Match the entries in Column I with Column II. Column I: (a) 4, 1, (1)/(4), (1)/(16); (b) 2, 3, 5, 7; (c) 13, 8

NCERT Exemplar · Q35

Q. Match the entries in Column I with Column II. Column I:

(a)

4, 1, \dfrac{1}{4}, \dfrac{1}{16}

;

(b)

2, 3, 5, 7

;

(c)

13, 8, 3, -2, -7

. Column II:

(i) A.P.;

(ii) sequence;

(iii) G.P.

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Concept understanding — Geometric Progression Arithmetic Progression

Comparing Arithmetic and Geometric Progressions

When you look at a sequence of numbers, the very first question to ask is: how does one term turn into the next? Two patterns cover almost every sequence you'll meet in Class 11 — adding a fixed number each time, or multiplying by a fixed number each time.

Two different rules of motion

An Arithmetic Progression (AP) moves by addition: each term is the previous term plus a constant common difference $d$ . If the first term is $a$ , the sequence is $a,\ a+d,\ a+2d,\ \dots$ , so the growth is a straight line — linear.
A Geometric Progression (GP) moves by multiplication: each term is the previous term times a constant common ratio $r$ . If the first term is $a$ , the sequence is $a,\ ar,\ ar^2,\ \dots$ , so the growth curves sharply upward (or shrinks toward zero) — exponential, not linear.

Note

This page is not the place to re-derive the individual formulas from scratch — the full nth-term and sum derivations for AP live under the Arithmetic Progression concept, and the equivalent derivations for GP live under the Geometric Progression concept. This page exists purely to help you tell the two patterns apart and choose the right one quickly.

The fastest way to tell them apart

Take any three consecutive terms $t_1, t_2, t_3$ of the sequence you're given:

Test for AP: compute $t_2-t_1$ and $t_3-t_2$ . If they're equal, you have an AP, and that equal value is $d$ .
Test for GP: compute $\dfrac{t_2}{t_1}$ and $\dfrac{t_3}{t_2}$ . If they're equal, you have a GP, and that equal value is $r$ .

If neither test succeeds, the sequence is neither an AP nor a GP on its own — it may still be built from a combination of the two (an arithmetico-geometric sequence), which needs its own separate technique.

Side-by-side

Feature	Arithmetic Progression	Geometric Progression
Rule to get the next term	Add $d$	Multiply by $r$
nth term	$a+(n-1)d$	$a\,r^{\,n-1}$
Sum of $n$ terms	$\dfrac{n}{2}[2a+(n-1)d]$	$a\,\dfrac{r^n-1}{r-1}$ (for $r\neq1$ )
Shape of growth	Straight line	Curve (exponential)
Allowed values of $d$ / $r$	Any real number, including $0$	Any real number except $0$

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The key idea is to classify each given list of numbers by checking whether it forms an Arithmetic Progression (A.P.), a Geometric Progression (G.P.), or is simply a sequence. The matches are: (a) → (iii) G.P., (b) → (ii) sequence, (c) → (i) A.P.

Why this approach works.

A sequence is any ordered list of numbers. An Arithmetic Progression has a constant difference between consecutive terms. A Geometric Progression has a constant ratio between consecutive terms. So for each list, we check the pattern of differences and ratios.

List (a): $4, 1, \frac{1}{4}, \frac{1}{16}$

Check differences: $1 - 4 = -3$ , $\frac{1}{4} - 1 = -\frac{3}{4}$ , $\frac{1}{16} - \frac{1}{4} = -\frac{3}{16}$ . Not constant, so not an A.P.

Check ratios: $\frac{1}{4} = \frac{1}{4}$ , $\frac{1/4}{1} = \frac{1}{4}$ , $\frac{1/16}{1/4} = \frac{1}{4}$ . Constant ratio $\frac{1}{4}$ . So it is a G.P.

Match: (a) → (iii).
List (b): $2, 3, 5, 7$

Differences: $3-2=1$ , $5-3=2$ , $7-5=2$ . Not constant, so not an A.P.

Ratios: $\frac{3}{2}=1.5$ , $\frac{5}{3}\approx 1.667$ , $\frac{7}{5}=1.4$ . Not constant, so not a G.P.

It is simply a sequence (a list of numbers with no special progression).

Match: (b) → (ii).
List (c): $13, 8, 3, -2, -7$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AHSEC Higher Secondary (HS) 1st Year Examination 2018Set ANNUAL4 marks
$a, b, c$
(i) $1.2.3 + 2.3.4 + 3.4.5 + \cdots + n(n+1)(n+2)$
(ii) $1^2-2^2+3^2-4^2+5^2-6^2+\cdots+1001^2-1002^2$
›Reveal solutionSolution

Option A proves a GP relation from given AP/GP conditions; Option B evaluates two series sums.

Option A: $a,b,c$ in AP $\implies c=2b-a$ . $a,b,d$ in GP $\implies b^2=ad \implies d=\dfrac{b^2}{a}$ .

To show $a,\ a-b,\ d-c$ are in GP, we must show $(a-b)^2 = a(d-c)$ .

$d-c = \dfrac{b^2}{a}-(2b-a) = \dfrac{b^2-2ab+a^2}{a} = \dfrac{(a-b)^2}{a}.$

$a(d-c) = a\cdot\dfrac{(a-b)^2}{a} = (a-b)^2.$

This exactly equals $(a-b)^2$ , confirming the GP condition. So $a,\,a-b,\,d-c$ are in GP.

Option B (i): $\sum_{r=1}^n r(r+1)(r+2)$ . Expand $r(r+1)(r+2)=r^3+3r^2+2r$ :

$\sum r^3+3\sum r^2+2\sum r = \left[\dfrac{n(n+1)}2\right]^2+3\cdot\dfrac{n(n+1)(2n+1)}6+2\cdot\dfrac{n(n+1)}2.$

Factoring out $\dfrac{n(n+1)}4$ : …

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