The sum of three decreasing numbers in A.P. i | vedclass

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(D) Let the three terms of the decreasing $A.P.$ be $a+d, a, a-d$,where $d > 0$.Sum: $(a+d) + a + (a-d) = 27$ $\Rightarrow 3a = 27$ $\Rightarrow a = 9

Vedclass · Accepted Answer

$17, 9, 1$

Vedclass · Answer

(D) Let the three terms of the decreasing $A.P.$ be $a+d, a, a-d$,where $d > 0$.Sum: $(a+d) + a + (a-d) = 27$ $\Rightarrow 3a = 27$ $\Rightarrow a = 9$.The terms are $9+d, 9, 9-d$.Adding $-1, -1, 3$ respectively gives: $(9+d-1), (9-1), (9-d+3) = (8+d), 8, (12-d)$.Since these are in $G.P.$,the square of the middle term equals the product of the extremes:$8^2 = (8+d)(12-d)$$64 = 96 - 8d + 12d - d^2$$d^2 - 4d - 32 = 0$$(d-8)(d+4) = 0$.Since the $A.P.$ is decreasing,$d$ must be positive,so $d = 8$.The numbers are $9+8, 9, 9-8$,which are $17, 9, 1$.

The sum of three decreasing numbers in $A.P.$ is $27$. If $-1, -1, 3$ are added to them respectively,the resulting series is in $G.P.$ The numbers are

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