Probably the easiest QOD ..

a1= a
a2=b
a3= (a+b)/2
a4=(a+3b)/4
a5=(3a+5b)/8
a6=(5a+11b)/16
a7=(11a+21b)/32
.
.
.

we take a4 as teh first term..
so, let the last term be m=n-3
then we get a series for both a's and b's coefficients whose difference is in AP wid comm diff 4.
a ka series is one term behind b ka series.
for series of a;
S_m = 3 + 5 + 11 + 21 + ................... + t_m
S_m = 3 + 5 + 11 + 21 + ........... + t_m-1 + t_m
---------------------------------------------------------------------------------------------------------------
0 = 3 + [2+6+10+....+ till m-1 terms] - t_m
=> t_m = 3 + [2+4(m-2)] = 4m-3

similarly for the series of b;
t'_m= 4m-6

the denominator is a GP whose nth term will be: 4X2^m-1

then a'_m = [(4m-6)a + (4m-3)b]/ 4.2^m-1

if n-> infi , m also -> infi

so, lim m->infi [(4m-6)a + (4m-3)b]/ 4.2^m-1

a+2b /3

but havent there been simpler QODs b4

hint :

2 a_n+2 + a_n+1 = a + 2b

now try finding proof for this and then use some logic

yup celestine thats the answer .. ..

The relation's crrctness cN BE EXAMINED BY takin an example!!

but yaar i cudnt get da Prooof how u achieved it [2]

hint pl cele!!!!!

SORRY HO GAYA!!!!!!!

NICE UN CELEE!!!!!!!!

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