NCERT Solutions for Class 9th Maths Chapter 8 Chapter 8 review (questions marked * are the harder set) — End-of-Chapter Exercises

Book page 194–195 Updated on2026-09-08

Q1.

Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.

Answer

Two terms of an AP give two equations in a and d.

a + 10d = 38 (1)
a + 15d = 73 (2)
(2) – (1): 5d = 35, so d = 7
From (1): a = 38 – 70 = –32
t₃₁ = a + 30d = –32 + 210 = 178

Shorter: t₃₁ is 15 steps past t₁₆, so t₃₁ = 73 + 15 × 7 = 73 + 105 = 178. ✓

Why it happens: in an AP the difference between the mth and nth terms is (m – n)d, with a cancelling out. Here t₁₆ – t₁₁ = 5d = 35 straight away. The full solve is worth doing once to see that a = –32 — the AP starts below zero and only becomes positive later.

Q2.

Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.

Answer

"The 7th term exceeds the 5th by 12" means t₇ – t₅ = 12, and those two terms are two steps apart.

t₇ – t₅ = (a + 6d) – (a + 4d) = 2d
2d = 12
d = 6
t₃ = a + 2d = 16 → a + 12 = 16 → a = 4

The AP is 4, 10, 16, 22, 28, 34, 40, …

Check: the 3rd term is 16 ✓; t₇ = 40 and t₅ = 28, and 40 – 28 = 12 ✓.

Why it happens: the phrase "exceeds by" is a difference of terms, and in an AP any such difference is a whole number of d's — never involving a. That is why the second condition fixes d on its own, leaving the first condition free to fix a. "Determine the AP" means state a and d, or list enough terms that the pattern is unambiguous.

Q3.

How many three-digit numbers are divisible by 7? (Hint: All three-digit numbers divisible by 7 form an AP. Find the smallest and largest such three-digit numbers.)

Answer

Find the two ends first.

Smallest: 100 ÷ 7 = 14.28…, so the first multiple of 7 at or above 100 is 7 × 15 = 105
Largest: 999 ÷ 7 = 142.71…, so the last one at or below 999 is 7 × 142 = 994

These form an AP with a = 105, d = 7, last term 994.

105 + (n – 1) × 7 = 994
(n – 1) × 7 = 889
n – 1 = 127
n = 128

There are 128 three-digit numbers divisible by 7.

Why it happens: the multiples of 7 are 7 × 15, 7 × 16, …, 7 × 142, so counting them is counting the whole numbers from 15 to 142 — that is 142 – 15 + 1 = 128. The "+1" is the fencepost correction, and it is the same correction that appears as the "– 1" in t_n = a + (n – 1)d. Both routes must agree, and they do.

Tip: counting the numbers from 15 to 142 as 142 – 15 = 127 forgets one end. Always test the idea on a tiny case: from 3 to 5 there are 3 numbers, not 2.

Q4.

How many multiples of 4 lie between 10 and 250? (Hint: All multiples of 4 form an AP. Find the smallest and largest multiples of 4 between 10 and 250.)

Answer

The multiples of 4 strictly between 10 and 250 run from 12 to 248.

Smallest: 4 × 3 = 12 (since 4 × 2 = 8 is below 10)
Largest: 4 × 62 = 248 (since 4 × 63 = 252 is above 250)
12 + (n – 1) × 4 = 248
(n – 1) × 4 = 236
n – 1 = 59
n = 60

There are 60 multiples of 4 between 10 and 250.

Why it happens: as in the previous question, the multiples of 4 here are 4 × 3 up to 4 × 62, so the count is 62 – 3 + 1 = 60. Note that 250 is not itself a multiple of 4, so the word "between" causes no trouble at that end; had the limit been 248, you would have to decide whether to include it.

Q5.

Find a GP for which the sum of the first two terms is – 4 and the fifth term is 4 times the third term.

Answer

Let the GP be a, ar, ar², … Translate both conditions.

a + ar = –4 i.e. a(1 + r) = –4 (1)
t₅ = 4 t₃ → ar⁴ = 4ar²
r² = 4 (dividing by ar², which is not 0)
r = 2 or r = –2

Case r = 2: a(1 + 2) = –4 gives 3a = –4, a = –4/3.

GP: –4/3, –8/3, –16/3, –32/3, …
Check: –4/3 – 8/3 = –12/3 = –4 ✓; t₅ = –64/3 and 4 t₃ = 4 × (–16/3) = –64/3 ✓

Case r = –2: a(1 – 2) = –4 gives –a = –4, a = 4.

