ARITHMETIC AND GEOMETRIC SEQUENCES WORD PROBLEMS

Problem 1 :

The product of three increasing numbers in GP is 5832. If we add 6 to the second number and 9 to the third number, then resulting numbers form an AP. Find the numbers in GP.

Solution :

Let three consecutive terms be a/r, a and ar

Product of three terms  = 5832

(a/r) ⋅ a ⋅ ar  =  5832

a3  =  5832

a3  =  183

a  =  18

Then a/r, a + 6, ar + 9 are in A.P 

2b  =  a + c

2 (a + 6)  =  (a/r) + (ar + 9)  ------(1)

By applying the value of a, we get

2 (18 + 6)  =  (18/r) + (18r + 9)

48r  =  18 + r(18r + 9)

48r  =  18 + 18r2 + 9r

18r2 + 9r - 48r + 18  =  0

18r2 - 39r + 18  =  0

Divide by 3, we get

6r2 - 13r + 6  =  0

6r2 -9r - 4r + 6  =  0

3r (2r - 3) -2 (2r - 3)  =  0

3r - 2 = 0                   2r - 3  =  0

r  =  2/3                       r  =  3/2

a/r  =  18/(2/3)

  =  27

a  =  18

ar  =  18  (2/3)

  =  12

a/r  =  18/(3/2)

  =  12

a  =  18

ar  =  18  (3/2)

  =  12

So, the required terms are 27, 18, 12.

Problem 2 :

Write the nth term of the sequence

3/1222, 5/2232, 7/3242 , . . . as a difference of two terms.

Solution :

By observing the denominator, we have the form n2 and (n+1)2

tn  =  (1/n2) - [1/(n + 1)2]

If n = 1

tn  =  (1/12) - [1/(1 + 1)2]

 =  (1/1) - (1/4)

=  (4 - 1)/1⋅4

=  3/12⋅ 22

If n = 2

tn  =  (1/22) - [1/(2 + 1)2]

 =  (1/4) - (1/9)

=  (9 - 4)/22⋅ 32

=  5/22⋅ 32

So, the required nth term is tn  =  (1/n2) - [1/(n + 1)2]

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