First term = 1 ,common diff = 4,which of the following represents the general term of the AP
A)k^2(k+1)^2 + 3
B)k^2(k-1)^2 - 3
C)k^2(k+1)^2 + 2
D)k^2(k-1)^2 + 5
1put n=0 in
a+(n-1)d.... we get -3 ..............................(1)
on puttin k=0 in given options.. we get -3...........(2)
fr option b)
hence d ans.....
1ans is b...
bhaiya, this is ram's method, so i deserve niether bash nor pink ...acc to him,general term of ap= a+(n-1)d
substitute 4 a and d, we get n-3, observing, only b matches the criteria...
but he deleted his ans coz he found inappropriate...if u think its ok, plz pink ramkumar's any random post...:)
1The ans is B,but is there any way to solve it rather than just putting values?
11bhai tu bahut tension leta hai [1][1][1]
A.P ko define kar diya gaya hai ques mein a=1 and d=4
so wats the idea of putting n=0?????????
1how can n =0 u can t put
dude
any one ?
watz so strange for zer0 ?
prob mayb , lil wrong !
not fully exposed !
11Mr archana
go to the forum
go to that topic in which u think the sum belongs
there is a button of new topic
click it
give a heading
and post ur question is a new topic
i think u r new[1]
1GOSH I WORKED ONLY FOR K=0
other values its B
SO , ITS LIKE FOR ZERO ITS D
n for all other , its B ?
EITHER ?? ZERO IS NOT GOOD ,
ohmy, is der any mistake of mine
for k=0 , itz d
n for non zero its
b ??
help me , i m teribly lost
1ajinkya..... i cant get u... on puttin 0.................................???????????
1dude , answer is d )
ap series is 1 , 5,9 13, 17 ......
so just pitting k as 0
+3
-3
+2
+5 ..........-------> d
ram , ur logik looks crkt
but i think u mistyped letter b to d
D !!!!!!! IS ANSWER
1hey see now.. sub n=0 in ans dat i gave...
then put 0 in one of d options.... u get -3 in both d cases .. he he..........
1b)?????
guess but not d actual one i got.. related ans i tol.