GP: 4, –8, 16, –32, 64, …
Check: 4 + (–8) = –4 ✓; t₅ = 64 and 4 t₃ = 4 × 16 = 64 ✓

Both are valid answers: –4/3, –8/3, –16/3, … and 4, –8, 16, –32, …

Why it happens: the condition t₅ = 4t₃ reduces to r² = 4, and a squared quantity has two square roots. Nothing in the problem rules out a negative ratio, so both must be reported. Dividing by ar² is legitimate here because a = 0 would make every term 0 and the sum could not be –4.

Q6.

Find all possible ways of expressing 100 as the sum of consecutive natural numbers.

Answer

Suppose the sum has k terms starting at a. Using the AP sum idea, the total is k terms whose average is (first + last)/2:

a + (a + 1) + … + (a + k – 1) = k × [2a + (k – 1)]/2 = 100
so k(2a + k – 1) = 200

Now k and (2a + k – 1) are two factors of 200. They always have opposite parity: if k is odd, k – 1 is even so 2a + k – 1 is even, and vice versa. So one factor must be odd. The odd factors of 200 = 2³ × 5² are 1, 5 and 25.

k	2a + k – 1	a	Valid?
1	200	100	the single number 100 — not a sum
5	40	(40 – 4)/2 = 18	yes
8	25	(25 – 7)/2 = 9	yes
25	8	negative	rejected
40	5	negative	rejected
200	1	negative	rejected

100 = 18 + 19 + 20 + 21 + 22
100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16

Both check out: 18 + 22 = 40, and 5 terms averaging 20 give 100; 9 + 16 = 25, and 8 terms averaging 12.5 give 100. So there are exactly two ways (three, if the single number 100 is allowed to count).

Why it happens: the parity argument is what makes the search finite. Because k and 2a + k – 1 can never both be even, every valid k pairs with an odd factor of 200 — and 200 has only three odd factors. This is also why a power of 2 such as 64 cannot be written as a sum of consecutive natural numbers at all: its only odd factor is 1.

Check it yourself: 18 + 19 + 20 + 21 + 22 = 100 and 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100. Add them and see.

Q7.

The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the 2nd hour, 4th hour and nth hour?

Answer

Doubling every hour is a GP with a = 30 and r = 2, counting from hour 0.

Start (hour 0): 30
End of 1st hour: 30 × 2 = 60
End of 2nd hour: 30 × 2² = 120
End of 3rd hour: 30 × 2³ = 240
End of 4th hour: 30 × 2⁴ = 480
End of nth hour: 30 × 2ⁿ

Why it happens: the exponent counts how many doublings have taken place, and after n hours there have been exactly n of them — so the exponent is n, not n – 1. The n – 1 appears in t_n = ar^n–1 only because there the first term is labelled n = 1; here the starting count is labelled hour 0. Match the exponent to the number of steps actually taken, and this never goes wrong.

Did you know? at this rate a single day of growth means 2²⁴ doublings — the 30 bacteria would become 30 × 1,67,77,216, over fifty crore.

Q8.

The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Answer

Write each term as a + (n – 1)d and add.

t₄ + t₈ = (a + 3d) + (a + 7d) = 2a + 10d = 24 → a + 5d = 12 (1)
t₆ + t₁₀ = (a + 5d) + (a + 9d) = 2a + 14d = 44 → a + 7d = 22 (2)
(2) – (1): 2d = 10, so d = 5
From (1): a + 25 = 12, so a = –13

The first three terms are –13, –8, –3.

Check: t₄ = –13 + 15 = 2 and t₈ = –13 + 35 = 22, sum 24 ✓. t₆ = –13 + 25 = 12 and t₁₀ = –13 + 45 = 32, sum 44 ✓.

Why it happens: dividing each equation by 2 is worth doing early — it turns 2a + 10d = 24 into a + 5d = 12, which is just t₆ = 12. In fact the sum of two terms equally spaced about a middle term is always twice that middle term, so the two conditions say t₆ = 12 and t₈ = 22 directly, and d = (22 – 12)/2 = 5 in one line.

Q9.

Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.

Answer

Require S_n = n(n + 1)/2 > 1000, that is n(n + 1) > 2000.

n = 44: 44 × 45 = 1980, which is not more than 2000 → S₄₄ = 990
n = 45: 45 × 46 = 2070, which is more than 2000 → S₄₅ = 1035

So the smallest such value is n = 45, giving S₄₅ = 1035 > 1000, while S₄₄ = 990 < 1000.

