Question

a gp consists of 2n terms . is the sum of the terms occupying the odd places is s1 and even places is s2 the common ratio of the progress is

Answer 1

Answer:

No. of terms in G.P. = 2n

Let G.P. sequence for s1 be “a1, a3, a5,….., a2n” and for s2 be “a2, a4, a6, ……, a2n”.  

∴ Sum of first odd terms = s1 = a1(nr - 1)/(r-1) ….. (i)

And,

Sum of first even terms = s2 = a2(nr - 1)/(r-1) ….. (ii)

Now, to find the common ratio of the progress, we need to divide the eq. (i) & (ii),

s1/s2 = [a1(nr - 1)/(r-1)] / [a2(nr - 1)/(r-1)]

⇒ s1/s2 = a1/a2 ….. (iii) [Cancelling the equal terms]

Since both the sequences of odd and even terms are in geometric progression, therefore, the first term,

a1 = a and a2 = ar  

Substituting the value of a1 and a2 in eq. (iii), we get

s1/s2 = a/ar = 1/r

or, s2/s1 = r

Hence, the common ratio of the progress, s2/s1 is r.

Answer 2

Answer:s2/s1

Step-by-step explanation:total terms =2n

Sum of odd terms=s1

Sum of even terms=s2

Using formula for sum of gp

S1=a1(r^n-1)/r-1.....eq1

S2=a2(r^n-1)/r-1.....eq2

Eq1/eq2

After simplifying

1/r=s1/s2

Therefore,r=s2/s1