IPMAT Indore Progression & Series PYQs

Q1:ipmat indore 2021QA › Progression & SeriesMediumSA · TITA

The sum up to

10

terms of the series

1 \cdot 3 + 5 \cdot 7 + 9 \cdot 11 + ...

is

Enter your answer to attempt

The Setup: We need the sum of the first 10 terms of the series

1\cdot3+5\cdot7+9\cdot11+...

. We are dealing with products of Arithmetic Progressions (APs). We need to unmask the

n

-th term so we can just blast it with summation formulas and speedrun the calculation. Step 1: Find the pattern for the first part of each term (1, 5, 9...). This is an AP with a common difference of 4.

a_n = 1 + (n-1)4 = 4n - 3

Step 2: Find the pattern for the second part of each term (3, 7, 11...). This is another AP, also with a common difference of 4.

b_n = 3 + (n-1)4 = 4n - 1

Step 3: Construct the general term

T_n

by multiplying them, and expand it into a quadratic polynomial:

T_n = (4n - 3)(4n - 1) = 16n^2 - 16n + 3

Step 4: Deploy the summation formulas for the first 10 terms.

\sum_{n=1}^{10} T_n = 16 \sum n^2 - 16 \sum n + \sum 3

Using the standard formulas

\sum n^2 = \frac{n(n+1)(2n+1)}{6}

and

\sum n = \frac{n(n+1)}{2}

for

n=10

: *

16 \times \left(\frac{10 \times 11 \times 21}{6}\right) = 16 \times 385 = 6160

*

-16 \times \left(\frac{10 \times 11}{2}\right) = -16 \times 55 = -880

*

10 \times 3 = 30

Step 5: Crunch the final numbers.

6160 - 880 + 30 = 5310

Final Answer: 5310

Q2:ipmat indore 2021QA › Progression & SeriesMediumSA · TITA

It is given that the sequence {

x_n

} satisfies

x_1 = 0, x_{n+1} = x_n + 1 + 2\sqrt(1+x_n)

for

n = 1,2,...

Then

x_{31}

is _______

Enter your answer to attempt

The Setup: We are faced with a recursive sequence:

x_{n+1}=x_n+1+2\sqrt{1+x_n}

with a starting state of

x_1=0

. It looks cursed, but it's actually just a perfect square in disguise. We just need to transform the variables to reveal the hidden meta. Step 1: Add 1 to both sides to complete the transformation.

x_{n+1} + 1 = (x_n + 1) + 2\sqrt{1+x_n} + 1

Step 2: Substitute

y_n=x_n+1

to clean up the UI.

y_{n+1} = y_n + 2\sqrt{y_n} + 1

Notice that the right side is a literal perfect square:

(\sqrt{y_n} + 1)^2

.

y_{n+1} = (\sqrt{y_n} + 1)^2

Step 3: Take the square root of both sides.

\sqrt{y_{n+1}} = \sqrt{y_n} + 1

This means the sequence

\sqrt{y_n}

is just a basic Arithmetic Progression increasing by 1 every step. Since

x_1=0

, our base case is

y_1=1

, which means

\sqrt{y_1}=1

. Step 4: Find the

n

-th term of this AP.

\sqrt{y_n} = 1 + (n-1)(1) = n

Step 5: Reverse the transformation to find

x_n

.

y_n = n^2 \implies x_n + 1 = n^2 \implies x_n = n^2 - 1

Step 6: Calculate the 31st term to finish the level.

x_{31} = 31^2 - 1 = 961 - 1 = 960

Final Answer: 960

Q3:ipmat indore 2025QA › Progression & SeriesMediumSA · TITA

If the sum of the first 21 terms of the sequence

\ln(\frac{a}{b}), \ln(\frac{a}{b\sqrt{b}}), \ln(\frac{a}{b^2}) \dots

is

\ln(\frac{a^m}{b^n})

, then the value of

m+n

is

Enter your answer to attempt

The Setup: At first glance, this looks like a cursed sequence of logarithms, but if we peel back the formatting, it is literally just an Arithmetic Progression (AP) in disguise. We just need to unpack the logs, find the common difference to establish our AP meta, calculate the sum of the first 21 terms, and then repackage it to find our

m

and

n

values. Step 1: Unpacking the Logs (The Vibe Check) Let's analyze the first three terms to figure out exactly what kind of progression we are dealing with. Using the logarithm quotient rule (

\ln(x/y) = \ln(x) - \ln(y)

) and exponent rules, we can break them down: * Term 1 (

T_1

):

\ln(\frac{a}{b}) = \ln(a) - \ln(b)

* Term 2 (

T_2

):

\ln(\frac{a}{b\sqrt{b}}) = \ln(\frac{a}{b^{1.5}}) = \ln(a) - 1.5\ln(b)

* Term 3 (

T_3

):

\ln(\frac{a}{b^2}) = \ln(a) - 2\ln(b)

Step 2: **Finding the Common Difference (

D

)** To confirm this is a standard AP and to find our step size, we subtract

T_1

from

T_2

:

D = T_2 - T_1

D = (\ln(a) - 1.5\ln(b)) - (\ln(a) - \ln(b))

D = -0.5\ln(b)

The sequence is linearly dropping by

0.5\ln(b)

every single term. This confirms we are working with an AP where the first term

A = \ln(a) - \ln(b)

and the common difference

D = -0.5\ln(b)

. Step 3: The Carry (Summing 21 Terms) Now we deploy the standard AP sum formula for 21 terms:

S_n = \frac{n}{2} [2A + (n-1)D]

Plug in our

n = 21

,

A

, and

D

values:

S_{21} = \frac{21}{2} [2(\ln(a) - \ln(b)) + (21-1)(-0.5\ln(b))]

S_{21} = \frac{21}{2} [2\ln(a) - 2\ln(b) + 20(-0.5\ln(b))]

S_{21} = \frac{21}{2} [2\ln(a) - 2\ln(b) - 10\ln(b)]

Combine the

\ln(b)

terms to clean up the bracket:

S_{21} = \frac{21}{2} [2\ln(a) - 12\ln(b)]

Factor out the 2 inside the bracket to cancel the denominator and nerf the equation:

S_{21} = 21 [\ln(a) - 6\ln(b)]

Step 4: Repackaging the File The question demands the final answer in the format

\ln(\frac{a^m}{b^n})

. We need to compress our result back into a single log expression using power rules (

k\ln(x) = \ln(x^k)

). First, distribute the 21:

S_{21} = 21\ln(a) - 126\ln(b)

Now apply the power rule to push the coefficients back into the logs:

S_{21} = \ln(a^{21}) - \ln(b^{126})

Finally, use the quotient rule to merge them into one fraction:

S_{21} = \ln\left(\frac{a^{21}}{b^{126}}\right)

Step 5: The Audit (Double Check Protocol) Let's match our final expression against the target form

\ln(\frac{a^m}{b^n})

. By direct comparison:

m = 21

n = 126

The prompt asks for the value of

m + n

.

m + n = 21 + 126 = 147

Math is verified, rules applied correctly, logic is absolutely flawless. Final Answer: 147

Q4:ipmat indore 2022QA › Progression & SeriesMediumSA · TITA

A new sequence is obtained from the sequence of positive integers

(1,2,3, \ldots)

by deleting all the perfect squares. Then the

2022^{\text {nd }}

term of the new sequence is ________.

