CBSE Class 11 Maths Word Problems on Arithmetic Progression ... - Free Printable

Let’s solve each question one by one, step by step. We’ll focus on Arithmetic Progression (A.P.) concepts — where numbers increase or decrease by a fixed amount called the “common difference”.

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Q1) The sides of a right-angled triangle are in A.P. Show that the sides are in the ratio 3:4:5.

Let the three sides be:
→ \( a - d \), \( a \), \( a + d \) (since they’re in A.P., we pick this symmetric form to make math easier)

Since it’s a right-angled triangle, the largest side is the hypotenuse → so \( a + d \) is the hypotenuse.

By Pythagoras theorem:

\[
(a - d)^2 + a^2 = (a + d)^2
\]

Expand all terms:

Left side:
\( (a^2 - 2ad + d^2) + a^2 = 2a^2 - 2ad + d^2 \)

Right side:
\( a^2 + 2ad + d^2 \)

Set them equal:

\[
2a^2 - 2ad + d^2 = a^2 + 2ad + d^2
\]

Subtract \( a^2 + d^2 \) from both sides:

\[
a^2 - 2ad = 2ad
\]

Bring all terms to one side:

\[
a^2 - 4ad = 0
\]

Factor:

\[
a(a - 4d) = 0
\]

So either \( a = 0 \) (not possible for a triangle side) or \( a = 4d \)

Now plug back into the sides:

→ First side: \( a - d = 4d - d = 3d \)
→ Second side: \( a = 4d \)
→ Third side: \( a + d = 4d + d = 5d \)

So sides are \( 3d : 4d : 5d \) → which simplifies to 3:4:5

✔ Proved.

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Q2) There are 20 rows of seats... first row 20 seats, second 21, third 22... total seats?

This is an A.P. with:

→ First term \( a = 20 \)
→ Common difference \( d = 1 \)
→ Number of terms \( n = 20 \)

Sum of A.P. formula:

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

Plug in:

\[
S_{20} = \frac{20}{2} [2(20) + (20 - 1)(1)] = 10 [40 + 19] = 10 × 59 = 590
\]

✔ Total seats = 590

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Q3) Interior angles of polygon in A.P. Sum = 120°? Wait — that can’t be right.

Wait — sum of interior angles of a polygon is usually much larger than 120°. Let me re-read.

It says: “The sum of angle is 120°” — but that must be a typo. Probably meant sum of angles = 1200°? Or maybe 120° per angle? No — let’s check context.

Actually, looking again: “The sum of angle is 120°” — that doesn’t make sense for a polygon unless it’s a triangle? But even then, sum is 180°.

Wait — perhaps it’s “the smallest angle is 120°”? That would make more sense.

But the problem says: “The sum of angle is 120° and common difference is 5°”

That must be wrong. Because even for a triangle, minimum sum is 180°. For quadrilateral, 360°, etc.

Perhaps it’s a misprint and should be sum = 1200°? Let’s assume that — because otherwise no solution exists.

Alternatively — maybe “sum of angles” refers to something else? Unlikely.

Wait — another possibility: Maybe it’s “each angle increases by 5°, starting from some value, and the total sum is S”, and we need to find number of sides.

But without correct sum, we can’t proceed.

Hold on — let me think differently.

Suppose the polygon has \( n \) sides → sum of interior angles = \( (n - 2) × 180° \)

Also, angles are in A.P. with common difference 5°, and let first angle be \( a \)

Then sum of angles = \( \frac{n}{2} [2a + (n - 1)×5] \)

Set equal:

\[
\frac{n}{2} [2a + 5(n - 1)] = (n - 2) × 180
\]

But we have two variables: \( a \) and \( n \). Need another condition.

Problem says: “The sum of angle is 120°” — still problematic.

Wait — perhaps it’s “the smallest angle is 120°”? That makes sense!

Let me assume that’s what was intended: “The smallest angle is 120° and common difference is 5°”

Then:

First angle \( a = 120° \)

Sum of angles = \( \frac{n}{2} [2×120 + (n - 1)×5] = \frac{n}{2} [240 + 5n - 5] = \frac{n}{2} (5n + 235) \)

Also, sum = \( (n - 2) × 180 \)

Set equal:

\[
\frac{n}{2} (5n + 235) = 180(n - 2)
\]

Multiply both sides by 2:

\[
n(5n + 235) = 360(n - 2)
\]

Expand:

Left: \( 5n^2 + 235n \)
Right: \( 360n - 720 \)

Bring all to left:

\[
5n^2 + 235n - 360n + 720 = 0
→ 5n^2 - 125n + 720 = 0
\]

Divide by 5:

\[
n^2 - 25n + 144 = 0
\]

Factor:

Find two numbers multiplying to 144, adding to 25 → 16 and 9

\[
(n - 16)(n - 9) = 0
\]

So \( n = 16 \) or \( n = 9 \)

Check which one works.

