Geometric Progressions Practice Problems: Level 01

Sol : Option D
Here a = 1/3 and r = -2/3
Hence the required sum =

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 ?

Sol : Option C
The total distance travelled by the ball =
This is an infinite G.P. with first term as 128/9 and the common difference = 3/4
Hence the required distance =

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.

Sol : Option C
Let ‘a’ be the first term and ‘r’ be the common ratio of the G.P.
We have = 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 = ⇒ r² = ⇒ r =
when r = , (i) ⇒ a = 48 x = 24 and the second term = ar = 24 x =12
When r = , 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.

Sol : Option B
The geometric mean between two terms ‘a’ and ‘b’ are given by
Hence the G.M. between 4/9 and 169/9 =

Q.8. Insert three geometric means between 2 and 81/8.

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= ⇒ 2r⁴ = ⇒ r⁴ = ⇒ 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.

Sol : Option C
Let the required numbers are ‘a’ and ‘b’.
We have = 75 ⇒ a + b = 150 ....(i)
Also = 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.

Sol : Option D
Let the first three terms of G.P. are ,a,ar
Given that x a x ar = 512 ⇒ a³ = 512 ⇒ a = 8
Now ,a+2 are in A.P.
⇒ (a+2)- = ar - (a-2)
⇒ 10 - = 8r - 10
⇒ 8r + = 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
When r = 2, the terms are 4, 8, 16
When r = 1/2, then the terms are 16, 8, 4.