Progressions and Series Questions for IIFT PDF
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Question 1: Let $\ a_{1},a_{2}…a_{52}\ $ be positive integers such that $\ a_{1}$ < $a_{2}$ < … < $a_{52}\ $. Suppose, their arithmetic mean is one less than arithmetic mean of $a_{2}$, $a_{3}$, ….$a_{52}$. If $a_{52}$= 100, then the largest possible value of $a_{1}$is
a) 48
b) 20
c) 23
d) 45
Question 2: A reputed paint company plans to award prizes to its top three salespersons, with the highest prize going to the top salesperson, the next highest prize to the next salesperson and a smaller prize to the third-ranking salesperson. If the company has 15 salespersons, how many different arrangements of winners are possible (Assume there are no ties)?
a) 1728
b) 2730
c) 3856
d) 1320
Question 3: If the positive real numbers a, b and c are in Arithmetic Progression, such that abc = 4, then minimum possible value of b is:
a) $2^{\frac{3}{2}}$
b) $2^{\frac{2}{3}}$
c) $2^{\frac{1}{3}}$
d) None of the above
Question 4: Let $a_{1},a_{2},a_{3},a_{4},a_{5}$ be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with $2a_{3}$
If the sum of the numbers in the new sequence is 450, then $a_{5}$ is
Question 5: The student mess committee of a reputed Engineering College has n members. Let P be the event that the Committee has students of both sexes and let Q be the event that there is at most one female student in the committee. Assuming that each committee member has probability 0.5 of being female, the value of n for which the events P and Q are independent is
a) 2
b) 3
c) 4
d) None of the above
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Question 6: If H$_1$ , H$_2$ , H$_3$ , …, H$_n$ , are ‘n’ Harmonic means between ‘a’ and ‘b’ (≠ a), then value of $\dfrac{H_{1}+a}{H_{1}-a}+\dfrac{H_{n}+b}{H_{n}-b}$ is equal to
a) n+1
b) 2n
c) 2n+3
d) n-1
Question 7: The sum of 4+44+444+…. upto n terms in
a) $\frac{40}{81}(8^{n}-1)-\frac{5n}{9}$
b) $\frac{40}{81}(8^{n}-1)-\frac{4n}{9}$
c) $\frac{40}{81}(10^{n}-1)-\frac{4n}{9}$
d) $\frac{40}{81}(10^{n}-1)-\frac{5n}{9}$
Question 8: If 2, a, b, c, d, e, f and 65 form an arithmetic progression, find out the value of ‘e’.
a) 48
b) 47
c) 41
d) None of the above
Question 9: Find the sum of the following series;
$\frac{2}{1!}+\frac{3}{2!}+\frac{6}{3!}+\frac{11}{4!}+\frac{18}{5!}+…$
a) 3e – 1
b) 3(e – 1)
c) 3(e + 1)
d) 3e + 1
Question 10: The sum of the series is:
$\frac{1}{1.2.3}+\frac{1}{3.4.5}+\frac{1}{5.6.7}+…$
a) $e^{2} – 1$
b) $\log_{e}{2}-1$
c) $2\log_{10}{2}-1$
d) None of the above
Answers & Solutions:
1) Answer (C)
Let ‘x’ be the average of all 52 positive integers $\ a_{1},a_{2}…a_{52}\ $.
$a_{1}+a_{2}+a_{3}+…+a_{52}$ = 52x … (1)
Therefore, average of $a_{2}$, $a_{3}$, ….$a_{52}$ = x+1
$a_{2}+a_{3}+a_{4}+…+a_{52}$ = 51(x+1) … (2)
From equation (1) and (2), we can say that
$a_{1}+51(x+1)$ = 52x
$a_{1}$ = x – 51.
We have to find out the largest possible value of $a_{1}$. $a_{1}$ will be maximum when ‘x’ is maximum.
(x+1) is the average of terms $a_{2}$, $a_{3}$, ….$a_{52}$. We know that $a_{2}$ < $a_{3}$ < … < $a_{52}\ $ and $a_{52}$ = 100.
Therefore, (x+1) will be maximum when each term is maximum possible. If $a_{52}$ = 100, then $a_{52}$ = 99, $a_{50}$ = 98 ends so on.
$a_{2}$ = 100 + (51-1)*(-1) = 50.
Hence, $a_{2}+a_{3}+a_{4}+…+a_{52}$ = 50+51+…+99+100 = 51(x+1)
$\Rightarrow$ $\dfrac{51*(50+100)}{2} = 51(x+1)$
$\Rightarrow$ $x = 74$
Therefore, the largest possible value of $a_{1}$ = x – 51 = 74 – 51 = 23.
2) Answer (B)
In this case first we have to select the top three salesmen out of 15 and then arrange them as first, second and third.
∴ Number of arrangements = 15C3 × 3! = 2730
3) Answer (B)
It has been given that a, b, and c are in an arithmetic progression.
Let a = x-p, b = x, and c = x+p
We know that a, b, and c are real numbers.
Therefore, the arithmetic mean of a,b,c should be greater than or equal to the geometric mean.
$\frac{a+b+c}{3} \geq \sqrt[3]{abc}$
$\frac{a+b+c}{3} \geq \sqrt[3]{4}$
$\frac{3x}{3}\geq\sqrt[3]{4}$
$x\geq\sqrt[3]{4}$
We know that $x$ = $b$.
Therefore,$b\geq\sqrt[3]{4}$or $b\geq 2^{\frac{2}{3}}$
Therefore, option B is the right answer.
