**Arithmetic and Geometric Sequences Word Problems :**

Here we are going to see some practice questions on arithmetic and geometric sequence.

**Question 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

a^{3} = 5832

a^{3} = 18^{3}

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 + 18r^{2} + 9r

18r^{2} + 9r - 48r + 18 = 0

18r^{2} - 39r + 18 = 0

Divide by 3, we get

6r^{2} - 13r + 6 = 0

6r^{2} -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 |

Hence the required terms are 27, 18, 12.

**Question 2 :**

Write the nth term of the sequence

3/1^{2}2^{2}, 5/2^{2}3^{2}, 7/3^{2}4^{2} , . . . as a difference of two terms.

**Solution :**

By observing the denominator, we have the form n^{2} and (n+1)^{2}

t_{n} = (1/n^{2}) - [1/(n + 1)^{2}]

If n = 1 t = (1/1) - (1/4) = (4 - 1)/1⋅4 = 3/1 |
If n = 2 t = (1/4) - (1/9) = (9 - 4)/2 = 5/2 |

Hence the required n^{th} term is t_{n} = (1/n^{2}) - [1/(n + 1)^{2}]

After having gone through the stuff given above, we hope that the students would have understood "Arithmetic and Geometric Sequences Word Problems".

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