Question

The sum of the first three terms of a geometric sequence is equal to 42. The sum of the squares of the same terms is equal to 1092. Find the three terms of the sequence. ​

Answer 1

sum1 = a + a r + a r2 = 42: the sum of the three terms given, r is the common ratio.

sum2 = a2 + a2r2 + ar2r4 = 1092: the sum of the squares of the three terms given .

sum1 = a + ar + ar2 = a(r3 - 1) / (r - 1) = 42 : apply formula for a finite sum of geometric series.

sum2 = a2 + a2r2 + ar2r4 = a2(r6 - 1) / (r2 - 1) = 1092: the sum of squares is a also a sum of geometric series.

sum2/sum12 = 1092 / 422 = [ a2(r6 - 1)/(r2 - 1)] / [a2(r3 - 1)2 / (r - 1)2]

(r2 - r + 1) / (r2 + r + 1) = 1092 / 422

r = 4 , r = 1/4 : solve for r

a = 2 : substitute r = 4 and solve for a

a = 32 : substitute r = 1/4 and solve for a

a = 2 , ar = 8 , ar2 = 32 : find the three terms for r = 4

a = 32 , ar = 8 , ar2 = 2 : find the three terms for r = 1/4

Answer 2

${\huge{\red{\mathfrak{answer}}}}$

This is the sum of a geometric series problem of the form a1 + a1 r + a1 r2 + a1 r3

Note if a1 = 2 and r = 4 We have 2 + 8 + 32 = 42 Then if we cube this expression we have:

4 + 64 + 1024 = 1092

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