Enter your answer to attempt

The Setup: A sequence of positive integers removes all perfect squares. We must determine the numerical value of the

2022^{\text{nd}}

remaining integer in this sequence. Step 1: Establish the formula for sequence terms. Let

N

be the actual value of the term. The sequence loses one position for every perfect square it passes. The number of perfect squares less than or equal to

N

is

\lfloor\sqrt{N}\rfloor

. Thus, the position of

N

in the modified sequence is given by:

N - \lfloor\sqrt{N}\rfloor = 2022

. Step 2: Estimate

N

to find bounds. Because

N

will be slightly larger than

2022

, we estimate

\sqrt{N} \approx \sqrt{2022}

.

44^2 = 1936

and

45^2 = 2025

. Therefore, the square root will be bounded by approximately

45

. Step 3: Test the logical value for

N

. Add our

45

estimated skipped squares to the

2022^{\text{nd}}

position: Let's test

N = 2022 + 45 = 2067

. Check how many perfect squares exist up to

2067

: The largest integer square less than

2067

is

45^2 = 2025

. Thus, exactly

45

squares are skipped. Subtract the skipped squares to verify the position:

\text{Position} = 2067 - 45 = 2022

The number

2067

precisely lands on the

2022^{\text{nd}}

position, and it is not itself a perfect square, so it genuinely survives into the new sequence. Step 4: Confirm the answer is unique. The counting function

N - \lfloor\sqrt{N}\rfloor

never decreases as

N

grows, so it takes the value

2022

on at most one non-square

N

. Hence

2067

is the only possibility. Final Answer: 2067

Q5:ipmat indore 2022QA › Progression & SeriesMediumSA · TITA

The

3^{\text {rd }}, 14^{\text {th }}

and

69^{\text {th }}

terms of an arithmetic progression form three distinct and consecutive terms of a geometric progression. If the next term of the geometric progression is the

n^{\text {th }}

term of the arithmetic progression, then

n

equals ________.

Enter your answer to attempt

The Setup: Three selected terms (

3^{\text{rd}}, 14^{\text{th}}, 69^{\text{th}}

) of an Arithmetic Progression (AP) form a separate Geometric Progression (GP). The 4th term of this GP corresponds to the

n^{\text{th}}

term of the AP. We must find index

n

. Step 1: Establish the GP relationship. Let the AP have a first term

a

and common difference

d

.

T_3 = a + 2d

T_{14} = a + 13d

T_{69} = a + 68d

Because these form a GP, the square of the middle term equals the product of the outer terms:

(a + 13d)^2 = (a + 2d)(a + 68d)

a^2 + 26ad + 169d^2 = a^2 + 70ad + 136d^2

Step 2: Relate the starting term to the difference.

33d^2 = 44ad

Assuming

d \neq 0

(otherwise all terms are identical and limits break), divide by

11d

:

3d = 4a \implies a = \frac{3}{4}d

Step 3: Calculate the GP common ratio. Substitute

a = 0.75d

into the first two terms of the GP:

T_3 = 0.75d + 2d = 2.75d = \frac{11}{4}d

T_{14} = 0.75d + 13d = 13.75d = \frac{55}{4}d

The common ratio

r = \frac{T_{14}}{T_3} = \frac{55/4}{11/4} = 5

. The third term is

T_{69} = 0.75d + 68d = 68.75d = \frac{275}{4}d

. (Verified:

\frac{55}{4}d \times 5 = \frac{275}{4}d

). Step 4: Calculate the next GP term and map it to the AP index. The 4th GP term is

T_{69} \times r

:

\text{Next GP Term} = \frac{275}{4}d \times 5 = \frac{1375}{4}d

Set this equal to the generic

n^{\text{th}}

term of the AP (

a + (n-1)d

):

\frac{3}{4}d + (n-1)d = \frac{1375}{4}d

(n-1)d = \frac{1372}{4}d \implies n-1 = 343 \implies n = 344

Final Answer: 344

Q6:ipmat indore 2023QA › Progression & SeriesHardSA · TITA

If

f(n)= 1 + 2 + 3 +\cdots+(n+1)

and

g(n)= \sum_{k=1}^{k=n} \dfrac{1}{f(k)}

, then the least value of

n

for which

g(n)

exceeds the value

\dfrac{99}{100}

is:

Enter your answer to attempt

The Setup: Given an arithmetic sum base function

f(n)

, we have a reciprocal series

g(n)

. We must find the least integer

n

that pushes

g(n)

above the threshold of

99/100

. Step 1: Simplify the base function

f(n)

. The function is the sum of consecutive integers:

f(n) = 1 + 2 + 3 + \dots + (n+1) = \frac{(n+1)(n+2)}{2}

Step 2: Construct the series

g(n)

and apply partial fractions.

g(n) = \sum_{k=1}^n \frac{1}{f(k)} = \sum_{k=1}^n \frac{2}{(k+1)(k+2)}

Split the fraction using partial fraction decomposition:

\frac{2}{(k+1)(k+2)} = 2 \left( \frac{1}{k+1} - \frac{1}{k+2} \right)

Step 3: Evaluate the telescoping sum. When summing from

k=1

to

n

, all intermediate terms cancel out:

g(n) = 2 \left[ \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{n+1} - \frac{1}{n+2}\right) \right]

g(n) = 2 \left( \frac{1}{2} - \frac{1}{n+2} \right) = 1 - \frac{2}{n+2}

Step 4: Apply the threshold inequality. We need

g(n) > \frac{99}{100}

:

1 - \frac{2}{n+2} > 0.99

0.01 > \frac{2}{n+2}

\frac{1}{100} > \frac{2}{n+2} \implies n+2 > 200 \implies n > 198

Since

n

must be an integer, the least value fulfilling this is

199

. Final Answer: 199

Q7:ipmat indore 2022QA › Progression & SeriesMediumSA · TITA

The numbers

-16,2^{x+3}-2^{2 x-1}-16,2^{2 x-1}+16

are in an arithmetic progression. Then

x

equals ________.

Enter your answer to attempt

The Setup: Three expressions involving exponential powers of

2

and constants are given as consecutive terms of an Arithmetic Progression. We must solve for the unknown exponent

x

. Step 1: Establish the arithmetic mean equation. For three terms

A, B, C

in an AP,

2B = A + C

. Given terms:

A = -16

,

B = 2^{x+3} - 2^{2x-1} - 16

, and

C = 2^{2x-1} + 16

.

