| 1. The sum of first 40 terms of an A. P whose first term is 2 and common difference is 4, will be: | Easy |
| A. 3200 B. 1600 C. 200 D. None of these |
View Answer
Answer: Option A
Explanation:
∵ S40 = 1/2 ×40[2(2)+(40-1)4]=3200
| 2. If the nth term of a series is 1/4 (3+n), then the sum of first 97 terms of this series is: | Easy |
| A. 25 B. 1261 C. 1470 D. None of these |
View Answer
Answer: Option B
Explanation:
Here a=T1 = 1, T2 = 5/4 so, d = 1/4
∵ S97 =1/2 ×97[2(1)+(97-1)(1/4)]=1261
| 3. If the third term of a G. P. is 4, then the product of the first five terms of this series is: | Easy |
| A. 43 B. 45 C. 44 D. None of these |
View Answer
Answer: Option B
Explanation:
Given ar2 = 4
∴Required product = a.ar.ar2.ar3.ar4
= a5r10 = (ar2)5 = 45
| 4. What is the 99th term of the series 2+7+14+23+34+……? | Easy |
| A. 10000 B. 9999 C. 9998 D. None of these |
View Answer
Answer: Option C
Explanation:
∵ Given series is [(22-2) +(32-2) +(42-2) +….]
-> Tn = (n+1)2-2
T99 = (99+1)2 – 2 = (100)2-2 = 9998
| 5. If x, 2x+2, 3x+3, …… are in G. P., then 4th term of this series is: | Medium |
| A. 27 B. -27 C. 13.5 D. None of these |
View Answer
Answer: Option D
Explanation:
∵ x, 2x+2, 3x+3 are in G.P
-> (2x+2)2 = x(3x+3)
-> 4(x+1)2 = 3x(x+1)
-> 4(x+1) = 3x
∵ x = -1 does not give a G.P, à x = -4
So, the G.P. is -4, -6, -9
∴ r = (-6)/(-4) = 3/2
∴ T4 = (-9) (3/2) = -13.5
| 6. The value of 91/3.91/9.91/27…. ∞ is | Easy |
| A. 1 B. 3 C. 9 D. None of these |
View Answer
Answer: Option B
Explanation:
Here required value is 9s, where s = 1/3+1/9+1/27+ …… to ∞
-> s = (1/3)/(1-(1/3) ) =1/2
∴ Required value = 91/2 = 3
| 7. 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 à ½(a+b) > √(ab)
-> a+b > 2√(ab) —– ❶
Similarly, b+c > 2√(bc) —– ❷
and c+a > 2√(ca) —– ❸
Multiplying these, we get
(a+b)(b+c)(c+a) > 8abc
| 8. Express the recurring decimal 0.125 125 125 … as a rational number. | Difficult |
| A. 125/900 B. 125/990 C. 125/999 D. None of these |
View Answer
Answer: Option C
Explanation:
0.125125125125….
= 0.125 + 0.000125 + 0.000000125 + ….
= 125/1000+125/1000000+125/1000000000+⋯to infinity
= 125/10^(3 ) [ 1+1/10^(3 ) +1/10^(6 ) +⋯…. to infinity]
= 125/(10×10×10) [1/{1-(1/10)^3 } ]
=1/8 ×1000/999 =125/999
| 9. Find the fourth term of the series 2,2 1/2,3 1/3,…… | Difficult |
| A. 9/4 B. 26/5 C. 5 D. None of these |
View Answer
Answer: Option C
Explanation:
The reciprocals of the terms of the given series are 1/2, 2/5, 3/10,….
Or 5/10, 4/10, 3/10, ….., which are in A.P. whose common difference = (4/10)-(5/10) = -1/10
∴ Fourth term of this A.P.
= a+(n-1)d = (5/10) + (4-1)(-1/10) = 2/10 = 1/5
∴ Fourth term of the given series = 1/(1/5) = 5
| 10. Find the sum of the first 20 terms of the series 1.32+2.52+3.72+……. | Difficult |
| A. 188090 B. 94045 C. 282135 D. None of these |
View Answer
Answer: Option A
Explanation:
The series formed by the first factors of the terms of the given series is 1, 2, 3, ……., whose nth term is n.
The series formed by the second factors of the terms of the given series is 32, 52, 72, …., whose nth term = [3 + (n-1)2]2 = (2n + 1)2
∴ nth term of the given series = n(2n+1)2
∴ Sum up to n terms
=∑[n(2n+1)2]=∑[n(4n2+4n+1)]
=4∑n3+4 ∑n2+∑n
= 4(n(n+1)/2]2 + 4[n (n+1) (2n+1)/6] + ½ n(n+1)
= [n(n+1)/6 ] [6n(n+1) + 4(2n+1) + 3]
= 1/6 n(n+1)(6n2+14n+7)
Putting n = 20, we get required sum = (1/6) (20)(21) (2400+280+7)
= 188090