Q. 44.2( 196 Votes )

# Which of the following are APs ? If they form an AP, find the common difference d andwrite three more terms.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix) 1, 3, 9, 27,…

(x)

(xi)

(xii)

(xiii)

(xiv)

(xv) 1^{2}, 5^{2}, 7^{2}, 73.....

Answer :

For a sequence to be an AP, the difference between two consecutive terms remains the same, that is called the common difference of AP

(i) if a_{k+1} – a_{k} is same for different values of k then the series is an AP.

We have, a_{1} = 2, a_{2} = 4, a_{3} = 8 and a_{4} = 16

a_{4} – a_{3} = 16 – 8 = 8

a_{3} – a_{2 }= 8 – 4 = 4

a_{2} – a_{1}= 4 – 2 = 2

Here, a_{k+1} – a_{k} is not same for all values of k.

Hence, the given series is not an AP.

(ii) As per the question:

a1 = 2,

a2 = 5/2,

a3 = 3

And

a4 = 7/2

a_{4} – a_{3} = – 3= 1/2

a_{3} – a_{2} = 3 – = 1/2

a_{2} – a_{1} = – 2 = 1/2

Now, we can observe that a_{k+1} – a_{k} is same for all values of k.

Hence, it is an AP.

And, the common difference = 1/2

Next three terms of this series are:

a_{5} = a + 4d

= 2 + 4* 1/2

= 4

a_{6} = a + 5d

= 2 + 5 * 1/2

=

a_{7} = a + 6d

= 2 + 6 * 1/2

= 5

Hence,

The next three terms of the AP are: 4, 9/2 and 5

(iii) a_{4} – a_{3} = - 7.2 + 5.2 = - 2

a_{3} – a_{2} = - 5.2 + 3.2 = - 2

a_{2} – a_{1} = - 3.2 + 1.2 = - 2

In this, a_{k+1} – a_{k} is same for all values of k.

Hence, the given series is an AP.

Common difference = - 2

Next three terms of the series are:

a_{5} = a + 4d

= -1.2 + 4 × (-2)

= -1.2 - 8 = -9.2

a_{6} = a + 5d

= -1.2 + 5 × (-2)

= -1.2 - 10 = -11.2

a_{7} = a + 6d

= -1.2 + 6 × (-2)

= -1.2 - 12 = -13.2

Next three terms of AP are: - 9.2, - 11.2 and – 13.2

(iv)a_{4} – a_{3} = 2 + 2 = 4

a_{3} – a_{2} = - 2 + 6 = 4

a_{2} – a_{1} = - 6 + 10 = 4

Here, a_{k+1} – a_{k} is same for all values of k

Hence, the given series is an AP.

Common difference = 4

Next three terms of the AP are:

a_{5} = a + 4d

= -10 + 4 × 4

= -10 + 16 = 6

a_{6} = a + 5d

= -10 + 5 × 4

= -10 + 20 = 10

a_{7} = a + 6d

= -10 + 6 × 4

= -10 + 24 = 14

Next three terms of AP are: 6, 10 and 14.

(v) a_{4} – a_{3} = 3 + 3√2 – 3 - 2√2 = √2

a_{3} – a_{2} = 3 + 2√2 – 3 - √2 = √2

a_{2} – a_{1} = 3 + √2 – 3 = √2

Here, a_{k+1} – a_{k} is same for all values of k

Hence, the given series is an AP.

Common difference = √2

Next three terms of the AP are

a_{6} = a + 5d = 3 + 5√2

a_{7} = a + 6d = 3 + 6√2

Next three terms of AP are: 3 + 4√2, 3 + 5√2 and 3 + 6√2

(v)

a_{4} – a_{3} = 0.2222 – 0.222 = 0.0002

a_{3} – a_{2} = 0.222 – 0.22 = 0.002

a_{2} – a_{1} = 0.22 – 0.2 = 0.02

Here, a_{k+1} – a_{k} is not same for all values of k

Hence, the given series is not an AP.

(vii) 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) 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. – 1/2

(ix) 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_{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 are:

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) Here, the exponent is increasing in each subsequent term.

a_{4}= a

^{4}, a

_{3}= a

^{3}, a

_{2}= a

^{2}, a

_{1}= a

a

_{4 }- a

_{3}= a

^{4}- a

^{3}

a

_{3}- a

_{2}= a

^{3}- a

^{2}

since, the difference is not same,

Since a_{k+1} – a_{k} is not same for all values of k.

Hence, it is not an AP.

(xii) Different terms of this AP can also be written as follows:

√2, 2√2, 3√2, 4√2, ………..

a4 – a3 = 4√2 - 3√2 = √2

a3 – a2 = 3√2 - 2√2 = √2

a2 – a1 = 2√2 - √2 = √2

Since ak+1 – ak is same for all values of k.

Hence, it is an AP.

Next three terms can be calculated as follows:

a5 = a + 4d = √2 + 4√2 = 5√2

a6 = a + 5d = √2 + 5√2 = 6√2

a7 = a + 6d = √2 + 6√2 = 7√2

Next three terms are; 5√2, 6√2 and 7√2

(xiii) a_{4} – a_{3} = √12 - √9 = 2√3 – 3 ( √12 =√ 2 x 2 x 3 = 2√3)

a_{3} – a_{2} = √9 - √6 = 3 - √6

a_{2} – a_{1} = √6 - √3

Since a_{k+1} – a_{k} is not same for all values of k.

Hence, it is not an AP

(xiv) 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.

(xv) 1^{2}, 5^{2}, 7^{2}, 73.....

a = 1

d = 5^{2} - 1 = 25 - 1 = 24

d = 7^{2} - 5^{2} = 49 - 25 = 24

d = 74 - 49 = 24

As, common difference is same. The sequence is in A.P.

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