Arithmetic Progression
NCERT Exercise 5.1
Part 4
Question: 4 – Which of the following are APs? If they form an AP, find the common difference and write three more terms.
From question 4 -(vii) to 4(xiv)
(vii) 0, - 4, - 8, - 12, ……..
Answer: Here; `a_4 – a_3 = - 12 + 8 = - 4`
`a_3 – a_2 = - 8 + 4 = - 4`
`a_2 – a_1 = - 4 – 0 = - 4`
Since `a_(k+1) – a_k` is same for all values of k.
Hence, this is an AP.
The next three terms can be calculated as follows:
`a_5 = a + 4d = 0 + 4(- 4) = - 16`
`a_6 = a + 5d = 0 + 5(- 4) = - 20`
`a_7 = a + 6d = 0 + 6(- 4) = - 24`
Thus, next three terms are; - 16, - 20 and – 24
(viii) – ½, - ½, - ½, - ½, ………..
Answer: Here, it is clear that d = 0
Since `a_(k+1) – a_k` is same for all values of k.
Hence, it is an AP.
The next three terms will be same, i.e. – ½
(ix) 1, 3, 9, 27, ………
Answer: `a_4 – a_3 = 27 – 9 = 18`
`a_3 – a_2 = 9 – 3 = 6`
`a_2 – a_1 = 3 – 1 = 2`
Since `a_(k+1) – a_k` is not same for all values of k.
Hence, it is not an AP.
(x) a, 2a, 3a, 4a, ……….
Answer: `a_4 – a_3 = 4a – 3a = a`
`a_3 – a_2 = 3a – 2a = a`
`a_2 – a_1 = 2a – a = a`
Since `a_(k+1) – a_k` is same for all values of k.
Hence, it is an AP.
Next three terms can be calculated as follows:
`a_5 = a + 4d = a + 4a = 5a`
`a_6 = a + 5d = a + 5a = 6a`
`a_7 = a + 6d = a + 6a = 7a`
Next three terms are; 5a, 6a and 7a.
(xi) a, a2, a3, a4, ……….
Answer: Here, exponent is increasing in each subsequent term.
Since `a_(k+1) – a_k` is not same for all values of k.
Hence, it is not an AP.
(xii) `sqrt2`, `sqrt8`, `sqrt(18)`, `sqrt(32)`
Answer: Different terms of this AP can also be written as follows:
`sqrt2`, `2sqrt2`, `3sqrt2`, `4sqrt2`, ………..
`a_4 – a_3 = 4sqrt2 - 3sqrt2 = sqrt2`
`a_3 – a_2 = 3sqrt2 - 2sqrt2 = sqrt2`
`a_2 – a_1 = 2sqrt2 - sqrt2 = sqrt2`
Since `a_(k+1) – a_k` is same for all values of k.
Hence, it is an AP.
Next three terms can be calculated as follows:
`a_5 = a + 4d = sqrt2 + 4sqrt2 = 5sqrt2`
`a_6 = a + 5d = sqrt2 + 5sqrt2 = 6sqrt2`
`a_7 = a + 6d = sqrt2 + 6sqrt2 = 7sqrt2`
Next three terms are; `5sqrt2`, `6sqrt2` and `7sqrt2`
(xiii) `sqrt3`, `sqrt6`, `sqrt9`, `sqrt12`, ………
Answer: `a_4 – a_3 = sqrt12 - sqrt9 = 2sqrt3 – 3`
`a_3 – a_2 = sqrt9 - sqrt6 = 3 - sqrt6`
`a_2 – a_1 = sqrt6 - sqrt3`
Since ak+1 – ak is not same for all values of k.
Hence, it is not an AP.
(xiv) 12, 32, 52, 72, …………
Answer: The given terms can be written as follows:
1, 9, 25, 49, …………
Here, `a_4 – a_3 = 49 – 25 = 24`
`a_3 – a_2 = 25 – 9 = 16`
`a_2 – a_1 = 9 – 1 = 8`
Since `a_(k+1) – a_k` is not same for all values of k.
Hence, it is not an AP.