Why it happens: S_n is roughly n²/2, so n² must be near 2000 and n near √2000 ≈ 44.7. That estimate tells you where to look; the two exact checks at n = 44 and n = 45 are what settle it. For a "smallest n" question you must always show that the value below fails — otherwise you have not proved it is the smallest.

Q10.

Which term of the GP: 2, 8, 32, … is 131072? Write the explicit formula as well as the recursive formula for the nth term.

Answer

a = 2 and r = 8 ÷ 2 = 4 (also 32 ÷ 8 = 4).

Explicit: t_n = 2 × 4^n–1
Recursive: t₁ = 2, t_n = 4 × t_n–1 for n ≥ 2

Solve t_n = 131072 by turning everything into powers of 2:

2 × 4^n–1 = 131072
4^n–1 = 65536
4^n–1 = 2^2(n–1) and 65536 = 2¹⁶
2(n – 1) = 16
n – 1 = 8
n = 9

So 131072 is the 9th term. Check: 2 × 4⁸ = 2 × 65536 = 131072. ✓

Why it happens: both 2 and 4 are powers of the same base, so the equation can be reduced to matching exponents of 2 — a step that removes all guesswork. Had the numbers not shared a base, you would have to build the terms up one at a time: 2, 8, 32, 128, 512, 2048, 8192, 32768, 131072 — the ninth. ✓

Q11.

The sum of the first three terms of a GP is 13/12 and their product is –1. Find the common ratio and the terms.

Answer

Choosing the three terms as a/r, a, ar makes the product collapse.

Product: (a/r) × a × (ar) = a³ = –1
a = –1

Now use the sum:

–1/r – 1 – r = 13/12
–(1/r + r) = 13/12 + 1 = 25/12
1/r + r = –25/12
Multiply by 12r: 12 + 12r² = –25r
12r² + 25r + 12 = 0

Factorise: 12r² + 16r + 9r + 12 = 4r(3r + 4) + 3(3r + 4) = (3r + 4)(4r + 3).

r = –4/3 or r = –3/4

With a = –1 and r = –3/4 the three terms are a/r = 4/3, a = –1, ar = 3/4:

4/3, –1, 3/4
Sum = 16/12 – 12/12 + 9/12 = 13/12 ✓
Product = (4/3) × (–1) × (3/4) = –1 ✓

The other root r = –4/3 gives the same three numbers in reverse order: 3/4, –1, 4/3.

Why it happens: writing the terms as a/r, a, ar puts the middle term at the centre, so r cancels in the product and one unknown disappears immediately. The two roots are reciprocals of each other because reversing a GP replaces r by 1/r — they are not two different progressions, only two ways of reading the same one.

Tip: the product being negative forces an odd number of negative terms. With r negative, the middle term is the negative one — which is exactly what a = –1 says.

Q12.

If the 4th, 10th and 16th terms of a GP are x, y and z respectively, prove that x, y, z are in GP.

Answer

Let the GP have first term a and common ratio r.

x = t₄ = ar³
y = t₁₀ = ar⁹
z = t₁₆ = ar¹⁵

Compare the two ratios of consecutive members of x, y, z:

y/x = ar⁹ ÷ ar³ = r⁶
z/y = ar¹⁵ ÷ ar⁹ = r⁶
So y/x = z/y = r⁶

The ratio of consecutive terms is the same constant, so x, y, z form a GP with common ratio r⁶. Equivalently y² = a²r¹⁸ = (ar³)(ar¹⁵) = xz, which is the same statement.

Why it happens: the positions 4, 10, 16 are themselves equally spaced — they form an AP with common difference 6. Picking equally spaced terms out of a GP always leaves a GP, because moving 6 places along multiplies by r⁶ every time. This is the multiplicative twin of a familiar fact about APs: equally spaced terms of an AP form an AP.

Tip: in a proof, do not stop at "the ratios look equal". Compute both ratios from the definition and show they are the same expression — that is what makes it a proof.

Q13.

The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Find the terms of the GP.

Answer

Let the terms be a, ar, ar².

a(1 + r + r²) = 26 (1)
a²(1 + r² + r⁴) = 364 (2)

The key is the identity 1 + r² + r⁴ = (1 + r + r²)(1 – r + r²). Substituting into (2):

a(1 + r + r²) × a(1 – r + r²) = 364
26 × a(1 – r + r²) = 364
a(1 – r + r²) = 14 (3)

Subtract (3) from (1), then add them:

(1) – (3): 2ar = 12 → ar = 6
(1) + (3): 2a(1 + r²) = 40 → a + ar² = 20

Put a = 6/r into a + ar² = 20:

6/r + 6r = 20
6 + 6r² = 20r
3r² – 10r + 3 = 0
(3r – 1)(r – 3) = 0
r = 3 or r = 1/3

With r = 3, ar = 6 gives a = 2, so the terms are 2, 6, 18. With r = 1/3 they are 18, 6, 2 — the same three numbers reversed.

