NIMCET 2013 Question Paper with solutions

Question : 12 of 120

Marks: +1, -0

In a GP consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of the GP is:

Solution:

Let the term of GP be

a,ar,ar^2,ar^3,ar^4,...

As per the question,

a=ar+ar^2⇒a(1)=a(r+r^2)⇒1=r+r^2

⇒1-r-r^2=0⇒r^2+r-1=0

Here,

a=1,b=1

and

c=-1

∴r={-b±√{b^{2}-4ac}}/{2×a}={-1±√{b^{2}-4ac}}/{2a}

={-1±√{(1)^{2}-4×1×(-1)}}/{2×1}={-1±√{1+4}}/{2}={-1±√{5}}/{2}

Since, r is positive
Hence,

r={-1+√5}/2

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