- Find the 20th and nth 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
a20 = ar20-1 = 5/2(1/2)n-1 = 5/(2)(2)n-1 = 5/2n
an = arn-1 = 5/2(1/2)n-1 = 5/(2)(2)n-1 = 5/2n
- Find the 12th 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.
∴ a8 = ar8–1 = ar7
⇒ ar7 = 192 a(2)7 = 192 a(2)7 = (2)6(3)
⇒a = [(2)6x3]/(2)7 = 3/2
a12 = ar12-1 = 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 q2 = ps.
Solution:
Let a be the first term and r be the common ratio of the G.P. According to the given condition,
a5 = ar5–1 = ar4 = p —————–(1)
a8 = a r8–1 = a r7 = q —————-(2)
a11 = a r11–1 = a r10 = s ————– (3)
Dividing equation (2) by (1), we obtain
[ar7]/[ar4] = q/p
r3 = q/p —————(4)
Dividing equation (3) by (2), we obtain
[ar10]/[ar7] = s/q
r3 = s/q —————-(5)
Equating the values of r3 obtained in (4) and (5), we obtain
q/p = s/q
q2 = 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 = arn–1
∴ a4 = ar3 = (–3) r3
a2 = a r1 = (–3) r
According to the given condition,
(–3) r3 = [(–3) r]2
⇒ –3r3 = 9 r2
⇒ r = –3 a7 = a r7–1 = a
r6 = (–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.
an = arn-1
(2)( √2)n-1 = 128
(2)(2)(n-1)/2 = 27
(2)[(n-1)/2]+1 = 27
n-1/2 + 1 = 7
n-1/2 = 6
n – 1 = 12
n = 13
Thus the 13th 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
an = arn-1
arn-1 = 729
(√3)( √3)n-1 = 729
(3)1/2( 3)(n-1)/2 =36
1+n-1/2 = 6
n = 12
Thus the 12th 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
an = arn-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
x2 = -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, a2, -a3…(if a≠-1)
Solution:
The given G.P. is 1, -a, a2, -a3…
Here first term = a1 = 1
Common ratio = r = – a
- Find the sum to n terms in the geometric progression x3, x5, x7…(if x ≠±1)
Solution:
The given G.P. is x3, x5, x7…
Here a = x3 and r = x2
Sequences and Series – Exercise 9.3[11-20] – Class XI
Sequences and Series – Exercise 9.3[21-32] – Class XI