If \( n = 9 \):

Last angle = \( a + (n-1)d = 120 + 8×5 = 160° \) → valid (less than 180° for convex polygon)

Sum = \( \frac{9}{2} [2×120 + 8×5] = \frac{9}{2} [240 + 40] = \frac{9}{2} × 280 = 9 × 140 = 1260° \)

Actual sum for 9-gon: \( (9-2)×180 = 7×180 = 1260° \) ✔

If \( n = 16 \):

Last angle = \( 120 + 15×5 = 195° \) → greater than 180° → not possible for convex polygon (usually assumed unless stated)

So likely answer is 9 sides

But wait — original problem said “sum of angle is 120°” — which contradicts everything.

Perhaps it’s a typo and should be “smallest angle is 120°” — which gives us n=9.

I’ll go with that assumption since otherwise problem is unsolvable.

✔ Answer: 9 sides

*(Note: If the problem truly meant sum = 120°, then no such polygon exists — so we assume typo.)*

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Q4) 1st and 2nd terms of A.P. are 1 and –3. Find nth term and sum of first n terms.

Given:

→ \( a_1 = 1 \)
→ \( a_2 = -3 \)

Common difference \( d = a_2 - a_1 = -3 - 1 = -4 \)

nth term formula: \( a_n = a + (n - 1)d \)

So:

\[
a_n = 1 + (n - 1)(-4) = 1 - 4n + 4 = 5 - 4n
\]

Sum of first n terms:

\[
S_n = \frac{n}{2} [2a + (n - 1)d] = \frac{n}{2} [2(1) + (n - 1)(-4)] = \frac{n}{2} [2 - 4n + 4] = \frac{n}{2} (6 - 4n)
\]

Simplify:

\[
S_n = \frac{n}{2} × 2(3 - 2n) = n(3 - 2n) = 3n - 2n^2
\]

✔ nth term: \( 5 - 4n \)
✔ Sum: \( 3n - 2n^2 \)

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Q5) First day attendance 750, declines by 50 every day. When will attendance be zero?

This is A.P.:

→ First term \( a = 750 \)
→ Common difference \( d = -50 \) (declining)

We want to find when \( a_n = 0 \)

Formula: \( a_n = a + (n - 1)d \)

Set to 0:

\[
750 + (n - 1)(-50) = 0
\]

\[
750 - 50(n - 1) = 0
\]

\[
50(n - 1) = 750
\]

\[
n - 1 = 15
\]

\[
n = 16
\]

So on the 16th day, attendance becomes 0.

✔ Answer: Day 16

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Q6) Sum of integers from 1 to 1000

This is A.P. with:

→ \( a = 1 \), \( l = 1000 \), \( n = 1000 \)

Sum = \( \frac{n}{2} (a + l) = \frac{1000}{2} (1 + 1000) = 500 × 1001 = 500500 \)

✔ Answer: 500500

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Q7) Sum of 24 terms of A.P. Given: \( a_1 + a_5 + a_{10} + a_{15} + a_{20} + a_{24} = 225 \)

In A.P., \( a_k = a + (k - 1)d \)

So:

\( a_1 = a \)
\( a_5 = a + 4d \)
\( a_{10} = a + 9d \)
\( a_{15} = a + 14d \)
\( a_{20} = a + 19d \)
\( a_{24} = a + 23d \)

Add them:

Total = \( 6a + (4 + 9 + 14 + 19 + 23)d \)

Calculate coefficients:

4+9=13; 13+14=27; 27+19=46; 46+23=69

So:

\[
6a + 69d = 225
\]

Divide entire equation by 3:

\[
2a + 23d = 75 \quad \text{(Equation 1)}
\]

Now, sum of 24 terms:

\[
S_{24} = \frac{24}{2} [2a + (24 - 1)d] = 12 [2a + 23d]
\]

From Equation 1: \( 2a + 23d = 75 \)

So:

\[
S_{24} = 12 × 75 = 900
\]

✔ Answer: 900

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Q8) Gate receipts Rs.9500 on first day... (problem cut off)

The problem is incomplete — it stops at “amounted to Rs.9500 on the first”. Without knowing how it changes (increases/decreases?) or over how many days, we cannot solve.

Assuming it’s similar to Q5 — maybe daily receipt decreases/increases by fixed amount?

But since it’s cut off, I’ll skip or note as incomplete.

However, if we assume it’s like Q5 — say, decreases by fixed amount until zero — but no info given.

Since user didn’t provide full question, I’ll leave it out.

But for completeness — if you meant same as Q5 pattern, please clarify.

For now, skipping Q8 due to incomplete data.

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Final Answers compiled below.

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Final Answer:
Q1) Sides are in ratio 3:4:5 (proved)
Q2) 590 seats
Q3) 9 sides (assuming smallest angle is 120°)
Q4) nth term: \( 5 - 4n \); Sum: \( 3n - 2n^2 \)
Q5) Day 16
Q6) 500500
Q7) 900
Q8) Incomplete question — cannot solve