Geometric progression

Question

The series S is in G.P with common ratio r. A new series is formed whose nth term is sum of (n+1)th term and nth term of series S . what is common ratio of the new series ?

1) r-1

2)r+1

3)r

4) The new series is not in G.P.

here G.P Is geometric progression

Answer 1

Answer 4) The new series is not in G.P.

Say, in series a=2

r=3

Series is { 6, 18, 54 }

Now according to the question series should be {6, __, 162}

if we take r=4

then a=1.27

but  it is not always case and  'a' is not an integer and also we are not getting exact value

So, I choose D) as answer

Comments

given ans is 3) option and explaination also given , same as ojas given , but m not getting ....

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I thought

acc. to them

say a G.P. series 3(1 , 2, 2^2  ,  2^3 , 2^4 ) [where a=3 , r=2]

So, new list will be 9(1 , 2 , 2^2 , 2^3 )

 

then according to them 3(2^3+2^4)  and 9(2^3) are equal? yes

But I think only trial and error method can get this answer

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i did with generral summation formula  but i stuck .... lets see any one reply ..

Answer 2

Ans: 3)

let the original series be, S : a, ar, ar^2, ar^3 ...  ar^(n-1), ar^n

So, the new series will be, N: a(1+r), a(1+r)r, a(1+r)r^2, a(1+r)r^3 ... a(1+r)r^(n-1)

Hence, it is a G.P. and the common ratio is r

Comments

explain more .... , i have also this soluation .

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since the nth term of the new series is the sum of nth and n+1 th element of the GP,

if we consider An as the nth element of the new series then An = ar^(n-1) + ar^(n)

so, A1 = a + ar = a(1+r)

A2 = ar + ar^2 = ar(1+r) = a(1+r)r

A3 = ar^2 + ar^3 = ar^2(1+r) = a(1+r)r^2

and An = ar^(n-1) + ar^(n) = a(r^n-1)(1+r) = a(1+r)(r^n-1).

now since the first term is a(1+r) and the ratio of any two consecutive terms is r, the series is in GP with common ratio r.