Saved Bookmarks
| 1. |
`{:("List-I","List-II"),((P)"A geometirc progression consists of an even" ,(1)7),("number of terms. if the sum of all the terms is",),("then its common ratio is",),((Q)Let S_(n).S_(2n).S_(3n)"be the sums of first n.2n.3n",(2)4),((n in N)"terms of an arithmetic progression then",),((S_(3n))/(S_(2n)-S_(n)) "is equal to",),((R)"Number of ordered pairs (a.b)where a.b" in N,(3)3),("such that" 6.a.b ("taken in that order") "are in harmonic progression is equal to",):}` |
|
Answer» Correct Answer - 2,3,1 (P) Let G,P.be `a,ar,ar^(2),.....,ar^(2n-1)` Now, `(a.(r^(2n)-1))/((r-1))=(5a(r^(2n)-1))/((r-1)(r+1))` `implies r=4` (Q) `S_(2n)-S_(n)` `=(2n)/(2)[2a+(2n-1)d]-(n)/(2)[2a+(n-1)d]` `(n)/(2)[2a+(3n-1)d]` `=(1)/(3).(3n)/(2).[2a+(3n-1)d]` `=(1)/(3)S_(3n)` `implies (S_(2n)-S_(n))/(S_(3n))=(1)/(3)` (R) 6,a,b in H.P. `impliesb=(6a)/(12-a)` `therefore in {3,4,6,8,9,10,11}implies7` |
|