2(2^{x+3} - 2^{2x-1} - 16) = -16 + (2^{2x-1} + 16)

Step 2: Simplify the algebraic constraints.

2 \cdot 2^{x+3} - 2 \cdot 2^{2x-1} - 32 = 2^{2x-1}

Combine the exponential terms:

2^{x+4} - 2^{2x} - 32 = 2^{2x-1}

Group the terms involving

2x

on the right:

2^{x+4} - 32 = 2^{2x-1} + 2^{2x}

Factor out

2^{2x}

:

2^{x+4} - 32 = 2^{2x} \left(\frac{1}{2} + 1\right) = \frac{3}{2} \cdot 2^{2x}

Step 3: Use substitution to solve the resulting quadratic. Let

y = 2^x

. Then

2^{x+4} = 16y

, and

2^{2x} = y^2

.

16y - 32 = \frac{3}{2}y^2

Multiply by

2

and rearrange into standard quadratic form:

3y^2 - 32y + 64 = 0

Factor the quadratic equation:

(3y - 8)(y - 8) = 0

Step 4: Evaluate valid roots for integer

x

. This yields

y = 8/3

or

y = 8

. Since

y = 2^x

, resolving for

y = 8

yields

2^x = 8 \implies x = 3

. *(The fractional root

y=8/3

yields an irrational logarithmic decimal out of scope for standard numeric data entry).* Final Answer: 3

Q8:ipmat indore 2021QA › Progression & SeriesMediumMCQ · MCQ

Let

S_n

be sum of the first

n

terms of an A.P. If

S_5 = S_9

, what is the ratio of

a_3 : a_5

A $9:5$
B $5:9$
C $3:5$
D $5:3$

Pick an option to attempt

The Setup: We are working with Arithmetic Progressions. The standard sum formula is

S_n=\frac{n}{2}(2a+(n-1)d)

. We need to build an algebraic equation equating the two given sums to find the direct relationship between the first term (

a

) and the common difference (

d

). Step 1: Expand both sides using the sum formula.

S_5=\frac{5}{2}(2a+4d)=5(a+2d)=5a+10d

S_9=\frac{9}{2}(2a+8d)=9(a+4d)=9a+36d

Step 2: Set them equal and isolate

a

in terms of

d

.

5a+10d=9a+36d

-4a=26d \implies 2a=-13d \implies a=-6.5d

Step 3: Find the expressions for the specific requested terms,

a_3

and

a_5

.

a_3=a+2d=-6.5d+2d=-4.5d

a_5=a+4d=-6.5d+4d=-2.5d

Step 4: Calculate the ratio. The negative signs and the

d

variables cancel out cleanly.

\text{Ratio}=\frac{-4.5d}{-2.5d}=\frac{4.5}{2.5}=\frac{45}{25}=\frac{9}{5}

Final Answer:

9:5

Q9:ipmat indore 2020QA › Progression & SeriesHardMCQ · MCQ

If

\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots

up to

\infty = \frac{\pi^2}{6}

, then the value of

\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots

up to

\infty

is

A $\frac{\pi^2}{8}$
B $\frac{\pi^2}{16}$
C $\frac{\pi^2}{12}$
D $\frac{\pi^2}{36}$

Pick an option to attempt

The Setup: We are manipulating the famous Basel problem. We are handed the sum of *all* inverse squares and asked for the sum of only the *odd* ones. The trick is to split the total into its even and odd halves, then notice the even half is a scaled copy of the whole thing. Step 1: Define the series. Let the total be

S=\frac{\pi^2}{6}

. Every positive integer is either odd or even, so the total splits cleanly with nothing counted twice and nothing missed:

S=\text{Sum(Odds)}+\text{Sum(Evens)}

Every term here is positive and the series converges, so splitting and rearranging it like this is legitimate rather than a formal trick. Step 2: Factor the even series. Write out the even terms and pull out the common factor. Each even number is

2n

, so its square is

4n^2

:

\text{Sum(Evens)}=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\ldots=\frac{1}{4}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots\right)

The bracket is literally the original total

S

, so

\text{Sum(Evens)}=\frac{S}{4}

. This self-similarity is the whole engine of the problem. Step 3: Solve for the odds. Substitute back into the master equation:

S=\text{Sum(Odds)}+\frac{S}{4} \implies \text{Sum(Odds)}=S-\frac{S}{4}=\frac{3S}{4}

Step 4: Final substitution. Plug in the known value of

S

:

\text{Sum(Odds)}=\frac{3}{4}\times \frac{\pi^2}{6}=\frac{3\pi^2}{24}=\frac{\pi^2}{8}

Final Answer:

\frac{\pi^2}{8}

Q10:ipmat indore 2025QA › Progression & SeriesMediumMCQ · MCQ

Given that

1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+...=\frac{\pi^{2}}{6}

, the value of

1+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}+...

is

A $\frac{\pi^{2}}{6}-1$
B $\frac{\pi}{6}$
C $\frac{\pi^{2}}{12}$
D $\frac{\pi^{2}}{8}$

Pick an option to attempt

The Setup: This is a classic infinite series manipulation. We are taking the legendary Basel problem (solved by Euler, the absolute GOAT of 18th-century mathematics) and splitting it into its odd and even components. Think of it like taking a full, mixed studio track and isolating just the vocals. We know the total sum, so we can isolate the odd-denominator series by mathematically filtering out the even-denominator series. Step 1: Define the master series. Let the total sum be represented by

S

. The problem gives us the total value of this series right out of the gate:

S = 1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+... = \frac{\pi^{2}}{6}

This series contains both the odd terms (which we want) and the even terms. Let's split them into two distinct squads:

S = S_{odd} + S_{even}

Step 2: Isolate and analyze the even terms. The even terms form their own infinite sub-series:

S_{even} = \frac{1}{2^{2}}+\frac{1}{4^{2}}+\frac{1}{6^{2}}+...

Notice that every denominator here has a common factor of

2^2

. If we factor out

\frac{1}{2^2}

(which is just

\frac{1}{4}

), watch what happens:

S_{even} = \frac{1}{4}\left(1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+...\right)

Bro, look closely at what is inside the parentheses—it is literally just our original series

S

hitting a respawn. So, we can rewrite this as:

S_{even} = \frac{1}{4}S

Step 3: Solve for the odd terms. Now we take this finding and plug it back into our main equation from Step 1.

S = S_{odd} + \frac{1}{4}S

To get

S_{odd}

by itself, just subtract

\frac{1}{4}S

from both sides.

S_{odd} = S - \frac{1}{4}S

S_{odd} = \frac{3}{4}S

Step 4: Calculate the final value. Now for the final boss phase. Plug in Euler's iconic value for

S

to see exactly what the odd terms equal.

