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Question
If pth, qth, and rth terms of an A.P. and G.P. are both a, b and c respectively, show that ab–c . bc – a . ca – b = 1
Solution
Let A and d be the first term and common difference respectively of an A.P. and x and R be the first term and common ratio respectively of the G.P.
∴ A + (p – 1)d = a .....(i)
A + (q – 1)d = b .....(ii)
And A + (r – 1)d = c ......(iii)
For G.P., we have
xRp–1 = a .....(iv)
xRq–1 = b .....(v)
And xRr–1 = c .....(vi)
Subtracting equation (ii) from equation (i) we get
(p – q)d = a – b ......(vii)
Similarly, (q – r)d = b – c ......(viii)
And (r – p)d = c – a ......(ix)
Now we have to prove that
ab–c . bc–a . ca–b = 1
L.H.S. ab–c . bc–a . ca–b
= `[x"R"^(p - 1)]^((q - r)d) * [x"R"^(q - 1)]^((r - p)d) * [x"R"^(r - 1)]^((p - q)d)` ....[From (i), (ii), (iii), (iv), (v), (vi), (vii), (viii), (ix)]
= `x^((q - r)d) * "R"^((p - 1) (q - r)d) * x^((r - p)d) * "R"^((q - 1) (r - p)d) * x^((p - q)d) * "R"^((r - 1)(p - q)d)`
= `x^((q - r)d + (r - p)d) "R"^((p - 1)(q - r)d + (q - 1)(r - p)d + (r - 1)(p - q)d)`
= `x^((q-r + r - p + p - q)d) * "R"^((pq - pr - q + r + qr - pq - r + p + pr + pr - qr - p + q)d)`
= `x^((0)d) * "R"^((0)d)`
= `x^0 * "R"^0`
= 1 R.H.S.
L.H.S. = R.H.S.
Hence proved.
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