4) Answer: 51
Sum of the sequence of even numbers is $2a_{3} + (2a_{3} – 2) + (2a_{3} – 4)$ $+ (2a_{3} – 6) + (2a_{3} – 8) = 450$
=> $10a_{3} – 20 = 450$
=> $a_{3} = 47$
Hence $a_{5} = 47 + 4 = 51$
5) Answer (B)
Let’s first estimate the sample space. In a committee of n members, there can be the following cases:
(0 males, n females), (1 male, n-1 females), (2 males, n-2 females) …(n males, 0 females).
Thus, the sample space has n+1 cases.
n(S) = n+1 cases
Barring 2 cases from the sample space, event P occurs for all other cases.
=> n(P) = n+1-2 = n-1 cases
=> p(P) = (n-1)/(n+1)
At most 1 female can be expressed as the sum of the cases exactly 0 females and exactly 1 female.
=> n(Q) = 2 cases
=> p(Q) = 2/(n+1)
The set $P \cap Q$ occurs when there is exactly one female member.
Hence, n($P \cap Q$) = 1
p($P \cap Q$) = 1/(n+1)
When two events are independent, the probability of both of them occurring is equal to the product of their individual probabilities.
Thus,
p($P \cap Q$) = p(P) * p(Q)
1/(n+1) = (n-1)/(n+1) * 2/(n+1)
=> n+1 = 2n -2
=> n= 3
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6) Answer (B)
Let us assume that n = 3 and a, b = 2, 6.
Therefore, the harmonic sequence will be: 2, H$_1$, H$_2$, H$_3$, 6
Hence, H$_2$ = $\dfrac{2*2*6}{2+6}$ = 3
H$_1$ = $\dfrac{2*2*3}{2+3}$ = $\dfrac{12}{5} = 2.4$
H$_3$ = $\dfrac{2*3*6}{3+6}$ = $4$
Therefore, $\dfrac{H_{1}+a}{H_{1}-a}+\dfrac{H_{n}+b}{H_{n}-b}$
$\dfrac{2.4+2}{2.4-2}+\dfrac{4+6}{4-6}$
$\Rightarrow$ $11-5$ = 6.
Option B: 2n = 2*3 = 6.
7) Answer (C)
Given that, S = 4+44+444+…
$\Rightarrow$ $S = \frac{4}{9}(9+99+999+…)$
$\Rightarrow$ $S = \frac{4}{9}(10-1+10^2-1+10^3-1+…+10^n-1)$
$\Rightarrow$ $S = \frac{4}{9}(10+10^2+10^3+…+10^n-n)$
$\Rightarrow$ $S = \frac{4}{9}(\frac{10(10^n – 1)}{10-1}-n)$
$\Rightarrow$ $S = \frac{40}{81}(10^n – 1) – \frac{4n}{9}$
Hence, option C is the correct answer.
Alternate method:
Solving for n = 2
Sum of series = 4+44 = 48
Substituting n = 2 in options
(A) $\frac{40}{81}(8^{2}-1)-\frac{5*2}{9}$ = $\frac{280-10}{9}$ = 30
(B) $\frac{40}{81}(8^{2}-1)-\frac{4*2}{9}$ = $\frac{280-8}{9}$ = $\frac{272}{9}$
(C) $\frac{40}{81}(10^{2}-1)-\frac{4*2}{9}$ = $\frac{440-8}{9}$ = 48
(D) $\frac{40}{81}(10^{2}-1)-\frac{5*2}{9}$ = $\frac{440-10}{9}$ = $\frac{430}{9}$
8) Answer (B)
Given that 2, a, b, c, d, e, f and 65 are in an AP.
65 = 2 + (8-1)d
d = 9.
Therefore, e = 2+(6-1)*9 = 2+45 = 47. Therefore, option B is the correct answer.
9) Answer (B)
$\frac{2}{1!}+\frac{3}{2!}+\frac{6}{3!}+\frac{11}{4!}+\frac{18}{5!}+…$ = 2 + 1.5 + 1 + 0.5 + 0.15 + smaller insignificant values $\approx$ 5.2
Option ‘A’ = 3e – 1 = 3*2.718-1 = 7.15 therefore cannot be our answer
Option ‘B’ = 3(e – 1) = 3*(2.718-1) = 5.154 therefore can be our answer
Option ‘C’ = 3(e + 1) = 3*(2.718+1) = 11.15 therefore cannot be our answer
Option ‘D’ = 3e + 1 = 9.15 therefore cannot be our answer
Therefore option ‘B’ is our answer
10) Answer (D)
$\frac{1}{1.2.3}+\frac{1}{3.4.5}+\frac{1}{5.6.7}+…$ = 0.166..+0.0166..+0.00476..+….. which is always <0.2 but a positive number.
Option A = $e^2$ – 1 = 7.34 – 1 = 6.34 therefore this cannot be the Answer
Option B = $\log_{e}{2}-1$ = $\frac{log_{10}{2}}{log_{10}{e}}$ – 1 = $\frac{log_{10}{2}}{log_{10}{2.72}}$ – 1
We know that the value of $\frac{log_{10}{2}}{log_{10}{2.72}}$ < 1 and therefore value of $\frac{log_{10}{2}}{log_{10}{2.72}}$ – 1 < 0
therefore this cannot be the Answer
Option C = $2\log_{10}{2}-1$ = $\log_{10}{4}-1$
We know that the value of $\log_{10}{4}$ < 1 and therefore value of $\log_{10}{4}-1$ < 0
therefore this cannot be the Answer
Therefore the Answer is option ‘D’
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