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1)a,b,c are three unequal numbers in GP Then a:b:c=
2)If a,b,c,d are in HP, then ab+bc+cd = ?
ans 1.:-
1:r:r^2
ans2:-
i dont know.do by ur own.
Let the terms of the H.P are (1/a)= a, (1/(a+d))= b , (1/(a+2d)) = c and (1/(a+3d)) =d where a, d are the first term and the common differences respectively.
Then ab+bc+bd =
= 1/a(a+d)+1/(a+d)(a+2d)+1/(a+2d)(a+3d)
= ((a+2d)(a+3d)+(a)(a+3d)+(a)(a+d))/(a(a+d)(a+2d)(a+3d))
= ((a^2+5ad+6d^2)+(a^2+3ad)+(a^2+ad))/(a(a+d)(a+2d)(a+3d))
= (3a^2+9ad+6d^2)/(a(a+d)(a+2d)(a+3d))
= ((3)(a+d)(a+2d))/(a(a+d)(a+2d)(a+3d))
= 3/a(a+d)
Since 1/(a+d) = d and 1/a =a
Then = 3ad
A Slight correction in last few lines
=3/(a)(a+3d)
since (a+3d)= d
=3ad
1)ar where a is first term and ris common ratio
2)3ad where a is first term and d is common difference
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