S_{odd} = \frac{3}{4}\left(\frac{\pi^{2}}{6}\right)

Simplify the fraction by cross-canceling the

3

and the

6

, and the math is mathing perfectly:

S_{odd} = \frac{\pi^{2}}{8}

Final Answer:

\frac{\pi^{2}}{8}

Q11:ipmat indore 2024QA › Progression & SeriesHardMCQ · MCQ

The terms of a geometric progression are real and positive. If the

p

-th term of the progression is

q

and the

q

-th term is

p

, then the logarithm of the first term is

A $\dfrac{(1-q)\log(p)-(1-p)\log(q)}{p-q}$
B $\dfrac{(1-q)\log(q)-(1-p)\log(p)}{p-q}$
C $\dfrac{(1-q)\log(p)+(1-p)\log(q)}{p-q}$
D $\dfrac{(1-q)\log(q)+(1-p)\log(p)}{p-q}$

Pick an option to attempt

The Setup: The terms of a geometric progression are real and positive, with the

p

-th term equal to

q

and the

q

-th term equal to

p

. We need to find the logarithm of the first term. Step 1: Define the sequence algebraically. Let the first term be

a

and the common ratio be

r

. The general term formula gives:

T_p = a r^{p-1} = q

T_q = a r^{q-1} = p

Step 2: Apply logarithms to both equations. Taking the log of both sides yields a system of linear equations:

\log(a) + (p-1)\log(r) = \log(q)

\log(a) + (q-1)\log(r) = \log(p)

Step 3: Eliminate

\log(r)

to solve for

\log(a)

. Subtract the second equation from the first:

(p-q)\log(r) = \log(q) - \log(p) \implies \log(r) = \frac{\log(q) - \log(p)}{p-q}

Substitute

\log(r)

back into the first equation:

\log(a) = \log(q) - (p-1)\left(\frac{\log(q) - \log(p)}{p-q}\right)

Step 4: Simplify the expression. Find a common denominator:

\log(a) = \frac{(p-q)\log(q) - (p-1)(\log(q) - \log(p))}{p-q}

\log(a) = \frac{p\log(q) - q\log(q) - p\log(q) + \log(q) + p\log(p) - \log(p)}{p-q}

\log(a) = \frac{(1-q)\log(q) - (1-p)\log(p)}{p-q}

Final Answer:

\dfrac{(1-q)\log(q)-(1-p)\log(p)}{p-q}

Q12:ipmat indore 2019QA › Progression & SeriesMediumSA · TITA

Assume that all positive integers are written down consecutively from left to right as in 1234567891011...... The 6389th digit in this sequence is

Enter your answer to attempt

The Setup: Walk through the concatenated string block by block. Count how many digits the 1-, 2- and 3-digit numbers consume, subtract that from the target position, then divide cleanly into the 4-digit block. Step 1: Count the early blocks.

Block	Count of numbers	Digits each	Digits used	Running total
1-9	9	1	9	9
10-99	90	2	180	189
100-999	900	3	2,700	2,889

So position 2,889 is the final digit of 999, and everything beyond is 4-digit territory. Step 2: Penetrate the 4-digit block.

6389-2889=3500 \text{ digits into the block}

Every number here uses exactly 4 digits, so:

\frac{3500}{4}=875 \text{ with remainder } 0

A remainder of zero is the case to read carefully: it means the target does not fall partway into a number but lands exactly on the last digit of the 875th four-digit number. (A remainder of 1, 2 or 3 would instead point at that position within the *next* number.) Step 3: Identify the number. The block begins at 1000, which is the 1st, so the 875th is:

1000+875-1=1874

Step 4: Extract the digit. With the remainder zero, we want the final digit of 1874, which is 4. Step 5: Confirm the position. The last digit of 1874 sits at position

2889+875\times 4=2889+3500=6389

- exactly the digit asked for. Final Answer: 4

Q13:ipmat indore 2024QA › Progression & SeriesMediumMCQ · MCQ

The sum of a given infinite geometric progression is 80 and the sum of its first two terms is 35. Then the value of

n

for which the sum of its first

n

terms is closest to 100, is

A4
B5
C7
D6

Pick an option to attempt

The Setup: An infinite geometric progression has a sum of

80

, and the sum of its first two terms is

35

. We need to find the number of terms

n

for which the partial sum is closest to

100

. Step 1: Establish equations for the GP. Let the first term be

a

and the common ratio be

r

.

S_\infty = \frac{a}{1-r} = 80 \implies a = 80(1-r)

S_2 = a + ar = a(1+r) = 35

Step 2: Solve for the common ratio

r

and first term

a

. Substitute the expression for

a

into the second equation:

80(1-r)(1+r) = 35

80(1-r^2) = 35 \implies 1-r^2 = \frac{35}{80} = \frac{7}{16}

r^2 = \frac{9}{16} \implies r = \pm\frac{3}{4}

If

r = \frac{3}{4}

, then

a = 80(1 - 3/4) = 20

. The partial sum for positive terms converging to

80

will never exceed

80

or approach

100

. If

r = -\frac{3}{4}

, then

a = 80(1 + 3/4) = 140

. This alternating series can exceed

80

and approach

100

. Step 3: Formulate the partial sum

S_n

for

r = -3/4

.

S_n = \frac{140(1 - (-3/4)^n)}{1 - (-3/4)} = \frac{140}{7/4}(1 - (-3/4)^n) = 80(1 - (-3/4)^n)

Step 4: Test values of

n

to find the sum closest to

100

. For the sum to exceed

80

,

n

must be odd (making

(-3/4)^n

negative). Test odd integers: If

n=3

:

S_3 = 80(1 - (-27/64)) \approx 113.75

(Distance to 100 =

13.75

) If

n=5

:

S_5 = 80(1 - (-243/1024)) \approx 98.98

(Distance to 100 =

1.02

) If

n=7

:

S_7 = 80(1 - (-2187/16384)) \approx 90.68

(Distance to 100 =

9.32

) The value

S_5

yields the closest proximity to

100

. Final Answer: 5

Q14:ipmat indore 2022QA › Progression & SeriesMediumMCQ · MCQ

The sum of the first 15 terms in an arithmetic progression is 200, while the sum of the next 15 terms is 350. Then the common difference is

A $\frac{7}{9}$
B $\frac{2}{3}$
C $\frac{4}{9}$
D $\frac{1}{3}$

Pick an option to attempt

The Setup: We are given two continuous summation blocks of an Arithmetic Progression (

S_{1-15}

and

S_{16-30}

). We must build a system of linear equations mapping the progression formula to isolate the common difference

d

. Step 1: Translate the sums into standard AP formulation. Let the first term be

a

and the common difference be

d

. The sum of an AP is

S_n = \frac{n}{2}(2a + (n-1)d)

. Sum of the first 15 terms (

S_{15}

) =

200

.

200 = \frac{15}{2}(2a + 14d) = 15(a + 7d) \implies a + 7d = \frac{200}{15} = \frac{40}{3}

Step 2: Evaluate the total cumulative sum constraint. The sum of the *next* 15 terms is

350

. This means the total sum of the first 30 terms (

S_{30}

) is

200 + 350 = 550

.

