1Answer guyz !
There exists a sequence such that
i)a1=a
ii)a2=b
and an+2=(an+1+an)/2
n≥0 ;
Find limn->∞ an
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10 Answers
1a1= a
a2=b
a3= (a+b)/2
a4=(a+3b)/4
a5=(3a+5b)/8
a6=(5a+11b)/16
a7=(11a+21b)/32
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.
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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;
Sm = 3 + 5 + 11 + 21 + ................... + tm
Sm = 3 + 5 + 11 + 21 + ........... + tm-1 + tm
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0 = 3 + [2+6+10+....+ till m-1 terms] - tm
=> tm = 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: 4X2m-1
then a'm = [(4m-6)a + (4m-3)b]/ 4.2m-1
if n-> infi , m also -> infi
so, lim m->infi [(4m-6)a + (4m-3)b]/ 4.2m-1
9hint :
2 an+2 + an+1 = a + 2b
now try finding proof for this and then use some logic
21The 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!!!!!!!!
[1]
9we are gvn
2 an+2 = an+1+ an
also 2 an+1 = an + an-1
........
.....
.....
.......
2a3 = a +b
now add all these and simplify u ll get