Let the first term a and the common ratio r o | vedclass

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(A) The first three terms of the geometric progression are $a, ar, ar^2$.Given the sum of their squares is $33033$:$a^2 + (ar)^2 + (ar^2)^2 = 33033$$a

Vedclass · Accepted Answer

$231$

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(A) The first three terms of the geometric progression are $a, ar, ar^2$.Given the sum of their squares is $33033$:$a^2 + (ar)^2 + (ar^2)^2 = 33033$$a^2(1 + r^2 + r^4) = 33033$Prime factorization of $33033 = 3 	imes 7 	imes 11^2 	imes 13 = 121 	imes 273$.Comparing $a^2(1 + r^2 + r^4) = 121 	imes 273$,we get $a^2 = 121 \Rightarrow a = 11$ and $1 + r^2 + r^4 = 273$.$r^4 + r^2 - 272 = 0$Let $x = r^2$,then $x^2 + x - 272 = 0$.$(x + 17)(x - 16) = 0$.Since $r$ is a positive integer,$r^2 = 16 \Rightarrow r = 4$.The sum of the first three terms is $a + ar + ar^2 = 11 + 11(4) + 11(4^2) = 11 + 44 + 176 = 231$.

Let the first term $a$ and the common ratio $r$ of a geometric progression be positive integers. If the sum of the squares of the first three terms is $33033$,then the sum of these three terms is equal to

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