Question

find the nth term of the sequence :
4; 6; 13; 27; 50; 84..

Answer 1

well the common difference of common difference are in AP..but still,,what next?

Answer 2

well this is Arithmetic or geometric seq. ?

Answer 3

this is none.. altough,,as i said,,the common differneces of the common difference are in arithmetic seq.

Answer 4

i.e. 1st common diff -> 2;7;14;23;24 =>no sequence
2nd -> 5;7;9;11 =>AP

Answer 5

but am unable to find the general term of the original seq.

Answer 6

well
i'm trying but this type is more confusing than what i had studied

Answer 7

hmmn,,well,,take your time .. ^^

Answer 8

i got to the solution..though i didnt take much help from you sir..but still ,,thank you for your time.. ^_^

Answer 9

Well
i would be happy if u told me the solution cause i tried but i couldn't find it

Answer 10

let nth term of 4,6,13,27,50,84...be a_n
of the 1st common diff. 2,7,14,23,34 .. be b_n
and 2nd 5,7,9,11 .. be c_n
{ _ denotes subscript }
we easily see that c_n=5+(n-1)2 = 3+2n
also, b_(n+1) - b_n = c_n = 3+2n
i.e. b_2 - b_1 = 3 + 2(1)
b_3 - b_2 = 3+2(2)
b_4 - b_3 = 3+2(3)
.
.
.
.
b_(n+1) - b_n = 3+2(n)
adding all eqns,,we easily see that we are left with
b_(n+1) - b_1 = 3n + 2((n)(n+1)/2)
we know b_1 = 2
thus,,b_(n+1) is calculated..
replace n+1 by n-1 on both LHS and RHS
thus we are able to find b_n

using the same approach,,now using b_n ,,we can calculate a_n ..which is what we require...
hope i was clear ..

Answer 11

Nice question. Quite challenging. :)

Answer 12

well my method was a bit lengthy..i would like to know shorter and more clever methods..if any..which there must be..hmmn..

Answer 13

I think your answer is the simplest method. Before this I was using summations an d everything shrink into 1 formula which is really confusing. So well done man. :)

Answer 14

^_^