NCERT Solutions for Class 9th Maths Chapter 8 In-text Questions — Visualising an AP
Book page 182–183 Updated on2026-09-08
Q1.
Exercise: Verify that the following sequences are arithmetic progressions and write their nth terms. What do you observe when you plot the ordered pairs emerging from them? (i) 2, 5, 8, 11, … (ii) –5, –1, 3, 7, …
Answer
(i) 2, 5, 8, 11, … — differences 5 – 2 = 3, 8 – 5 = 3, 11 – 8 = 3. Constant, so it is an AP with a = 2, d = 3.
What the plot shows: plotting (1, 2), (2, 5), (3, 8), (4, 11) — and likewise (1, –5), (2, –1), (3, 3), (4, 7) — the points lie exactly on a straight line, just as in Fig. 8.4.
Left: an AP — equal steps up for equal steps right, so the points are collinear. Right: a GP for comparison — the same points bend upwards.
Why it happens: moving one step to the right raises y by d every single time, and a graph whose rise is the same for every unit of run is a straight line. In the language of Chapter 2, tn = a + (n – 1)d is a linear expression in n, and its graph is a line of slope d. So "AP" and "collinear points" are two descriptions of one fact.
Q2.
Exercise: Using the formula tn = a + (n – 1) × d, find the nth term of the following arithmetic progressions. (i) 1/2, 5/2, 9/2, 13/2, … (ii) 1.5, 3.5, 5.5, 7.5, …
Answer
(i) Here a = 1/2 and d = 5/2 – 1/2 = 4/2 = 2 (also 9/2 – 5/2 = 2 and 13/2 – 9/2 = 2).
Why it happens: both progressions have the same common difference, d = 2, and differ only in their first term — 0.5 against 1.5 — so the second is the first shifted up by 1. That is the point of the formula tn = a + (n – 1)d: fractions, decimals and negative numbers all go into it unchanged, because nothing in the derivation assumed the terms were whole numbers. Only the size of the step matters, never what kind of number the terms are.
Tip: always simplify to a single expression in n at the end. (4n – 3)/2 is easier to substitute into than 1/2 + (n – 1) × 2.
Q3.
Exercise: Find recursive rules for the APs in the previous exercises.
Answer
A recursive rule for an AP needs two things: the first term, and the instruction "add d".
AP
a
d
Recursive rule
2, 5, 8, 11, …
2
3
t1 = 2, tn = tn–1 + 3 for n ≥ 2
–5, –1, 3, 7, …
–5
4
t1 = –5, tn = tn–1 + 4 for n ≥ 2
1/2, 5/2, 9/2, 13/2, …
1/2
2
t1 = 1/2, tn = tn–1 + 2 for n ≥ 2
1.5, 3.5, 5.5, 7.5, …
1.5
2
t1 = 1.5, tn = tn–1 + 2 for n ≥ 2
Why it happens: the general AP has explicit rule tn = a + (n – 1)d and recursive rule t1 = a, tn = tn–1 + d. They agree because a + (n – 1)d minus a + (n – 2)d is exactly d. Notice the last two rows: identical recursive steps, different starting terms — which is why the starting term can never be left out. Without it the rule describes infinitely many different sequences.
Tip: the condition n ≥ 2 matters. The rule tn = tn–1 + d is meaningless at n = 1, because there is no t0.