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Arithmetic Progression Assignment

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Added on  2021-04-24

Arithmetic Progression Assignment

   Added on 2021-04-24

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Ans.1. (a) Here the first term is a= 2.6 and Common Ratio (r) =?We have, ar= 4.68, ar^2 =8.424, ar^3= 15.1632 r=ar/a= 4.68/2.6= 1.8r=ar^2/ar=8.424/4.68=1.8r=ar^3/ar^2=15.1632/8.424=1.8Therefore the First term is 2.6 and Common Ratio= 1.8The recurrence System is (ar(n1))=¿2.6*(1.8)^(n-1) ; n=1,.....(b) The closed form for this sequence is 2.6*(1.8)^(n-1) ; n=1,.....(c) Tenth Term= a*r^(10-1) = 2.6*(1.8)^(9)=515.734(d)Let the number be ‘n’.a*r^(n-1) ≤ 39000or, 2.6*(1.8)^(n-1)≤39000or, (1.8)^(n-1)≤15000or, (n-1)ln(1.8)≤ln(15000)or, n-1≤16.35or, n≤17.35Since ‘n’ should be a natural number, so it should be n=17Ans.2. The sequence converges to -26Reason:limn{52(0.91)n26}¿limn{52(0.91)n}26 =0-26 =-26Ans.3. It is an Arithmetic ProgressionHere 1st Term a= 6 (From Feb 2016)And Common Difference = 4
Arithmetic Progression Assignment_1
Number of terms = 14Therefore, the required number of people=12+ n=114{6+(n1)4}= 12 + n=114{6} + n=114{(n1)4}= 12+6*14 + n=114{n4} - n=114{4}=12 +14*6+4*14*15/2-14*4=460Ans.4. n=1138{n33n2} = n=138{n33n2} - n=111{n33n2}= n=138{n3} - n=138{3n2} - n=111{n3} + n=111{3n2}=(38*39/2)^2 -3*38*39*77/6 – (11*12/2)^2 + 3*11*12*23/6=549081-57057- 4356+ 1518=489186Ans.5. We have the 8th term as 14C8 *((3*a^2)^8)*(9*a^3)^(-6)Therefore the required coefficient is 14C8*3^(-4)Ans.8.(a) We have f(x)=(14 + x)^(-1/4)Therefore differentiating both sides w.r.t. x we get,f(x)=(-1/4)(14 + x)^(-5/4)or, g(x)=(-4)f(x)= (14 + x)^(-5/4)
Arithmetic Progression Assignment_2
Now, we are havingf(x)=2 -2x+522x^2 - 1522x^3 +....Differentiating, both sides w.r.t. x and multiplying by -4 we get,g(x)=(-4)f(x)= 42+202x -+902 x^2+.... Which is the required Taylor Series for g(x)= (14 + x)^(-5/4).(b)We have f(x)=(14 + x)^(-1/4)Therefore integrating both sides w.r.t. x we get,F(x)=(4/3)(14 + x)^(3/4)or, h(x)=F(x)=(4/3) (14 + x)^(3/4)Now, we are havingf(x)=2 -2x+522x^2 - 1522x^3 +....Integrating, both sides w.r.t. x and multiplying by ¾ we get,h(x)=F(x)= ¾ 2x – 3/8 2 x^2 + 5/3 2 x^3 – 15/8 2 x^4 , which is the required Taylor Series for h(x)= (14 + x)^(3/4).Ans.9.The required equation is:z^2 – (3/4 +2i +3/4 -2i)z + (3/4 + 2i)(3/4-2i)=0or, z^2 –(6/4)z + (73/16)=0Ans.10.z^4 = -23-2i = -2(3 + i) =-4(32+i2) =(2i)^2 (cos(π/6)+sin(π/6)) =((2i)^2)*(e^(iπ/6))
Arithmetic Progression Assignment_3

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