Arithmetic Progression Assignment

Added on - 24 Apr 2021

  • 10

    Pages

  • 1152

    Words

  • 11

    Views

  • 0

    Downloads

Trusted by +2 million users,
1000+ happy students everyday
Showing pages 1 to 3 of 10 pages
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= 15.1632r=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 1stTerm a= 6 (From Feb 2016)And Common Difference = 4
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 8thterm as14C8*((3*a^2)^8)*(9*a^3)^(-6)Therefore the required coefficient is14C8*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)
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)= ¾2x – 3/82 x^2 + 5/32 x^3 – 15/82 x^4 ,which is the required Taylor Series forh(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))
desklib-logo
You’re reading a preview
Preview Documents

To View Complete Document

Click the button to download
Subscribe to our plans

Download This Document