| 1. If the third term of a G. P. is 6, then the product of the first five terms of this series is: | Easy |
| A. 63 B. 65 C. 64 D. None of these |
View Answer
Answer: Option B
Explanation:
Given, ar2 = 6
∴Required product = a.ar.ar2.ar3.ar4
= a5r10 = (ar2)5 = 65
| 2. If x, 2x+2, 3x+3, …… are in G. P., what is the fifth term of this series? | Easy |
| A. -20 B. -20.25 C. -13.5 D. None of these |
View Answer
Answer: Option B
Explanation:
x, 2x+2, 3x+3 are in G.P
-> (2x+2)2 = x(3x+3)
-> 4(x+1)2 = 3x(x+1)
-> x2+5x+4 = 0
-> (x+4)(x+1) = 0
∵ x = -1 does not give a G.P, à x = -4
So, the G.P. is -4, -6, -9…..
∴ r = (-6)/(-4) = 3/2
∴ T5 = (-4) (3/2)4 = -20.25
| 3. The value of 251/3.251/9.251/27…. ∞ is: | Easy |
| A. 4 B. 9 C. 5 D. None of these |
View Answer
Answer: Option C
Explanation:
Here required value is 25x, where x = 1/3+1/9+1/27+ …… to ∞, which
is the sum of infinite terms of a G.P. with r<1 = a/(1-r)
-> x = (1/3)/(1-(1/3) ) =1/2
∴ Required value = 251/2 = 5
| 4. The first three terms of a geometric progression are √5, ∛5 and 6√5. Which term of the series is equal to the common ratio? | Easy |
| A. 6th term B. 5th term C. 4th term D. None of these |
View Answer
Answer: Option B
Explanation:
Write the terms as 51/2, 51/3, 51/6 or 53/6, 52/6, 51/6
Common ratio of this G.P. = 51/3/51/2 = 5‑1/6
Fourth term = Third term * Common ratio =51/6 * 5-1/6 = 50 = 1
Fifth term = Fourth term * Common ratio = 1* 5-1/6 = 5-1/6
Hence, the fifth term is equal to the common ratio.
Alternatively
Once you find the common ratio, check the various options.
Note: If nth term of a G.P. equals the common ratio, then (n-1)th term must be 1, and vice versa.
| 5. When a ball is dropped from a height of h feet and bounces on a horizontal surface, it rebounds to a height of 7h/10 feet. If the ball falls from a height of 21 feet, calculate the total distance it will travel if it continues striking the surface indefinitely. | Difficult |
| A. 119 feet B. 238 feet C. 357 feet D. None of these |
View Answer
Answer: Option A
Explanation:
The ball travels a distance h feet while falling down, then goes up (7/10)h, then comes down (7/10)h, and so on.
Total distance travelled by the ball
= h+2[7h/10+(7/10)2h+(7/10)3h+…..∞]
=h+2*7h/10[1+ 7/10 + (7/10)2 + ….. ∞]
Now, 1+ 7/10 + (7/10)2 + ….. ∞ = 1/(1-(7/10)) = 10/3 (sum of infinite terms of a G.P.)
-> h+2*7h/10[1+ 7/10 + (7/10)2 + ….. ∞] = h+(14h/10)(10/3)
= 17h/3 =(17*21)/3 = 119
| 6. The second term of a G.P. is 100 and the common ratio is 1/k, where k is a positive integer. The product of the first five terms of this series is more than the product of the first four terms; however, it is less than the product of the first six terms. What is the value of k? | Difficult |
| A. 3 B. 4 C. 5 D. None of these |
View Answer
Answer: Option B
Explanation:
The common ratio is given to be positive (r = 1/k, and k is a positive integer), and one of the terms (the second term = 100) is positive, meaning that all terms of the series are positive.
Let Pk be the product of the first k terms of this GP, and Tk be the kth term.
Given, P5 > P4 -> P4 * T5 > P3 * T4 (because Pk = Pk-1 * Tk) -> T5>1
Now, T5 = ar4 = t2 *r3 = 100r3 (where a is the first term, and r is the common ratio of the G.P.)
-> 100r3 > 1
-> k3 < 100 (because r = 1/k) —–❶
Also, P6 < P5 -> P5 * T6 < P4 * T5 -> T6 < 1
Now, T6 = ar5 = t2 *r4 = 100r4
-> 100r4 < 1
-> k4 > 100 —–❷
From ❶ k < 5, and from ❷ k ≥ 4, and it is given that k is a positive integer -> k = 4
| 7. Two numbers, a and b, are in the sequence 16, a, b, 6. The first three numbers form a geometric sequence and the last three numbers form an arithmetic sequence. What is the value of a2-b2? | Medium |
| A. 12 or 63 B. 15 or 56 C. 15 or 63 D. None of these |
View Answer
Answer: Option C
Explanation:
Given, 16, a, b are in G.P.
-> (a/16) = (b/a)
-> a2 = 16b
-> b = a2/16 —–❶
Also given, a, b, 6 are in A.P.
-> (6-b) = (b-a)
-> 2b = a+6 —–❷
-> 2(a2/16) – a = 6 (putting value of b from ❶ in ❷)—–❸
Solving ❸, a = -4 or 12, and b = 1 or 9 respectively (from ❷)
Therefore, a2-b2 = 15 or 63
| 8. If 0.187 187 187 … can be written as 187/x, then what is x? | Medium |
| A. 900 B. 1000 C. 999 D. None of these |
View Answer
Answer: Option C
Explanation:
0.187187187187…….
= 0.187 + 0.000187 + 0.000000187 + ….
= 187/1000+187/1000000+187/1000000000+⋯to infinity
= 187/103 [ 1+1/103) +1/106 +⋯…. to infinity]
= 187/1000 [1/{1-(1/10)3} ] =187/999
| 9. If a, b, c are three numbers, then the value of (a+b)(b+c)(c+a) is: | Medium |
| A. =8abc B. >8abc C. <8abc D. Can’t be determined |
View Answer
Answer: Option B
Explanation:
Consider two numbers a and b
∵A.M > G.M -> ½(a+b) > √(ab)
-> a+b > 2√(ab) —– ❶
Similarly, b+c > 2√(bc) —– ❷
and c+a > 2√(ca) —– ❸
Multiplying ❶, ❷ and ❸, we get (a+b)(b+c)(c+a) > 8abc
| 10. If a, b, c, d and e are five positive real numbers such that abcde = 1, what is the minimum value of (1+a)(1+b)(1+c)(1+d)(1+e)? | Medium |
| A. 64 B. 32 C. 16 D. None of these |
View Answer
Answer: Option B
Explanation:
A.M. ≥ G.M. (Arithmetic Mean is greater than or equal to Geometric Mean)
-> (1+a)/2 ≥ √a
-> 1+a ≥ 2√a
Similarly, 1+b ≥ 2√b, 1+c ≥ 2√c, 1+d ≥ 2√d, 1+e ≥ 2√e
-> (1+a)(1+b)(1+c)(1+d)(1+e) ≥ 2√a*2√b*2√c*2√d*2√e
-> (1+a)(1+b)(1+c)(1+d)(1+e) ≥ 32√(abcde) ≥ 32
Therefore, the minimum value is 32.