0 votes 0 votes 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 Quantitative Aptitude sequences&series + – minal asked Jun 6, 2016 minal 2.3k points 2.2k views answer comment Share See all 0 reply Please log in or register to add a comment.
0 votes 0 votes 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 srestha answered Jun 6, 2016 srestha 5.2k points comment Share See all 3 Comments See all 3 3 Comments reply minal 2.3k points commented Jun 7, 2016 reply Share given ans is 3) option and explaination also given , same as ojas given , but m not getting .... 0 votes 0 votes srestha 5.2k points commented Jun 8, 2016 i edited by srestha Jun 8, 2016 reply Share 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 0 votes 0 votes minal 2.3k points commented Jun 8, 2016 reply Share i did with generral summation formula but i stuck .... lets see any one reply .. 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes 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 Ojas Mohril 1 answered Jun 6, 2016 Ojas Mohril 1 28 points comment Share See all 2 Comments See all 2 2 Comments reply minal 2.3k points commented Jun 7, 2016 reply Share explain more .... , i have also this soluation . 0 votes 0 votes Ojas Mohril 1 28 points commented Jun 11, 2016 reply Share 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. 0 votes 0 votes Please log in or register to add a comment.