Q. 1: If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is
A) 2: 3
B) 3: 2
C) 3: 4
D) 4: 3
(a+6d)^2 = (a+2d)(a+16d)
a^2 + 12 ad + 36d^2 = a^2 + 18 ad + 32d^2
Since, d is positive,
We get the ratio of a:d = 2:3
Option (A)
a + b + c + d = 7
Since, each kid gets 1 eraser, so a + b + c + d = 3
Now, no child can get more than 3 erasers.
There can be two cases. 2, 1, 0, 0 which can be represented in 4!/2! Ways = 12 ways
And 1,1,1,0 which can be represented in 4!/3! Ways = 4 ways
Answer: 16
Option (A)
Q. 3:
A) 2
B) 1/3
C) 6
D) 2/3
Correct Answer:- A
Explanation:-
g(3) = 3^2 – 2(3) – 1 = 2.
Q. 4: Let a1, a2,……..a3n be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + ….+a3n = 1830, then what is the smallest positive integer m such that m (a1 + a2 + …. + an ) > 1830?
A) 8
B) 9
C) 10
D) 11
a = 3
a + d = 7 => d=4
Applying formula of sum for AP
(3n/2) [6 + (3n-1)4] = 1830
On solving, we get n = 10
m>61/7
Max positive integer value of m = 9
Option (B)
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