Question

The sum of the first three terms of a geometric sequence is 52. If the third term is changed to 20, then these three terms form an arithmetic sequence.
(a) Find the possible values of the common ration r.
(b) if r>2, find the eighth term of the geometric sequence.

Answer 1

Let a be the first term of the geometric sequence
the geometric sequence is: a, ra , r*ra
and the arithemetic sequence is : a, ra ,20
by the question:
a+ra+r*ra =52
(r*r+r+1)a=52....(1)
20-ra =ra-a
2ra-a =20
(2r-1)a=20....(2)

A)(2)/(1),
(2r-1)/(r*r+r+1)=20/52
20r*r+20r+20=104r-52
20r*r-84r+72 =0
4(r-3)(5r-6)=0
r=3 or 1.2
B) Are you asking the eighth terms

As r >2 so the only possible ratio is 3
so sub r=3 into (2),
(2*3-1)a =20
5a=20
a=4
So the first term is 4
so the eighth term is 4* (3)^(8-1) = 8748

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