Check: 2 + 6 + 18 = 26 ✓
4 + 36 + 324 = 364 ✓

Why it happens: squaring the terms of a GP with ratio r gives a GP with ratio r², which is why the second condition involves 1 + r² + r⁴. The factorisation 1 + r² + r⁴ = (1 + r + r²)(1 – r + r²) — an identity of the kind met in Chapter 4 — lets you divide equation (2) by equation (1) and drop from degree four to degree two in one move.

Q14.

Suppose P1 = 1, P2 = 2 and for n > 2, Pn = P1 + P2 + … + Pn–1 + 1. Find the values of P1, P2, …, P8. Can you find a simpler recursive formula for Pn? Can you give an explicit formula?

Answer

Each term is 1 more than the sum of everything before it.

P₁ = 1, P₂ = 2
P₃ = 1 + 2 + 1 = 4
P₄ = 1 + 2 + 4 + 1 = 8
P₅ = 1 + 2 + 4 + 8 + 1 = 16
P₆ = 1 + 2 + 4 + 8 + 16 + 1 = 32
P₇ = 64 P₈ = 128

So the values are 1, 2, 4, 8, 16, 32, 64, 128.

A simpler recursive formula. Compare two consecutive terms, for n > 3:

P_n = (P₁ + … + P_n–2 + P_n–1) + 1
P_n–1 = (P₁ + … + P_n–2) + 1
Subtracting: P_n – P_n–1 = P_n–1
so P_n = 2 P_n–1

This also holds at n = 3, since P₃ = 4 = 2 × 2 = 2P₂. So P₁ = 1, P_n = 2P_n–1 for n ≥ 2 — a GP with a = 1 and r = 2.

Explicit formula: P_n = 2^n–1
Check: P₁ = 2⁰ = 1, P₅ = 2⁴ = 16, P₈ = 2⁷ = 128 ✓

Why it happens: the rule as printed forces you to add up every earlier term, so computing P₂₀ would mean nineteen additions. Subtracting consecutive statements of the rule cancels the whole long sum at once and leaves a two-term relation. Behind it is a neat fact about powers of 2: 1 + 2 + 4 + … + 2^k = 2^k+1 – 1, so adding 1 to that sum gives exactly the next power of 2.

Tip: "find a simpler recursive formula" is an invitation to subtract the rule from itself one step later. It works whenever the rule contains a running sum.

Q15.

Suppose W1 = 1, W2 = 2 and for n > 2, Wn = W1 + W2 + … + Wn–2 + 2. Find the values of W1, W2, …, W8. Do you recognise this sequence?

Answer

Note that this sum stops at W_n–2, not W_n–1 — it leaves out the term immediately before.

W₁ = 1, W₂ = 2
W₃ = W₁ + 2 = 1 + 2 = 3
W₄ = W₁ + W₂ + 2 = 1 + 2 + 2 = 5
W₅ = 1 + 2 + 3 + 2 = 8
W₆ = 1 + 2 + 3 + 5 + 2 = 13
W₇ = 1 + 2 + 3 + 5 + 8 + 2 = 21
W₈ = 1 + 2 + 3 + 5 + 8 + 13 + 2 = 34

The sequence is 1, 2, 3, 5, 8, 13, 21, 34 — the Virahānka–Fibonacci sequence.

Here is why, and not merely that it looks familiar. Subtract the rule at n – 1 from the rule at n:

W_n = (W₁ + … + W_n–3 + W_n–2) + 2
W_n–1 = (W₁ + … + W_n–3) + 2
Subtracting: W_n – W_n–1 = W_n–2
that is W_n = W_n–1 + W_n–2

With W₁ = 1 and W₂ = 2 this is exactly the rule V_n = V_n–1 + V_n–2 of page 179, so the two sequences are the same one.

Why it happens: stopping the sum two places back instead of one is what produces the Virahānka rule. The same subtraction trick as in the previous question turns a rule involving all earlier terms into a rule involving just two — and it shows that a sequence can be defined in more than one way while remaining the same sequence. Checking eight terms suggests the answer; the subtraction proves it for every n.

Did you know? the identity behind this is 1 + 2 + 3 + 5 + … + V_n–2 + 2 = V_n: the sum of the Virahānka numbers up to any point, plus 2, is the term two places later.

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