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Arithmetic Progressions
EXERCISE 5.1
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. (i) 2, 4, 8, 16, . . . (ii) 5 7 2, , 3, , 2 2 . . . (iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . (iv) – 10, – 6, – 2, 2, . . . (v) 3, 3 2  , 3 2 2  , 3 3 2 ,  . . . (vi) 0.2, 0.22, 0.222, 0.2222, . . . (vii) 0, – 4, – 8, –12, . . . (viii) – 1 2 , – 1 2 , – 1 2 , – 1 2 , . . . 56 (ix) 1, 3, 9, 27, . . . (x) a, 2a, 3a, 4a, . . . (xi) a, a2, a3, a4, . . . (xii) 2, 8, 18 , 32, . . . (xiii) 3, 6, 9 , 12 , . . . (xiv) 12, 32, 52, 72, . . . (xv) 12, 52, 72, 73, . . .

Step-by-Step Solution

Key Concept: An arithmetic progression (AP) is a sequence of numbers in which the difference between any two successive terms is constant. This constant difference is called the common difference \(d\). For a sequence \(a_1, a_2, a_3, \dots\) to be an AP, we must have \(a_{k+1} - a_k = d\) for all \(k\). If the sequence is an AP, the next three terms are obtained by repeatedly adding \(d\) to the last given term.
1. Check each given sequence
- Compute the successive differences.
- If all the differences are equal, the sequence is an AP; otherwise it is not.
- For an AP, record the common difference \(d\) and generate three more terms by adding \(d\) to the last term repeatedly.

2. Individual items
- (i) 2, 4, 8, 16, …
\[\Delta_1 = 4-2 = 2,\; \Delta_2 = 8-4 = 4,\; \Delta_3 = 16-8 = 8\]
The differences are not equal ⇒ Not an AP.

- (ii) 5, \frac{7}{2}, 3, \frac{2}{2}, … (interpreted as 5, 3.5, 3, 1)
\[\Delta_1 = 3.5-5 = -1.5,\; \Delta_2 = 3-3.5 = -0.5,\; \Delta_3 = 1-3 = -2\]
Differences are not equal ⇒ Not an AP.

- (iii) –1.2, –3.2, –5.2, –7.2, …
\[\Delta = (-3.2)-(-1.2) = -2,\; (-5.2)-(-3.2) = -2,\; (-7.2)-(-5.2) = -2\]
All differences are \(-2\). Hence it is an AP with \(d = -2\).
Next three terms: \(-9.2, -11.2, -13.2\).

- (iv) –10, –6, –2, 2, …
\[\Delta = (-6)-(-10) = 4,\; (-2)-(-6) = 4,\; 2-(-2) = 4\]
Common difference \(d = 4\). Next three terms: \(6, 10, 14\).

- (v) 3, 3\frac{2}{+}, 3\frac{2}{2}+ , 3\frac{3}{2}+ , … (the notation is ambiguous; the successive differences are not constant) ⇒ Not an AP.

- (vi) 0.2, 0.22, 0.222, 0.2222, …
\[\Delta_1 = 0.22-0.2 = 0.02,\; \Delta_2 = 0.222-0.22 = 0.002,\; \Delta_3 = 0.2222-0.222 = 0.0002\]
Differences are not equal ⇒ Not an AP.

- (vii) 0, –4, –8, –12, …
\[\Delta = -4-0 = -4,\; -8-(-4) = -4,\; -12-(-8) = -4\]
AP with \(d = -4\). Next three terms: \(-16, -20, -24\).

- (viii) –\frac{1}{2}, –\frac{1}{2}, –\frac{1}{2}, –\frac{1}{2}, …
All terms are equal, so the common difference \(d = 0\). It is an AP.
Next three terms: \(-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}\).

- (ix) 1, 3, 9, 27, …
Ratios are constant (3) but differences are 2, 6, 18 … not equal ⇒ Not an AP.

- (x) a, 2a, 3a, 4a, …
\[\Delta = 2a-a = a,\; 3a-2a = a,\; 4a-3a = a\]
AP with common difference \(d = a\). Next three terms: \(5a, 6a, 7a\).

- (xi) a, a^2, a^3, a^4, …
Successive differences are \(a^2-a, a^3-a^2, a^4-a^3\); they are equal only for special values of \(a\) (e.g., \(a=1\) or \(a=0\)). In general, Not an AP.

- (xii) 2, 8, 18, 32, …
\[\Delta_1 = 6,\; \Delta_2 = 10,\; \Delta_3 = 14\]
Differences are not equal ⇒ Not an AP.

- (xiii) 3, 6, 9, 12, …
\[\Delta = 3\]
AP with \(d = 3\). Next three terms: \(15, 18, 21\).

- (xiv) 12, 32, 52, 72, …
\[\Delta = 20\]
AP with \(d = 20\). Next three terms: \(92, 112, 132\).

- (xv) 12, 52, 72, 73, …
Differences: 40, 20, 1 – not constant ⇒ Not an AP.

3. Summary
- APs: (iii), (iv), (vii), (viii), (x), (xiii), (xiv)
- For each AP, the common difference \(d\) and three additional terms are listed above.
- All other listed sequences are not arithmetic progressions.

Correct Answer: (iii) AP, d = -2, next terms: -9.2, -11.2, -13.2; (iv) AP, d = 4, next terms: 6, 10, 14; (vii) AP, d = -4, next terms: -16, -20, -24; (viii) AP, d = 0, next terms: -1/2, -1/2, -1/2; (x) AP, d = a, next terms: 5a, 6a, 7a; (xiii) AP, d = 3, next terms: 15, 18, 21; (xiv) AP, d = 20, next terms: 92, 112, 132. All other sequences are not APs.
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