# Progressions and Series Questions for IIFT PDF

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## Progressions and Series Questions for IIFT PDF

Download important IIFT Progressions ans Series Questions PDF based on previously asked questions in IIFT and other MBA exams. Practice Progressions ans Series questions and answers for IIFT exam.

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

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.

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

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.
$\frac{a+b+c}{3} \geq \sqrt{abc}$
$\frac{a+b+c}{3} \geq \sqrt{4}$
$\frac{3x}{3}\geq\sqrt{4}$

$x\geq\sqrt{4}$
We know that $x$ = $b$.
Therefore,$b\geq\sqrt{4}$or $b\geq 2^{\frac{2}{3}}$
Therefore, option B is the right answer.

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 $a_{5} = 47 + 4 = 51$

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

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.

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) $\frac{40}{81}(10^{2}-1)-\frac{5*2}{9}$ = $\frac{440-10}{9}$ = $\frac{430}{9}$

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.

$\frac{2}{1!}+\frac{3}{2!}+\frac{6}{3!}+\frac{11}{4!}+\frac{18}{5!}+…$ = 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

$\frac{1}{1.2.3}+\frac{1}{3.4.5}+\frac{1}{5.6.7}+…$ = 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’

We hope this Progressions and Series questions and answers for IIFT PDF will be helpful to you.