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प्रश्न
If a, b, c are in G.P. and a, x, b, y, c are in A.P., prove that `1/x + 1/y = 2/b`
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उत्तर
a, b and c are in G.P.
`=>` b2 = ac
a, x, b, y and c are in A.P.
`=>` 2x = a + b `=> x = (a + b)/2`
2b = x + y `=> b = (x + y)/2`
2y = b + c `=> y = (b + c)/2`
Now,
`1/x + 1/y = 2/(a + b) + 2/(b + c)`
= `(2b + 2c + 2a + 2b)/(ab + ac + b^2 + bc)`
= `(2a + 2c + 4b)/(ab + b^2 + b^2 + bc)`
= `(2a + 2c + 4b)/(ab + 2b^2 + bc)`
= `(2(a + c + 2b))/(b(a + 2b + c))`
= `2/b`
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The fifth, eight and eleventh terms of a geometric progression are p, q and r respectively. Show that : q2 = pr.
Q 1.2
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Show that : x2, b2, y2 are in A.P.
Find the sum of G.P. :
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Q 3.3
Q 7
