//program to get 0.9,0.99,0.999,0.9999.....nth terms.

#include<stdio.h>

#include<conio.h>

#include<math.h>

void main()

{

float n,j, k=0.9;

clrscr();

printf("enter number of terms\n");

scanf("%f",&n);

for(j=1;j<=n;j++)

{

printf("%f,",k);

k=k+0.9/(pow(10,j));

}

getch();

}

-------------------------------------------------------------------------------------------------------------------------

->and to display all the terms we have used loop

or

we can apply following technique:

//getting 0.9,0.99,0.999.....nth term

#include<stdio.h>

#include<conio.h>

#include<math.h>

void main()

{

float k=9,n,p=10,j,dis,m=1;

clrscr();

printf("enter number of terms\n");

scanf("%f",&n);

for(j=1;j<=n;j++)

{

dis=k/(pow(10,j));

printf("%f,",dis);

m++;

k=pow(p,m)-1;

}

getch();

}

->take 9 as a fixed value

->we enter no. of terms for 'n'.

->to display 0.9, we divide it (9) by 10

#include<stdio.h>

#include<conio.h>

#include<math.h>

void main()

{

float n,j, k=0.9;

clrscr();

printf("enter number of terms\n");

scanf("%f",&n);

for(j=1;j<=n;j++)

{

printf("%f,",k);

k=k+0.9/(pow(10,j));

}

getch();

}

-------------------------------------------------------------------------------------------------------------------------

**logics in mind for this:-****->0.9,0.99,0.999............. can be written as****0.9,0.9+0.09,0.99+0.009.......****-> for second term 0.99, it can be written as 0.9+0.09****->further it can be as(taking only two terms)****0.9+( 0.9+0.9/10)...****->it can be as****0.9+(0.9+0.9/10**^{1)...}^{->now we have to convert it in a formula as}**k=k+0.9/(pow(10,j)).here k=0.9 is taken**^{ }->and to display all the terms we have used loop

or

we can apply following technique:

//getting 0.9,0.99,0.999.....nth term

#include<stdio.h>

#include<conio.h>

#include<math.h>

void main()

{

float k=9,n,p=10,j,dis,m=1;

clrscr();

printf("enter number of terms\n");

scanf("%f",&n);

for(j=1;j<=n;j++)

{

dis=k/(pow(10,j));

printf("%f,",dis);

m++;

k=pow(p,m)-1;

}

getch();

}

->take 9 as a fixed value

->we enter no. of terms for 'n'.

->to display 0.9, we divide it (9) by 10

^{1}^{->to get 0.99, we take here 99, and for this we use 100-1 concept. we do this using k=pow(p,m)-1}^{ here pow finds power of 10(p).}^{->in next execution we get 0.99 and so on..}^{ }^{-> this is how loop continues and we get all terms}^{ }^{we can use different technique for same program.}**similarly we can also get****1)9,99,999,9999......****2)2,22,222,2222....****3)5,55,555,5555......****4)0.99999,0.9999,0.999,0.99,0.9****for this,****there are 5 nines so it can be written as****(100000-1)/10**^{5},(10000-1)/10^{4}........**(****10**^{5}**-1)/10**^{5},(**10**^{4}**-1)/10**^{4}........**now make a formula to get this one. And it is easy task...**

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