| 1. The sum of 3rd and 15th elements of an A.P. is equal to the sum of the 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero? | Easy |
| A. 11th B. 12th C. 10th D. Can’t be determined |
View Answer
Answer: Option B
Explanation:
Let the A.P. be a, a+d, a+2d, ……
Now, Tn = a+(n-1)d
-> T3+T15 = T6+T11+T13
-> a+2d+a+14d = a+5d+a+10d+a+12d
-> a+11d = 0
-> T12 = 0
| 2. If a, b, c and d are four positive real numbers such that abcd = 1, what is the minimum value of (1+a)(1+b)(1+c)(1+d)? | Medium |
| A. 12 B. 14 C. 16 D. None of these |
View Answer
Answer: Option C
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+a)(1+b)(1+c)(1+d) ≥ 2√a*2√b*2√c*2√d
-> (1+a)(1+b)(1+c)(1+d) ≥ 16√abcd
Therefore, minimum value is 16
| 3. What is the value of the following expression?(1/22-1) + (1/42-1) + (1/62-1) + ………. (1/202-1) | Difficult |
| A. 9/21 B. 10/21 C. 11/21 D. None of these |
View Answer
Answer: Option B
Explanation:
(1/22-1) + (1/42-1) + (1/62-1) + ………. (1/202-1) =
1/((2-1)(2+1)) + 1/((4-1)(4+1)) + 1/((6-1)(6+1)) + ………. 1/((20-1)(20+1))
= 1/(1*3) + 1/(3*5) + 1/(5*7) + ……… 1/(19*21)
Applying 1/ab = 1/(b-a) (1/a – 1/b), the given expression can be written as:
1/((3-1))(1/1 – 1/3) + 1/((5-3))(1/3 – 1/5) + 1/((7-5))(1/5 – 1/7) + ……… 1/((21-19))(1/19 – 1/21)
= 1/((2) )(1/1 – 1/3) + 1/((2) )(1/3 – 1/5) + 1/((2) )(1/5 – 1/7) + ……… 1/((2) )(1/19 – 1/21)=1/2(1 – 1/21) = 10/21
| 4. If the sum of the first 11 terms of an A.P. equals that of the first 19 terms, then what is the sum of the first 30 terms? | Easy |
| A. 124 B. 232 C. 164 D. None of these |
View Answer
Answer: Option D
Explanation:
From the given information,
11/2(2a+10d) = 19/2(2a+18d)
-> 22a +110d = 38a+342d
-> 16a+232d = 0
-> 2a+29d = 0
Now, ∑30 = 30/2(2a+29d) = 15*0 = 0
| 5. Consider the sequence of numbers a1, a2, a3……. to infinity where a1 = 81.33, a2 = -19 and aj= aj-1 – aj-2 for j≥3. What is the sum of the first 3602 terms of this sequence? | Difficult |
| A. 81.33 B. 19 C. 62.33 D. None of these |
View Answer
Answer: Option C
Explanation:
aj= aj-1 – aj-2 for j≥3
Thus,
a3 = a2 – a1
a4 = a3 – a2 = a2 – a1 – a2 = -a1
a5 = a4 – a3 = -a1 – (a2 – a1) = -a2
a6 = a5 – a4 = -a2 – (-a1) = a1 – a2
a7 = a6 – a5 = a1 – a2 + a2 = a1
Sum of first 6 terms = a1+a2+(a2-a1)-a1-a2+a1-a2 = 0
à Sum of first 3600 terms = 0 (because the sum of every 6 terms is 0)
à Sum of first 3602 terms = a1+a2 = 81.33 – 19 = 62.33
| 6. The first three terms of a geometric progression are √5, 3√5, and 6√5 What is the fourth term? | Easy |
| A. 1 B. 0 C. 6 D. None of these |
View Answer
Answer: Option A
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
| 7. When a ball is dropped from a height of h feet on to a horizontal surface it bounces back to a height 9h/10 foot. If the ball is allowed to fall from a height of 16 feet, find the total distance that it will travel if it continues striking the surface indefinitely. | Difficult |
| A. 76 feet B. 304 feet C. 228 feet D. None of these |
View Answer
Answer: Option B
Explanation:
The ball travels a distance h feet while falling down, then goes up (9/10)h, then comes down (9/10)h, and so on.
Total distance travelled by the ball
= h+2[9h/10+(9/10)2h+(9/10)3h+…..∞]
=h+2*9h/10[1+ 9/10 + (9/10)2 + ….. ∞]
Now, 1+ 9/10 + (9/10)2 + ….. ∞ = 1/(1-(9/10)) = 10 (sum of infinite terms of a G.P.)
-> h+2*9h/10[1+ 9/10 + (9/10)2 + ….. ∞] = h+18h/10*10
= 19h = 19*16 = 304 feet
| 8. The sums of n terms of two arithmetic series are in the ratio (7n-5):(5n+17). What is the ratio of the sixth terms of the two series? | Medium |
| A. 2:3 B. 1:1 C. 3:2 D. None of these |
View Answer
Answer: Option B
Explanation:
Let the first series have a1 as the first term and d1 as the common difference, and the second series have a2 as the first term and d2 as the common difference. Then, sum of n terms of the two series will be (n/2)( (2a1+(n-1)d1) and (n/2)(2a2+(n-1)d2)
From the given information,
(2a1+(n-1)d1)/(2a2+(n-1)d2) = (7n-5)/(5n+17) —–❶
Now, ratio of 6th terms will be (a1+5d1)/(a2+5d2)
Now, (a1+5d1)/(a2+5d2) = 2a1+10d1/2a2+10d2 (Multiplying both numerator and denominator by 2) —–❷
From ❶, if n = 11, then (2a1+10d1)/(2a2+10d2) = 72/72 = 1 —–❸
From ❷ & ❸, (a1+5d1)/(a2+5d2) = 1
| 9. Two positive numbers, a and b, are in the sequence 4, a, b, 12. The first three numbers form a geometric sequence and the last three numbers form an arithmetic sequence. What is the value of (b-a)? | Medium |
| A. 3 or 8 B. 6 or -4 C. 9 or 4 D. None of these |
View Answer
Answer: Option A
Explanation:
Given, 4, a, b are in G.P.
-> (a/4) = (b/a)
-> a2 = 4b
-> b = a2/4 —–❶
Also given, a, b, 12 are in A.P.
-> (12-b) = (b-a)
-> 2b = a+12 —–❷
-> 2(a2/4) – a = 12 (putting value of b from ❶ in ❷)—–❸
Solving ❸, a = 6 or -4, and b = 9 or 4 respectively (from ❷)
Therefore, b-a = 3 or 8
| 10. What is the fourth term of a geometric progression if the second term is 14/3 and the fifth term is -112/81? | Easy |
| A. 56/27 B. 28/27 C. 56/54 D. None of these |
View Answer
Answer: Option A
Explanation:
Given, ar = 14/3 —– ❶
ar4 = – 112/81 —– ❷
Dividing ❷ by ❶, we get ar4/ar = (-112/81)/(14/3)
-> r3 = -8/27
-> r = -2/3
Now, fourth term = ar3 and fifth term = ar4
-> Fourth term = Fifth term/r = (-112/81)/((-2)/3) = 56/27