**Find the 20**^{th}and n^{th}terms of the G.P.^{5}/_{2},^{5}/_{4},^{5}/_{8}, …

Solution:

The given G.P. is ^{5}/_{2}, ^{5}/_{4}, ^{5}/_{8}, …

Here a = first term = ^{5}/_{2}

r = common ratio = (^{4}/_{5})/(^{5}/_{2}) = ^{1}/_{2}

a_{20} = ar^{20-1} = ^{5}/_{2}(^{1}/_{2})^{n-1} = ^{5}/_{(2)(2)}^{n-1} = ^{5}/_{2}^{n}

a_{n} = ar^{n-1} = ^{5}/_{2}(^{1}/_{2})^{n-1} = ^{5}/_{(2)(2)}^{n-1} = ^{5}/_{2}^{n}

**Find the 12**^{th}term of G.P. whose 8th term is 192 and the common ratio is 2.

Solution:

Common ratio, r = 2

Let a be the first term of the G.P.

∴ a_{8} = ar^{8–1} = ar^{7}

⇒ ar^{7} = 192 a(2)^{7} = 192 a(2)^{7} = (2)^{6}(3)

⇒a = [(2)^{6}x3]/(2)^{7} = ^{3}/_{2}

a_{12} = ar^{12-1} = ^{3}/_{2}(2)^{11} = (3)(2) = 3072

**The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q**^{2}= ps.

Solution:

Let a be the first term and r be the common ratio of the G.P. According to the given condition,

a_{5} = ar^{5–1} = ar^{4} = p —————–(1)

a_{8} = a r^{8–1} = a r^{7} = q —————-(2)

a_{11} = a r^{11–1} = a r^{10} = s ————– (3)

Dividing equation (2) by (1), we obtain

[ar^{7}]/[ar^{4}] = ^{q}/_{p}

r^{3} = ^{q}/_{p} —————(4)

Dividing equation (3) by (2), we obtain

[ar^{10}]/[ar^{7}] = ^{s}/_{q}

r^{3} = ^{s}/_{q} —————-(5)

Equating the values of r^{3} obtained in (4) and (5), we obtain

^{q}/_{p} = ^{s}/_{q}

q^{2} = ps

Thus the given result is proved.

**The 4th term of a G.P. is square of its second term, and the first term is –3. Determine its 7th term.**

Solution:

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

∴ a = –3

It is known that, an = ar^{n–1}

∴ a_{4} = ar^{3} = (–3) r^{3}

a_{2} = a r^{1} = (–3) r

According to the given condition,

(–3) r^{3} = [(–3) r]^{2}

⇒ –3r^{3} = 9 r^{2}

⇒ r = –3 a_{7} = a r^{7–1} = a

r^{6} = (–3) (–3)^{6} = – (3)^{7} = –2187

Thus, the seventh term of the G.P. is –2187.

**Which term of the following sequences:**

**(a)2, 2√2, 4, …is 128?**

**(b) √3, 3, 3√3,…is 729?**

**(c) ^{1}/_{3},^{ 1}/_{9}, ^{1}/_{27},…^{1}/_{19683} ?**

Solution:

(a) The given sequence 2, 2√2, 4, …is 128?

Here a = 2 and r = (2√2)/2 = √2

Let nth term of the given sequence be 128.

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

(2)( √2)^{n-1} = 128

(2)(2)^{(n-1)/2} = 2^{7}

(2)^{[(n-1)/2]+1} = 2^{7}

^{n-1}/_{2} + 1 = 7

^{n-1}/_{2} = 6

n – 1 = 12

n = 13

Thus the 13^{th} term of the given sequence is 128.

(b) the given sequence is √3, 3, 3 √3

a = √3 and r = ^{3}/_{√3} = √3

Let the nth term of the given sequence be 729

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

ar^{n-1} = 729

(√3)( √3)^{n-1} = 729

(3)^{1/2}( 3)^{(n-1)/2} =3^{6}

^{1+n-1}/_{2} = 6

n = 12

Thus the 12^{th} term of the given sequence 729.

(c) The given sequence is ^{1}/_{3}, ^{1}/_{9},^{ 1}/_{27},…

Here, a = ^{1}/_{3} and r = ^{1}/_{9} + ^{1}/_{3} = ^{1}/_{3}

Let the nth term of the given sequence be ^{1}/_{19683}

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

(^{1}/_{3}).(^{ 1}/_{3})^{n-1} = ^{1}/_{19683}

(^{1}/_{3})^{n} = (^{1}/_{3})^{9}

n = 9

Thus, the 9th term of the given sequence is ^{1}/_{19683}

**For what values of x, the numbers**^{2}/_{7}, x, –^{7}/_{2}are in G.P.?

Solution:

The given numbers are –^{2}/_{7}, x, –^{7}/_{2}

Common ratio =

^{x}/_{(-2/7)}=^{-7x}/_{2}

Also, common ratio = (^{-7}/_{2})/_{x} = –^{7}/_{2x}

^{-7}/_{2x} = –^{7}/_{2x}

x^{2} = ^{-2×7}/_{-2×7} = 1

x = √1

x = ±1

Thus for x = ±1 the given numbers will be in G.P.

**Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015**

Solution:

The given G.P. is 0.15, 0.015, 0.0015

Here a = 0.15 and r = ^{0.015}/_{0.15} = 0.1

**Find the sum to n terms in the geometric progression √7 , √21, 3√7, …**

Solution:

The given G.P. is √7 , √21, 3√7, …

Here, a = √7 and r =^{√21}/_{7} = √3

**Find the sum to n terms in the geometric progression 1, -a, a**^{2}, -a^{3}…(if a≠-1)

Solution:

The given G.P. is 1, -a, a^{2}, -a^{3}…

Here first term = a_{1} = 1

Common ratio = r = – a

**Find the sum to n terms in the geometric progression x**^{3}, x^{5}, x^{7}…(if x ≠±1)

Solution:

The given G.P. is x^{3}, x^{5}, x^{7}…

Here a = x^{3} and r = x^{2}

Sequences and Series – Exercise 9.3[11-20] – Class XI

Sequences and Series – Exercise 9.3[21-32] – Class XI