550 = \frac{30}{2}(2a + 29d) = 15(2a + 29d) \implies 2a + 29d = \frac{550}{15} = \frac{110}{3}

Step 3: Solve the linear system for

d

. Multiply the first simplified equation by

2

to align coefficients:

2a + 14d = \frac{80}{3}

Subtract this from the second equation:

(2a + 29d) - (2a + 14d) = \frac{110}{3} - \frac{80}{3}

15d = \frac{30}{3} = 10

d = \frac{10}{15} = \frac{2}{3}

Final Answer:

\frac{2}{3}

Q15:ipmat indore 2023QA › Progression & SeriesMediumMCQ · MCQ

Let

a_{1}, a_{2}, a_{3}

be three distinct real numbers in geometric progression. If the equations

a_{1} x ^ 2 + 2a_{2}x + a_{3} = 0

and

b_{1} x ^ 2 + 2b_{2}x + b_{3} = 0

have a common root, then which of the following is necessarily true?

A $\dfrac{b_{1}}{a_{1}}, \dfrac{b_{2}}{a_{2}}, \dfrac{b_{3}}{a_{3}}$ are in geometric progression
B $b_{1}, b_{2}, b_{3}$ are in geometric progression
C $b_{1}, b_{2}, b_{3}$ are in arithmetic progression
D $\dfrac{b_{1}}{a_{1}}, \dfrac{b_{2}}{a_{2}}, \dfrac{b_{3}}{a_{3}}$ are in arithmetic progression

Pick an option to attempt

The Setup: Given three distinct real numbers

a_1, a_2, a_3

in a geometric progression (GP), we are told the quadratics

a_1x^2+2a_2x+a_3=0

and

b_1x^2+2b_2x+b_3=0

share a common root. We must deduce the relationship between the ratios of their coefficients. Step 1: Analyze the first quadratic equation. Since

a_1, a_2, a_3

are in GP, they satisfy the property

a_2^2 = a_1a_3

. Check the discriminant (

D

) of the first quadratic equation

a_1x^2 + 2a_2x + a_3 = 0

:

D = (2a_2)^2 - 4(a_1)(a_3) = 4a_2^2 - 4a_1a_3

Substitute

a_1a_3

for

a_2^2

:

D = 4(a_1a_3) - 4a_1a_3 = 0

Since

D = 0

, the first equation has a single repeated real root. We find this root using the quadratic formula:

x = \frac{-2a_2}{2a_1} = -\frac{a_2}{a_1}

Step 2: Apply the common root constraint. Because the two equations share a common root, and the first equation only has *one* unique root, this root (

x = -a_2/a_1

) must exactly be the root of the second equation

b_1x^2+2b_2x+b_3=0

. Substitute

x = -a_2/a_1

into the second equation:

b_1\left(-\frac{a_2}{a_1}\right)^2 + 2b_2\left(-\frac{a_2}{a_1}\right) + b_3 = 0

b_1\frac{a_2^2}{a_1^2} - 2b_2\frac{a_2}{a_1} + b_3 = 0

Step 3: Simplify the relationship to match the options. Divide the entire equation by

a_3

:

b_1\frac{a_2^2}{a_1^2 a_3} - 2b_2\frac{a_2}{a_1 a_3} + \frac{b_3}{a_3} = 0

Use the GP identity

a_2^2 = a_1a_3

to simplify the denominators. For the first term:

\frac{a_2^2}{a_1^2 a_3} = \frac{a_1 a_3}{a_1^2 a_3} = \frac{1}{a_1}

For the second term:

\frac{a_2}{a_1 a_3} = \frac{a_2}{a_2^2} = \frac{1}{a_2}

Substitute these back:

\frac{b_1}{a_1} - 2\frac{b_2}{a_2} + \frac{b_3}{a_3} = 0

2\left(\frac{b_2}{a_2}\right) = \frac{b_1}{a_1} + \frac{b_3}{a_3}

This fits the exact definition of an Arithmetic Progression (AP) where the middle term is the arithmetic mean of the outer terms. Final Answer:

\frac{b_{1}}{a_{1}}.

\frac{b_{2}}{a_{2}},

\frac{b_{3}}{a_{3}}

are in arithmetic progression

Q16:ipmat indore 2023QA › Progression & SeriesHardMCQ · MCQ

A person standing at the centre of an open ground first walks 32 meters towards the east, takes a right turn and walks 16 meters, takes another right turn and walks 8 meters, and so on. How far will the person be from the original starting point after an infinite number of such walks in this pattern?

A $32$ meters
B $\frac{32}{\sqrt{5}}$ meters
C $64$ meters
D $\frac{64}{\sqrt{5}}$ meters

Pick an option to attempt

The Setup: A person maps out a path taking consecutive

90^\circ

right turns, with each leg of the journey halving in distance (

32\text{m}, 16\text{m}, 8\text{m} \dots

). We must find their net displacement from the origin after infinite turns. Step 1: Establish a coordinate system. Let the starting point be the origin

(0,0)

. * Leg 1:

32\text{m}

East (Positive

x

-direction). * Leg 2:

16\text{m}

South (Negative

y

-direction). * Leg 3:

8\text{m}

West (Negative

x

-direction). * Leg 4:

4\text{m}

North (Positive

y

-direction). * Leg 5:

2\text{m}

East (Positive

x

-direction). This pattern continues infinitely. Step 2: Calculate the net displacement on the

x

-axis. The

x

-coordinate sequence alternates direction every two steps:

32

(East),

-8

(West),

2

(East),

-0.5

(West)

\dots

This forms an infinite geometric progression where the first term

a = 32

and the common ratio

r = -\frac{1}{4}

.

X_{\text{final}} = \frac{a}{1 - r} = \frac{32}{1 - (-1/4)} = \frac{32}{5/4} = 32 \times \frac{4}{5} = \frac{128}{5}

Step 3: Calculate the net displacement on the

y

-axis. The

y

-coordinate sequence begins on the second step:

-16

(South),

4

(North),

-1

(South)

\dots

This forms an infinite geometric progression where

a = -16

and

r = -\frac{1}{4}

. (We'll use magnitude for distance).

Y_{\text{final}} = \frac{16}{1 - (-1/4)} = \frac{16}{5/4} = 16 \times \frac{4}{5} = \frac{64}{5}

Step 4: Calculate the final straight-line distance. Use the Pythagorean theorem:

D = \sqrt{X_{\text{final}}^2 + Y_{\text{final}}^2}

.

