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NCERT Solutions for Class 9th Maths Chapter 8 Exercise Set 8.2 — Sum of the First n Natural Numbers

Book page 185–186 Updated on2026-09-08

Q1.
Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ….
Answer

First term a = 3; common difference d = 8 – 3 = 5 (and 13 – 8 = 5, 18 – 13 = 5).

tn = a + (n – 1)d = 3 + (n – 1) × 5 = 5n – 2
t10 = 5 × 10 – 2 = 48
t26 = 5 × 26 – 2 = 128
Why it happens: the 26th term is 16 steps beyond the 10th, so it must exceed it by 16 × 5 = 80. And 48 + 80 = 128 — the two answers are consistent with each other, which is a stronger check than re-substituting into the same formula twice.
Q2.
Which term of the AP : 21, 18, 15, … is – 81? Also, is 0 a term of this AP? Give reasons for your answer.
Answer

Here a = 21 and d = 18 – 21 = –3, so the AP falls.

tn = 21 + (n – 1)(–3) = 21 – 3n + 3 = 24 – 3n
24 – 3n = –81
3n = 105
n = 35

So –81 is the 35th term. Check: 24 – 3 × 35 = 24 – 105 = –81. ✓

Is 0 a term? Solve 24 – 3n = 0.

3n = 24
n = 8

Yes — 0 is a term, the 8th term, because n = 8 is a natural number. Listing confirms it: 21, 18, 15, 12, 9, 6, 3, 0, –3, …

Why it happens: every term of this AP is a multiple of 3 (21 is, and subtracting 3 preserves that). 0 is a multiple of 3 and lies below the first term, so the falling sequence must land on it — and the equation says exactly where. A number like 10 is not a multiple of 3, so 24 – 3n = 10 would give n = 14/3, not a position, and 10 could never appear.
Tip: a negative common difference is perfectly normal. Do not force d to be positive by subtracting the wrong way round.
Q3.
Find the nth term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP.
Answer

a = 11 and d = 8 – 11 = –3 (also 5 – 8 = –3, 2 – 5 = –3).

tn = 11 + (n – 1)(–3)
= 11 – 3n + 3
= 14 – 3n
Check: t1 = 11, t2 = 8, t3 = 5, t4 = 2 ✓

Recursive rule:

t1 = 11,   tn = tn–1 – 3   for n ≥ 2
Why it happens: the two rules answer different questions. "What is the 40th term?" is a job for the explicit rule — 14 – 120 = –106, in one line. "What comes next?" is a job for the recursive rule — subtract 3. They must agree, and they do: 14 – 3n minus 14 – 3(n – 1) equals –3, which is precisely the step the recursive rule takes.
Q4.
An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term. (Hint: If ‘a’ is the first term and ‘d’ the common difference, then we arrive at the equations a + 2d = 12 and a + 49d = 106. Solve this pair of linear equations for ‘a’ and ‘d’.)
Answer

Write both facts using tn = a + (n – 1)d. The last term is the 50th, so n = 50 gives a + 49d.

a + 2d = 12    (1)
a + 49d = 106  (2)

Subtract (1) from (2) — this eliminates a:

47d = 94
d = 2
From (1): a + 4 = 12, so a = 8

Now the 29th term:

t29 = a + 28d = 8 + 28 × 2 = 8 + 56 = 64
Why it happens: an AP is completely pinned down by two numbers, a and d. So any two facts about it give two linear equations in a and d — and a pair of linear equations in two unknowns is what Chapter 2 taught you to solve. Subtracting is the efficient move here because a appears with coefficient 1 in both equations.
Check it yourself: with a = 8 and d = 2 the AP is 8, 10, 12, 14, … Its 3rd term is 12 ✓ and its 50th is 8 + 98 = 106 ✓.
Q5.
How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?
Answer

The 2-digit multiples of 3 run from 12 to 99, and they form an AP with d = 3.

12, 15, 18, …, 99
tn = 12 + (n – 1) × 3 = 3n + 9
3n + 9 = 99
3n = 90
n = 30

So there are 30 such numbers. For the sum, pair the ends as in Section 8.5:

S = 12 + 15 + … + 96 + 99
S = 99 + 96 + … + 15 + 12
2S = 111 + 111 + … + 111  (111 added 30 times)
2S = 30 × 111 = 3330
S = 1665
Why it happens: the reversal trick from Section 8.5 was never really about 1, 2, 3, … — it works for any AP, because when one line rises by d the other falls by d, so every column has the same total (first term + last term). Here that total is 12 + 99 = 111, and there are 30 columns. In words: sum = (number of terms) × (average of first and last term) = 30 × 55.5 = 1665.
Check it yourself: another route — the 2-digit multiples of 3 are 3 × 4, 3 × 5, …, 3 × 33, so the sum is 3 × (4 + 5 + … + 33) = 3 × (S33 – S3) = 3 × (561 – 6) = 3 × 555 = 1665. ✓
Q6.
Harish started work at an annual salary of ₹5,00,000 and received an increment of ₹20,000 each year. After how many years did his income reach ₹7,00,000?
Answer

His annual salaries form an AP with a = ₹5,00,000 and d = ₹20,000.

tn = 500000 + (n – 1) × 20000
500000 + (n – 1) × 20000 = 700000
(n – 1) × 20000 = 200000
n – 1 = 10
n = 11

His salary is ₹7,00,000 in his 11th year of work — that is, after 10 years of increments.

Why it happens: the first year carries no increment, so the number of increments is always one less than the year number. Reading (n – 1) = 10 as "10 raises have happened" is the whole content of the question; answering "11 years" to "after how many years" would silently count the starting year as a year of increase.
Check it yourself: ₹5,00,000 + 10 × ₹20,000 = ₹5,00,000 + ₹2,00,000 = ₹7,00,000. ✓
Q7.
A child arranges marbles in rows so that the first row has 1 marble, the second has 2 marbles, the third has 3, and so on up to 25 rows. How many marbles does the child use in all?
Answer

The total is 1 + 2 + 3 + … + 25 — the sum of the first 25 natural numbers.

S25 = n(n + 1)/2 with n = 25
= (25 × 26)/2
= 650/2
= 325 marbles
Why it happens: the arrangement is literally the triangular array of Fig. 8.1 with 25 rows, so the count is the 25th triangular number. Pairing gives the same answer without the formula: 1 + 25 = 26, 2 + 24 = 26, …, 12 + 14 = 26 — that is 12 pairs worth 26 each, plus the unpaired middle marble row of 13. So 12 × 26 + 13 = 312 + 13 = 325. ✓
Tip: the last row has 25 marbles, not 325. The question asks for the total, so it is a sum, not a term.
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