Q.2. Find the sum of the G. P.: 8/10, 8/100, 8/1000, 8/10000, … to n terms.
A. [image]
B. [image]
C. [image]
D. [image]
Sol : Option A
He given G.P. is [image]
Here a = 8/10 and common ratio = 1/10
The required sum = [image]
Q.3. Find the sum of the following sequence: 12, 132, 1332, 13332 ......... to n terms.
A. [image]
B. [image]
C. [image]
D. [image]
Sol : Option D
We have S = 12 + 132 + 1332 +13332 + ……..n terms
= 12 ( 1 + 11 + 111 + 1111 + …… n terms)
= [image] (9+99+999+......n terms)
=[image] [(10-1)+(100-1)+(1000-1)+......n terms]
= [image] [(10+100+1000+......n terms)-n]
= [image]
= [image]
Q.4. Find the sum of the following infinite G. P. [image]
A. 1/2
B. 1
C. 1/3
D. 1/5
Sol : Option D
Here a = 1/3 and r = -2/3
Hence the required sum = [image]
Q.5. The distance travelled (in m) by a ball dropped from a height are 128/9, 32/3, 8, 6... How much distance will it travel before coming to rest ?
A. [image]
B. 120cm
C. [image]
D. [image]
Sol : Option C
The total distance travelled by the ball = [image]
This is an infinite G.P. with first term as 128/9 and the common difference = 3/4
Hence the required distance = [image]
Q.6. The sum of an infinite G. P. with positive terms is 48 and sum of its first two terms is 36. Find the second term.
A. 10
B. 18
C. 12
D. 20
Sol : Option C
Let ‘a’ be the first term and ‘r’ be the common ratio of the G.P.
We have [image] = 48 ⇒ a = 48(1-r)...(i)
Also it is given that a +ar = 36
⇒ a(1+r) = 36
⇒ 48(1-r) (1 + r) = 36 (from (i))
⇒ 1 - r2 = [image] ⇒ r2 = [image] ⇒ r = [image]
when r = [image] , (i) ⇒ a = 48 x [image] = 24 and the second term = ar = 24 x [image] =12
When r = [image] , the terms of the G.P. will become negative.
So the second term is 12.
Q.7. Find the G. M. between 4/9 and 169/9.
A. [image]
B. [image]
C. [image]
D. [image]
Sol : Option B
The geometric mean between two terms ‘a’ and ‘b’ are given by [image]
Hence the G.M. between 4/9 and 169/9 = [image]
Q.8. Insert three geometric means between 2 and 81/8.
A. 3, 9/2, - 27/4
B. - 3, 9/2, 27/4
C. 3, 9/2, 27/4
D. 3, 9/2, 27/8
Sol : Option C
Let x, y and z be the three geometric means between 2 and 81/8
Then 2, x, y, z, 81/8 are in G.P. If ‘r’ is the common ratio, then
ar4=[image] ⇒ 2r4 = [image] ⇒ r4 = [image] ⇒ [image] If r = 3/2, then x = 2 × 3/2 = 3, y = 2 × 9/4 = 9/2, z = 2 × 27/8 = 27/4
If r = - 3/2, then x = 2 × (-3/2) = - 3, y = 2 × 9/4 = 9/2, z = 2 × (- 27/8) = - 27/4
Hence the G.M. between 2 and 81/8 are 3, 9/2, 27/4 or -3, 9/2, -27/4
Q.9. The arithmetic mean between two numbers is 75 and their geometric mean is 21. Find the numbers.
A. 133 and 17
B. 63 and 87
C. 3 and 147
D. 73 and 77
Sol : Option C
Let the required numbers are ‘a’ and ‘b’.
We have [image] = 75 ⇒ a + b = 150 ....(i)
Also [image] = 21 or ab = 441
We know that (a-b)2 = (a+b)2 - 4ab = 1502 - 4 x 441 = 22500 -1764 = 20736
⇒ a – b = 144 ….(ii)
Adding (i) and (ii), we get 2a = 294 ⇒ a = 147
(i) ⇒ b = 3.
So the numbers are 147 and 3.
Q.10. The product of first three terms of a G. P. is 512. If we add 2 to its second term, the three terms form an A. P. Find the terms of the G. P.
A. 4, 8, 16
B. 16, 8, 4
C. 12, 24, 48
D. Option A or B
Sol : Option D
Let the first three terms of G.P. are [image] ,a,ar
Given that [image] x a x ar = 512 ⇒ a3 = 512 ⇒ a = 8
Now [image] ,a+2 are in A.P.
⇒ (a+2)- [image] = ar - (a-2)
⇒ 10 - [image] = 8r - 10
⇒ 8r + [image] = 20
⇒ 8r2 - 20r + 8 =0
⇒ 2r2 - 5r + 2 =0
⇒ 2r2 - 4r - r + 2 =0
⇒ 2r(r-2) - (r-2) =0
⇒(2r-1)(r-2) = 0
⇒ r = 2 or [image]
When r = 2, the terms are 4, 8, 16
When r = 1/2, then the terms are 16, 8, 4.