 × 2 = 8

G_{3} = 8 × 2 = 16

G_{4} = 16 × 2 = 32

G_{5} = 32 × 2 = 64

G_{6} = 64 × 2 = 128

Hence six G.M. are 4, 8, 16, 32, 64, 128.

Question 8.

For which value of x, numbers x, x + 3, x + 9 are in G.P. ?

Solution:

x, x + 3, x + 9 are in G.P.

Then, (x + 3)^{2} = x × (x + 9)

⇒ x^{2} + 9 + 6x = x^{2} + 9x

⇒ x^{2} + 6x + 9 – x^{2} – 9x = 0

⇒ – 3x + 9 = 0

⇒ \(\frac { -9 }{ -3 }\) = 3

Hence, x = 3

Question 9.

Find four terms which are in G.P., whose third term is 4 more than first term and second term is 36 more than 4^{th} term.

Solution:

Let first four terms of G.P. are a, ar, ar^{2}. ar^{3} where a is first term and r is common ratio.

According to question,

Third term = First term + 4

⇒ ar^{2} = a + 4

⇒ a(r^{2} – 1) = 4

According to question,

Second term = Fourth term + 36

⇒ ar = ar^{3} + 36

⇒ ar – ar^{3} = 36

⇒ ar(1 – r^{2}) = 36

Divide equation (ii) by (i), we get

⇒ – r = 9

⇒ r = – 9

Hence, common ratio of G.P.,

r = – 9

Putting r = – 9 in equation (i),

a[(-9)^{2} – 1] = 4

⇒ a (81 – 1) = 4

Question 10.

If 4^{th} term of any G.P. is p, 7^{th} term is q and 10^{th} term is r, then prove that q^{2} – pr.

Solution:

Let a is first term and R is common ratio of G.P., then according to question

T_{4} = aR^{4 – 1}

p = aR^{3 }….. (i)

T_{7} = aR^{7 – 1
}q = aR^{6 }….. (ii)

T_{10} = aR^{10 – 1}

r = aR^{9 }^{ }….. (iii)

Multiplying equation (i) and (iii), we get

a^{2 }R^{12} = pr ^{ }….. (iv)

Squaring of equation (ii),

q^{2} = a^{2 }R^{12 }Hence, from equation (iv) and (v)

q^{2} = pr ….. (v)

Hence Proved.

Question 11.

If (p + q)^{th} term in GP. is x and (p – q)^{th }term is y, then find p^{th} term.

Solution:

Let a is first term and r is common ratio of G.P.

So, (p + q)^{th} term = ar^{p} + q – 1 = x … (i)

and (p – q)^{th} term = ar^{p} – q – 1 = y … (ii)

Multiplying equation (i) and (ii), we get

a^{2 }(r^{p} + q – 1 + p – q) = x × y

or a^{2 }r^{2p-2 }= xy

ora^{2 }r^{2(p – 1) }= xy

or (ar^{p – 1})^{2} = xy

or ar^{p – }^{1} = √xy

Hence, pth term = ar^{p – 1
}= √xy

Question 12.

If a, b, c are in G.P. and a^{x} = b^{y} = c^{z} , then prove that

Solution:

Let a^{x}= b^{y} = C^{z }= k

and a, b, c are in G.P.

b^{2} = ac

Putting the value of a, b, c in equation (i),

Question 13.

If n G.M. inserted between a and b then prove that product of all geometric means will be

\({ \left( \sqrt { xy } \right) }^{ n }\)

Solution:

Let G_{1}, G_{2}, G_{3}, G_{4}, ……G_{n }are in G.M. in between x and y then

Question 14.

If x, y, z are in G.P., arithmetic mean of x, y is A_{1} and A.M. of y, z is A_{2} then prove that :

Solution:

x, y, z are in G.P.

∴ y^{2} = xz …… (i)

Arithmetic mean of x and y is A_{1
}

Arithmetic mean of y and z is A_{2
}

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