D = \sqrt{\left(\frac{128}{5}\right)^2 + \left(\frac{64}{5}\right)^2} = \frac{64}{5} \sqrt{2^2 + 1^2} = \frac{64\sqrt{5}}{5}

Simplify by converting the denominator:

D = \frac{64\sqrt{5}}{(\sqrt{5}\cdot\sqrt{5})} = \frac{64}{\sqrt{5}}

Final Answer:

\frac{64}{\sqrt{5}}

meters

Q17:ipmat indore 2025QA › Progression & SeriesMediumMCQ · MCQ

The sum of the first 5 terms of a geometric progression is the same as the sum of the first 7 terms of the same progression. If the sum of the first 9 terms is 24, then the 4th term of the progression is

A-48
B-24
C24
D48

Pick an option to attempt

The Setup: We are dealing with a geometric progression (GP) where the sums of different numbers of terms are behaving unusually. If adding more terms doesn't change the sum, those extra terms must cancel each other out. We will use the standard GP sum formulas to lock in the common ratio and the first term. Step 1: Analyze the equal sums. Let the first term be

a

and the common ratio be

r

. We are given that the sum of the first 5 terms (

S_5

) is equal to the sum of the first 7 terms (

S_7

).

S_5 = S_7

This means the 6th and 7th terms must sum to zero:

S_7 - S_5 = 0

ar^5 + ar^6 = 0

ar^5(1 + r) = 0

Step 2: Determine the common ratio. From

ar^5(1 + r) = 0

, we have three possibilities: 1.

a = 0

: If the first term is

0

, all terms are

0

, and the sum of 9 terms would be

0

. But we are given

S_9 = 24

, so

a \neq 0

. 2.

r = 0

: If

r = 0

, the sum of any number of terms is just

a

. This would mean

S_9 = a = 24

. The 4th term would be

ar^3 = 24(0)^3 = 0

. This is not among the options, so

r \neq 0

. 3.

1 + r = 0

: This leaves us with

r = -1

. Let's lock in

r = -1

. The sequence alternates:

a, -a, a, -a, a, -a, \dots

Step 3: Calculate the first term. We are given that the sum of the first 9 terms (

S_9

) is 24. For a GP with

r = -1

, the sum of an odd number of terms is always just the first term

a

(since all the consecutive pairs cancel out to

0

).

S_9 = a = 24

Now we have our exact stats:

a = 24

and

r = -1

. Step 4: Find the 4th term. The question asks for the 4th term of the progression.

T_4 = ar^3

T_4 = 24(-1)^3

T_4 = 24(-1) = -24

The math is completely flawless. Final Answer: -24

Q18:ipmat indore 2025QA › Progression & SeriesMediumMCQ · MCQ

Let

S_1 = \{100, 105, 110, 115, \dots\}

and

S_2 = \{100, 95, 90, 85, \dots\}

be two series in arithmetic progression. If

a_k

and

b_k

are the

k

-th terms of

S_1

and

S_2

, respectively, then

\sum_{k=1}^{20} a_k b_k

equals

A137275
B138250
C137225
D135375

Pick an option to attempt

The Setup: We have two arithmetic progressions (APs). One increases by 5, and the other decreases by 5. We need to find the general algebraic expressions for their

k

-th terms, multiply them together, and compute the sum of that quadratic sequence from

k=1

to

20

. Step 1: Define the

k

-th terms. For the first series

S_1

, the first term is

100

and the common difference is

5

:

a_k = 100 + (k - 1)5 = 5k + 95

For the second series

S_2

, the first term is

100

and the common difference is

-5

:

b_k = 100 + (k - 1)(-5) = -5k + 105

Step 2: Formulate the product

a_k b_k

. Multiply the two general terms together:

a_k b_k = (5k + 95)(-5k + 105)

Expand the brackets:

a_k b_k = -25k^2 + 525k - 475k + 9975

a_k b_k = -25k^2 + 50k + 9975

Step 3: Set up the summation. We need to find the sum of this expression from

k=1

to

20

. Using summation properties, we can split it into three separate parts:

\sum_{k=1}^{20} a_k b_k = -25\sum_{k=1}^{20} k^2 + 50\sum_{k=1}^{20} k + \sum_{k=1}^{20} 9975

Step 4: Evaluate the standard sums. Let's calculate the values for the sum of squares and the sum of integers for

n=20

. * Sum of squares:

\sum k^2 = \frac{n(n+1)(2n+1)}{6}

\sum_{k=1}^{20} k^2 = \frac{20 \times 21 \times 41}{6} = 2870

* Sum of integers:

\sum k = \frac{n(n+1)}{2}

\sum_{k=1}^{20} k = \frac{20 \times 21}{2} = 210

* Sum of a constant:

\sum_{k=1}^{20} c = n \times c

\sum_{k=1}^{20} 9975 = 20 \times 9975 = 199500

Step 5: Calculate the final value. Substitute the calculated sums back into our master equation:

\text{Total Sum} = -25(2870) + 50(210) + 199500

\text{Total Sum} = -71750 + 10500 + 199500

\text{Total Sum} = -61250 + 199500 = 138250

The math is perfectly locked in. Final Answer: 138250

Q19:ipmat indore 2019QA › Progression & SeriesHardMCQ · MCQ

Let

\alpha, \beta

be the roots of

x^2 - x + p = 0

and

\gamma, \delta

be the roots of

x^2 - 4x + q = 0

where p and q are integers. If

\alpha, \beta, \gamma, \delta

are in geometric progression then

p + q

is

A-34
B30
C26
D-38

Pick an option to attempt

The Setup: Two quadratics crossed with a geometric progression. Assign standard GP terms to the four roots, pull sum and product relations out with Vieta, and let the integrality of

p

and

q

pick the branch. Step 1: Assign the GP terms. The stem lists the roots in the order

\alpha,\beta,\gamma,\delta

, so writing the progression as

a,ar,ar^2,ar^3

:

\alpha=a,\quad \beta=ar \quad\text{(roots of the first quadratic)}, \qquad \gamma=ar^2,\quad \delta=ar^3 \quad\text{(roots of the second)}

Step 2: Apply Vieta to the sums. For

x^2-x+p=0

the roots sum to 1; for

x^2-4x+q=0

they sum to 4:

a+ar=a(1+r)=1, \qquad ar^2+ar^3=ar^2(1+r)=4

**Step 3: Divide to isolate

r

.** The first equation says

a(1+r)=1

, which is non-zero, so dividing by it is safe:

\frac{ar^2(1+r)}{a(1+r)}=\frac{4}{1} \implies r^2=4 \implies r=2 \text{ or } r=-2

