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Arithmetic Progressions
EXAMPLES
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

Which of the following list of numbers form an AP? If they form an AP, write the next two terms : (i) 4, 10, 16, 22, . . . (ii) 1, – 1, – 3, – 5, . . . (iii) – 2, 2, – 2, 2, – 2, . . . (iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . . .

Step-by-Step Solution

Key Concept: A sequence of numbers forms an Arithmetic Progression (AP) if the difference between any two successive terms is constant. This constant difference is called the common difference (d). If a sequence is an AP, the next term can be obtained by adding d to the last known term.
1. Check each list for a constant common difference
- Compute successive differences $d_i = a_{i+1} - a_i$.
- If all $d_i$ are equal, the list is an AP; otherwise it is not.

(i) 4, 10, 16, 22, …
\[ d_1 = 10-4 = 6,\; d_2 = 16-10 = 6,\; d_3 = 22-16 = 6 \]
All differences are $6$, so the sequence is an AP with common difference $d = 6$.
Next two terms:
\[ a_5 = 22 + 6 = 28,\; a_6 = 28 + 6 = 34 \]

(ii) 1, –1, –3, –5, …
\[ d_1 = (-1)-1 = -2,\; d_2 = (-3)-(-1) = -2,\; d_3 = (-5)-(-3) = -2 \]
All differences are $-2$, so the sequence is an AP with $d = -2$.
Next two terms:
\[ a_5 = -5 + (-2) = -7,\; a_6 = -7 + (-2) = -9 \]

(iii) –2, 2, –2, 2, –2, …
\[ d_1 = 2-(-2) = 4,\; d_2 = (-2)-2 = -4,\; d_3 = 2-(-2) = 4 \]
The differences alternate between $4$ and $-4$; they are not constant. Hence this list does not form an AP.

(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, …
\[ d_1 = 1-1 = 0,\; d_2 = 1-1 = 0,\; d_3 = 2-1 = 1,\; d_4 = 2-2 = 0,\; d_5 = 2-2 = 0,\; d_6 = 3-2 = 1 \]
The successive differences are $0,0,1,0,0,1,\dots$ – not a single constant value. Therefore this list does not form an AP.

2. Summarize the results
- (i) Forms an AP; next two terms are $28$ and $34$.
- (ii) Forms an AP; next two terms are $-7$ and $-9$.
- (iii) Does not form an AP.
- (iv) Does not form an AP.

Correct Answer: (i) Yes, AP with $d=6$; next terms $28,\;34$. (ii) Yes, AP with $d=-2$; next terms $-7,\;-9$. (iii) Not an AP. (iv) Not an AP.
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