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

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

t_n  =  (1/n²) - [1/(n + 1)²]

So, the required n^th term is t_n  =  (1/n²) - [1/(n + 1)²]

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