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Exercise 5.3 · Q3

Q.Which number should be added to the numbers 3, 8, 13 so that the resulting numbers are in G.P.?

(a) 4
(b) 2
(c) 5
(d) −2-2
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As printed, 3, 8, 13 are already in AP so no constant addition can make a GP; solved on the standard 3, 7, 13 form of this question, x=5x=5 gives the valid GP 8,12,188,12,18 — matching option (c).

[!FORMULA] If xx is added to a,b,ca,b,c to form a GP: (b+x)2=(a+x)(c+x)(b+x)^2=(a+x)(c+x)

a,b,ca,b,c = the three given numbers, xx = the number to be added.

  1. Apply the condition to the printed numbers 3,8,133,8,13: (8+x)2=(3+x)(13+x)(8+x)^2=(3+x)(13+x).
  2. Expand the left side: 64+16x+x264+16x+x^2. Expand the right side: 3×13+3x+13x+x2=39+16x+x23\times13+3x+13x+x^2=39+16x+x^2.
  3. Equating: 64+16x+x2=39+16x+x2  ⇒  64=3964+16x+x^2=39+16x+x^2 \;\Rightarrow\; 64=39 — a contradiction, so no finite xx satisfies this for exactly 3, 8, 13.
  4. Reason: 3,8,133,8,13 are already an AP (8−3=13−8=58-3=13-8=5). Adding the same constant to every term of an AP keeps it an AP with the SAME common difference; a nonzero-common-difference AP can only equal a GP if that difference is 0 — so this specific triple can never be converted to a GP by adding a constant. …

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