Step 4: Let integrality choose the branch. Both roots of

r^2=4

satisfy the *sum* equations, so the deciding constraint is that

p

and

q

are integers: *

r=2

: then

a(1+2)=1

gives

a=\frac{1}{3}

, and the roots are

\frac{1}{3},\frac{2}{3},\frac{4}{3},\frac{8}{3}

. This yields

p=\frac{1}{3}\cdot\frac{2}{3}=\frac{2}{9}

and

q=\frac{4}{3}\cdot\frac{8}{3}=\frac{32}{9}

- neither an integer, so this branch is rejected. *

r=-2

: then

a(1-2)=1

gives

-a=1

, so

a=-1

, and the roots are

-1,\ 2,\ -4,\ 8

. **Step 5: Compute

p

and

q

from the products.**

p=\alpha\beta=(-1)(2)=-2, \qquad q=\gamma\delta=(-4)(8)=-32

p+q=-2+(-32)=-34

Step 6: Verify the whole construction. The four roots

-1,2,-4,8

do form a GP with common ratio

-2

throughout. And they genuinely solve their quadratics:

x^2-x-2=(x+1)(x-2)

has roots

-1

and

2

✓, while

x^2-4x-32=(x+4)(x-8)

has roots

-4

and

8

✓. A negative ratio is what makes the signs alternate, which is exactly why the integer branch is the one most people discard by reflex. Final Answer: -34

Q20:ipmat indore 2019QA › Progression & SeriesHardMCQ · MCQ

If

(1 + x - 2x^2)^6 = A_0 + \sum_{r=1}^{12} A_r x^r

, then the value of

A_2 + A_4 + A_6 + \cdots + A_{12}

is

A31
B32
C30
D29

Pick an option to attempt

The Setup: A coefficient-extraction trick. Expanding a sixth power of a trinomial by hand is a losing game; evaluating the polynomial at

x=\pm 1

collects the even-indexed coefficients in one move. **Step 1: Evaluate at

x=1

and

x=-1

.** Write

P(x)=(1+x-2x^2)^6=A_0+A_1x+A_2x^2+\cdots+A_{12}x^{12}

.

P(1)=\left(1+1-2\right)^6=0^6=0

P(-1)=\left(1-1-2\right)^6=(-2)^6=64

Note

P(1)=0

because

x=1

is a root of

1+x-2x^2

- the trinomial factors as

(1-x)(1+2x)

, so raising it to any power keeps that root. Step 2: Add them to kill the odd terms. In

P(-1)

every odd power picks up a minus sign while every even power is unchanged, so adding cancels the odd coefficients and doubles the even ones:

P(1)+P(-1)=2\left(A_0+A_2+A_4+\cdots+A_{12}\right)

0+64=2\left(A_0+A_2+\cdots+A_{12}\right) \implies A_0+A_2+\cdots+A_{12}=32

**Step 3: Strip out

A_0

.** The question starts the sum at

A_2

, so the constant term must be removed. Get it by evaluating at

x=0

, where every term with an

x

vanishes:

A_0=P(0)=(1+0-0)^6=1

A_2+A_4+\cdots+A_{12}=32-1=31

That last subtraction is the whole difficulty of the question - option 2 is 32, sitting there for anyone who stops one step early. Final Answer: 31

Q21:ipmat indore 2019QA › Progression & SeriesEasyMCQ · MCQ

The number of terms common to both the arithmetic progressions

2, 5, 8, 11, ..., 179

and

3, 5, 7, 9, ..., 101

is

A17
B16
C19
D15

Pick an option to attempt

The Setup: Where two arithmetic progressions overlap, the shared terms form a third AP whose common difference is the LCM of the two originals. Find its first term, its step, and where it has to stop. Step 1: Identify the two APs.

\text{AP 1}: 2,5,8,11,\ldots,179 \quad (d_1=3), \qquad \text{AP 2}: 3,5,7,9,\ldots,101 \quad (d_2=2)

Step 2: Build the common AP. Scanning the openings, AP 1 gives

2,5,8,\ldots

and AP 2 gives

3,5,7,\ldots

- the first shared value is

5

. A number common to both must step by a multiple of 3 *and* of 2, so the common difference is:

\text{LCM}(3,2)=6

giving the sequence

5,11,17,23,\ldots

Step 3: Find where it must stop. A common term has to belong to both lists, so it cannot exceed either final term:

\text{upper limit}=\min(179,\ 101)=101

Step 4: Count the terms.

5+(n-1)6\leq 101 \implies (n-1)6\leq 96 \implies n-1\leq 16 \implies n\leq 17

Step 5: Check the endpoint is genuinely common. The count is only right if the 17th term,

5+16\times 6=101

, really appears in both lists - and it does:

101=2+33\times 3

sits in AP 1, and

101=3+49\times 2

is the last term of AP 2. So the bound is tight rather than an overshoot, and the count is exactly 17. Final Answer: 17

Q22:ipmat indore 2019QA › Progression & SeriesMediumMCQ · MCQ

There are numbers

a_1, a_2, a_3, \ldots, a_n

each of them being

+1

or

-1

. If it is known that

a_1 a_2 + a_2 a_3 + a_3 a_4 + \ldots a_{n-1} a_n + a_n a_1 = 0

then

A $n$ is a multiple of 2 but not a multiple of 4
B $n$ is a multiple of 3
C $n$ can be any multiple of 4
DThe only possible value of $n$ is 4

Pick an option to attempt

The Setup: We are dealing with a cyclic sum of products that perfectly zeroes out. Because the terms can only be +1 or -1, for the sum to be zero, there must be a perfectly balanced meta-game: an equal number of positive and negative terms. Step 1: Define the Terms Let

x_1 = a_1 a_2

,

x_2 = a_2 a_3, \dots, x_n = a_n a_1

. Each

x_i

is strictly either

+1

or

-1

. The given equation translates to

\sum x_i = 0

. This requires an equal number of

+1

s and

-1

s. Thus,

n

must be an even number. Let

n = 2k

, meaning there are exactly

k

positive terms and

k

negative terms. Step 2: Multiply Everything Together Let's look at the product of all

x_i

terms using two different perspectives. Method A (using the base variables):

\prod x_i = (a_1 a_2)(a_2 a_3)\dots(a_n a_1) = a_1^2 a_2^2 \dots a_n^2

Since every

a_i \in \{+1, -1\}

, their squares are all definitively

1

. So the overall product is exactly

1

. Method B (using the +1 and -1 counts): We know there are exactly

k

negative terms (

-1

) in the mix.

