How to find the common ratio of geometric sequence when the sum of 5th term to 8th term is twice the first four terms?

2 Answers
Narad T.
Jul 15, 2017

The common ratio is #=root(4)2=1.19#

Explanation:

Let the first term be #=a#

and the common ration #=r#

The geometric sequence is

#u_1=a#

#u_2=ar#

#u_3=ar^2#

#u_4=ar^3#

#u_5=ar^4#

#u_6=ar^5#

#u_7=ar^6#

#u_8=ar^7#

Therefore,

#2(u_1+u_2+u_3+u_4)=u_5+u_6+u_7+u_8#

So,

#2(a+ar+ar^2+ar^3)=ar^4+ar^5+ar^6+ar^7#

#2(1+r+r^2+r^3)=r^4+r^5+r^6+r^7#

#2(1+r+r^2+r^3)=r^4(1+r+r^2+r^3)#

#r^4=2#

#r=root(4)2=1.19#

Ratnaker Mehta
Jul 15, 2017

# root(4)2, or, -1.#

Explanation:

Let the Geom. Seq. be #a,ar, ar^2,...,ar^(n-1),...#, and let, #S_n#

be the sum of its first #n# terms, where, #a in RR-{0}, r in RR-{1}.#

We have,

#S_n=sum_1^n ar^(n-1)=a+ar+...+ar^(n-1)=(a(r^n-1))/(r-1), rne1#

From what is given, we find,

#sum_5^8 ar^(n-1)=2S_4.#

#:. sum_1^8ar^(n-1)-sum_1^4ar^(n-1)=2S_4, i.e.,#

#S_8-S_4=2S_4, or, S_8=3S_4.#

#:. (a(r^8-1))/(r-1)=3*(a(r^4-1))/(r-1).#

#:. r^8-1=3r^4-3,.....[because, ane0, rne1].#

#:. r^8-3r^4+2=0.#

#:. (r^4-1)(r^4-2)=0.#

Since, #rne1 rArr r^4ne1, :., r^4=2.#

#:." The Common Ratio "r=root(4)2.#

As respected George C. Sir has rightly pointed out, #r=-1#.

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