Question

Question 10 The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Class X1 - Maths -Sequences and Series Page 199

Answer 1

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Answer 2

Let three numbers in GP are a , ar , ar².
Given, a + ar + ar² = 56
a(1 + r + r²) = 56 --------------------(1)

A/C to question,
(a - 1) , (ar - 7), ( ar² - 21) are in AP
so, common difference is always constant.
e.g (ar² - 21) - (ar - 7) = (ar - 7) - (a - 1)
ar² - ar - 21 + 7 = ar - a - 7 + 1
ar² - 2ar + a - 8 = 0
a( r² - 2r + 1) = 8 --------------------(2)

dividing equation (1) by equation (2) , we get
a( 1 + r + r²)/a(r² - 2r + 1) = 56/8
(1 + r + r²)/(1 - 2r + r²) = 7
1 + r + r² = 7 - 14r + 7r²
6r² - 15r + 6 = 0
2r² - 5r + 2 = 0
2r² - 4r - r + 2 = 0
2r( r - 2) -(r - 2) = 0
(2r - 1)(r - 2) = 0
r = 2, 1/2

if r = 2 then, from equation (1)
a + 2a + 4a = 56
7a = 56
a = 8

Then, numbers are a = 8
ar = 8 × 2 = 16
ar² = 8 × 2² = 32
hence, 8 , 16 , 32

similarly if r = 1/2 then from equation (1)
a + a/2 + a/4 = 56
7a/4 = 56
a = 32

then, numbers are 32 , 32 × 1/2 , 32 × 1/2²
32 , 16 , 8