CAT Previous Year Questions | Progression
CAT 2022 Slot 2 Question 1-
Question 1 of 15
On day one, there are 100 particles in a laboratory experiment. On day n, where n≥2, one out of every n particles produces another particle. If the total number of particles in the laboratory experiment increases to 1000 on day m, then m equals
For any natural number n, suppose the sum of the first n terms of an arithmetic progression is (n+2n2). If the nth term of the progression is divisible by 9 , then the smallest possible value of n is
Consider a sequence of real numbers x1,x2,x3,… such that xn+1=xn+n−1 for all n≥1. Ifx1=−1 then x100 is equal to
Three positive integers x, y and z are in arithmetic progression. If y − x > 2 and xyz = 5(x + y + z), then z − x equals
For a sequence of real numbers x1, x2, ..., xn, if x1 − x2 + x3 − ... + (−1)n + 1xn = n2 + 2n for all natural numbers n, then the sum x49 + x50 equals
Let teh m-th and n-thterms of a Geometric progression be 3/4 and 12, respectively, when m < n. If the common ratio of the progression is an integer r, then the smallest possible value of r + n − m is
A gentleman decided to treat a few children in the following manner. He gives half of his total stock of toffees and one extra to the first child, and then the half of the remaining stock along with one extra to the second and continues giving away in this fashion. His total stock exhausts after he takes care of 5 children. How many toffees were there in his stock initially?
Let a1 , a2 be integers such that a1 – a2 + a3 – a4 + ........ +(−1)n-1 an = n , for n ≥ 1. Then a51 + a52 + ........ + a1023 equals
The number of common terms in the two sequences: 15, 19, 23, 27, ...... , 415 and 14, 19, 24, 29, ...... , 464 is
If (2n+1) + (2n+3) + (2n+5) + ... + (2n+47) = 5280 , then what is the value of 1+2+3+ ... +n ?
If a1 + a2 + a3 + ... + an = 3(2n+1 − 2), then a11 equals
Let a1, a2, ... , a52 be positive integers such that a1 ＜ a2 ＜ ... ＜ a52. Suppose, their arithmetic mean is one less than the arithmetic mean of a2, a3, ..., a52. If a52 = 100, then the largest possible value of a1 is
Let t1, t2,… be real numbers such that t1+ t2 +... + tn = 2n2 + 9n + 13, for every positive integer n ≥ 2. If tk=103, then k equals
The value of the sum 7 x 11 + 11 x 15 + 15 x 19 + ..... + 95 x 99 is
Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is A.1/6 B.3/6 C.3/2 D.5/2 Ex