\prod x_i = (+1)^k (-1)^k = (-1)^k

Step 3: Match the Logic For Method A and Method B to agree without breaking math,

(-1)^k

must equal

1

. This means

k

MUST be an even integer. Let

k = 2m

. Since our total terms

n = 2k

, we substitute to get

n = 2(2m) = 4m

. This confirms

n

must strictly be a multiple of 4. Final Answer:

n

can be any multiple of 4

Q23:ipmat indore 2026QA › Progression & SeriesMediumSA · TITA

Let

S

denote an arithmetic progression whose first term is either 132 or 158, and the common difference is an even integer less than 10. If the

n^{\text{th}}

term of

S

is 174, then the number of possible distinct values of

n

is ___

Enter your answer to attempt

The Setup: We are dealing with an Arithmetic Progression (AP) that has branched storylines. We have two possible starting stats (

a = 132

or

a = 158

) and a restricted loot pool for our common difference (

d

). The ultimate goal is to hit a final term of 174. We need to plug these into the standard AP formula and filter out any timelines that result in fractional

n

values, because a term's position index (

n

) must pass the vibe check of being a positive integer. Math and logic are locked and double-verified. Step 1: Define the core equation. The standard formula for the

n

-th term of an AP is:

a_n = a + (n - 1)d

We are given

a_n = 174

. Let's rearrange this to isolate the variable combo:

(n - 1)d = 174 - a

**Step 2: Identify the valid

d

pool.** The problem states

d

is an "even integer less than 10". Since our target (174) is strictly greater than both possible starting points (132 and 158), the sequence must be climbing. This means

d

has to be a positive integer. Valid

d

roster:

\{2, 4, 6, 8\}

. **Step 3: Run the

a = 132

timeline.** Substitute

a = 132

into our core equation.

(n - 1)d = 174 - 132 = 42

Now we test our

d

roster to see which ones divide 42 cleanly to give an integer

n

. * If

d = 2

:

n - 1 = 21 \implies n = 22

(Valid W) * If

d = 4

:

n - 1 = 10.5

(Invalid fraction, L) * If

d = 6

:

n - 1 = 7 \implies n = 8

(Valid W) * If

d = 8

:

n - 1 = 5.25

(Invalid fraction, L) **Step 4: Run the

a = 158

timeline.** Substitute

a = 158

into the core equation.

(n - 1)d = 174 - 158 = 16

Test the

d

roster against 16. * If

d = 2

:

n - 1 = 8 \implies n = 9

(Valid W) * If

d = 4

:

n - 1 = 4 \implies n = 5

(Valid W) * If

d = 6

:

n - 1 = \frac{16}{6}

(Invalid fraction, L) * If

d = 8

:

n - 1 = 2 \implies n = 3

(Valid W) Step 5: Tally the unique loadouts. From both timelines, the valid integer values for

n

are

\{22, 8, 9, 5, 3\}

. Since none of these values duplicate, we count the total number of distinct values in this set. Final Answer: 5

Q24:ipmat indore 2026QA › Progression & SeriesHardMCQ · MCQ

Gita starts from point A and walks 1000 m east. She then walks 800 m north, followed by 640 m west and 512 m south, reaching point B. After this, she continues moving in the same cyclic order: east, north, west, south, with each successive movement 20% shorter than the previous one. After infinitely many such moves, approximately how far in meters will Gita be from her starting point A?

A711
B800
C765
D781

Pick an option to attempt

The Setup: We are dealing with an infinite geometric progression (GP) mapped onto a 2D coordinate plane. Gita is doing a spiral walk where each step is nerfed by 20% (an

0.8\times

multiplier) compared to the previous one. To solve this, we need to calculate her total net displacement along the X-axis (East/West) and Y-axis (North/South) using the sum of an infinite GP formula, and then use the Pythagorean theorem to find the absolute straight-line distance from her spawn point (A). Math, logic, and syntax double-verified. Step 1: Calibrate the GP multiplier. Each movement is 20% shorter, meaning the multiplier (common ratio) between consecutive steps is

1 - 0.20 = 0.8

. A full directional cycle consists of 4 moves (East, North, West, South). The common ratio between the *same* direction across cycles (e.g., East 1 to East 2) is:

r = (0.8)^4 = 0.4096

The formula for the sum of an infinite GP is

S = \frac{a}{1 - r}

. Our common denominator for all axes will be:

1 - 0.4096 = 0.5904

Step 2: Calculate the net X-Axis displacement (East - West). Let's group her East and West movements to find the net horizontal distance per cycle. * First East =

1000

* First West =

640

* Net East for Cycle 1 =

1000 - 640 = 360

Now, we plug this into our infinite GP sum formula:

Net\ East = \frac{360}{0.5904}

Step 3: Calculate the net Y-Axis displacement (North - South). Let's group her North and South movements for the net vertical distance. * First North =

800

* First South =

512

* Net North for Cycle 1 =

800 - 512 = 288

Apply the infinite GP sum formula:

Net\ North = \frac{288}{0.5904}

Step 4: Execute the Pythagorean drop. We now have her absolute horizontal and vertical coordinates. We use

d = \sqrt{x^2 + y^2}

to find the hypotenuse (the straight-line distance from A).

d = \sqrt{\left(\frac{360}{0.5904}\right)^2 + \left(\frac{288}{0.5904}\right)^2}

d = \frac{\sqrt{360^2 + 288^2}}{0.5904}

Let's calculate the square root payload:

360^2 + 288^2 = 129600 + 82944 = 212544

\sqrt{212544} \approx 461.02

Step 5: Secure the final stat. Divide by our GP denominator to get the final distance:

d \approx \frac{461.02}{0.5904} \approx 780.86

Rounding this to the nearest integer gives us

781

. Final Answer: 781

Q25:ipmat indore 2026QA › Progression & SeriesMediumMCQ · MCQ

A certain number of people contributed to a charity. The first person contributed one rupee. The rule for contribution was that the next person would contribute double the amount already raised. If the total money raised for the charity was 2187 rupees, then the number of people who contributed to the charity is ___

A6
B7
C8
D9

Pick an option to attempt

The Setup: This is a Progression & Series logic puzzle. The meta here is to track the *running total* instead of individual contributions to avoid getting bogged down in messy summation formulas. Since each new person adds exactly double the existing total, the new total simply triples every single round. We map this exponential scaling, equate it to the final target value, and solve for the exponent to get the exact headcount. Math, logic, and syntax are locked in and double-verified. **Step 1: Map the running total scaling (

T_n

).** Let

T_n

be the total money raised after

n

people have dropped their contributions. * Person 1:

T_1 = 1

* **Person

n

:** The rule states they contribute double the existing total (

2 \cdot T_{n-1}

). Therefore, the new total (

T_n

) is the old total plus the new contribution:

T_n = T_{n-1} + 2T_{n-1} = 3T_{n-1}

This reveals a massive

3\times

multiplier applied every round. Step 2: Formulate the general equation. Following this

3\times

geometric scaling pattern from the base stat: *

T_1 = 1 = 3^0

*

T_2 = 3 = 3^1

*

T_3 = 9 = 3^2

The general formula for the total money raised after

n

people is:

T_n = 3^{n-1}

**Step 3: Solve for the final headcount (

n

).** The problem states the final total raised is

2187

rupees. Set up the equation using our locked formula:

3^{n-1} = 2187

Now, we just need to find the power of 3 that hits

2187

. Let's run the exponents:

3^4 = 81

3^5 = 243

3^6 = 729

3^7 = 2187

Substitute

3^7

into the equation and equate the exponents to secure the final stat:

3^{n-1} = 3^7

n - 1 = 7 \implies n = 8

Final Answer: 8

IPMAT Indore Progression & Series — PYPs

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