Let the first term a and the common ratio r of a geometric progression be positive integers. If the sum of squares of it

Question

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

Accepted Answer

Given: The first three terms of the GP are $a$ , $ar$ , and $ar^2$ , where $a$ and $r$ are positive integers.

Find: The sum $a + ar + ar^2$ .

From the condition on the sum of squares,

a^2 + (ar)^2 + (ar^2)^2 = 33033

So,

a^2(1 + r^2 + r^4) = 33033

Factorizing the number used in the solution,

33033 = 11^2 \cdot 3 \cdot 7 \cdot 13

Hence the solution takes

a^2 = 11^2 \implies a = 11

Substituting $a = 11$ ,

1 + r^2 + r^4 = \frac{33033}{11^2} = 273

So,

r^2(1 + r^2) = 272

And since

272 = 16 \cdot 17

we get

r^2 = 16 \implies r = 4

because $r > 0$ .

Now the sum of the three terms is

a + ar + ar^2 = a(1 + r + r^2)

Substituting $a = 11$ and $r = 4$ ,

1 + r + r^2 = 1 + 4 + 16 = 21

Therefore,

a(1 + r + r^2) = 11 \cdot 21 = 231

Therefore, the sum of the three terms is $231$ . The solution working gives option C, although the source the solution states option A.

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