Question

In a gp of real number it is given that t1 t2 t3 t4 = 30 and t1 squares t2 square t3 square t4 square =340 determin the first term and common ratios

Answer 1

Answer :

When; a=2 and r= 2

Then; t1=2, t2 =4, t3=8, t4 =16

When; a=16 and r= 1/2

Then; t1=16, t2 = 8, t3 = 4, t4 = 2

Explanation :-

Let the GP be- a, ar, ar², ar³

Sum will be- a(1−r⁴)/1−r=30

Squares of terms- a²,a²r²,a²r⁴,a²r6

Sum will be = a²(1−r8)/1−r² = 340

=> a(1+r²)/(1+r) = 30

=> a²(1+r²)/(1+r⁴) = 340

=> 45 /(1+r²)(1+r)² = 17/(1+r⁴)

=> 45 + 45r⁴ = 17(1+r²) (1+r²+2r)

=> 45 + 45r⁴ = 17(1+r²+2r+r²+r⁴+2r³)

=> 14 +14r⁴ = 17r² + 17r + 17r³

The two real solutions will be

=> r = 2 or r = 1/2

So, a = 2, r = 2 or a = 16, r = 1/2...

Hope it will help you....