Arithmetic Progression NCERT Exercise 5.1 part four Class Ten Mathematics

Arithmetic Progression

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, ……..

Solution: 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, ………..

Solution: 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, ………

Solution: `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, ……….

Solution: `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, ……….

Solution: 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)`

Solution: 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`, ………

Solution: `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, …………

Solution: 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.



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