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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