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Question

In a G.P. of real numbers, it is given that T1+T2+T3+T4=30 and T21+T22+T23+T24=340. Determine the G.P.

A
2,4,8,16,32,...
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B
16,8,4,...
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C
1,2,4,...
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D
8,2,0.5,...
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Solution

The correct options are
A 2,4,8,16,32,...
C 16,8,4,...
T1+T2+T3+T4=30
T12+T22+T32+T42=340
t1=a,ratio=r
a(1+r+r2+r3)=30a(r41)r1=30(1)
a2(1+r2+r4+r6)=340a2(r81)r21=340(2)
by dividing eq.2(eq.1)2
a2(r81)r21a2(r41)2(r1)2=340900=1745
a2(r81)×(r1)2(r21)×(r41)2=r4+1r41×r1r+1=1745
r4+1r3+r2+r+1×1r+1=1745
r4+1r4(r3+r2+r+1)r3r+1r=1+1r4(1+1r+1r2+1r3)(1+1r)=1745
We can see clearly 2 satisfy the equation
So r=2 and so as equation is symmetric in r and 1r, r can also be 1r
By (1) if r=2a(r41)r1=30
a(241)=30
a=2
So, 2,4,8,16,..
if r=12a((12)41)121=30
a(1516)12=30
a=16
hence we get 16,8,4,2,...

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