Question

Given that (2-3k) (2k+3) and (k+7), where k is a constant, are three consecutive terms of a linear sequence (AP), find its common difference

Answer 1

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The nth term of a arithmetic sequence is

:

a(n) = a(1) + d(n-1), where a(1) is the first term and d is the common difference

:

we are given

:

2-3k = a(1) + d(n-1)

3+2k = a(1) + d(n+1-1) = a(1) + dn

7+k = a(1) + d(n+2-1) = a(1) + d(n+1)

:

by the definition of an arithmetic sequence, we know that

:

3+2k - (2-3k) = 7+k - (3+2k)

:

5k + 1 = -k + 4

:

6k = 3

;

k = 3/6 = 1/2

:

***************************

d = 5(1/2) + 1 = 7/2 = 3.5

so that the common difference is 3.5

Answer 2

Answer:

d = 7/2

Step-by-step explanation:

We have,

( 2k+3) - ( 2- 3k) = (k+7) - ( 2k+3)

or, 2k+3 - 2+3k = k+7 - 2k - 3

or, 5k + 1 = - k + 4

or, 5k + k = 4 - 1

or, 6k = 3

or, k = 3/6

or, k = 1/2

So, d = (2k + 3) - ( 2- 3k) [ As, d = t2 - t1]

= 2k + 3 - 2 + 3k

= 5k + 1

= 5 × 1/2 + 1

= 5/2 + 1

= 7/2 Ans

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