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The product of first three terms of G.P. is 1000. If 6 is added to its second term and 7 is added to its third term, the terms become in A.P. Find G.P.

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Let the numbers in G.P. be \(\frac{a}{r}\), a, ar … (i)

Product \(\frac{a}{r}\), a. ar = 1000

⇒ a3 = 1000

⇒ a = 10

According to question

A.P. a1 = \(\frac{a}{r}\) = \(\frac{10}{r}\)

a2 = a + 6 = 10 + 6 = 16

a3 = ar + 7 = 10r + 7

Also, a3 = a1 + 2(a2 − a1)        [∴ a, b, c are in A.P.]

⇒ 10r + 7\(\frac{10}{r}\) + 2\(\big[10 − \frac{10}{r}\big]\)

⇒ 10r2 + 7r = 10 + 32r − 20

⇒ 10r2 − 25r + 10 = 0

⇒ (r − 2)(10r − 5) = 0

⇒ r = 2 or \(\frac{1}{2}\)

Substituting the value of a and r in eq (i), we get G.P. : 5,10, 20… when r = 2 and G.P. : 20,10, 5... when r = \(\frac{1}{2}\)

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