Question

find the nth and the sum of the first n term of progression 8,98,998,9998....in term of n.​

Answer 1

Answer:

nth term = 10^n - 2 ; Sum of n terms = $\frac{10(10^{n} - 1)}{9} - 2^{n}$

Step-by-step explanation:

sequence : 8,98,998,9998 ... n terms

this can be written as : (10-2) , (100-2) , (1000-2) , (10000-2) ...

1st term = 10 - 2

2nd term = 100 - 2 = 10^2 - 2

3rd term = 10^3 - 2

so, nth term = 10^n - 2

sum of n terms = 8+98+998+9998+ ... = (10-2)+(100-2)+(1000-2)...n terms

separating the terms, we get :

S(n) = (10+100+1000+10000+....) - (2+2+2+2+...)

we see, the first term is a geometric progression, with first term = 10, and ratio = 10

so, S(n) = sum of GP - 2(n)

->sum of GP : s(n) = a*(r^n-1) / (r-1)   =  10*(10^n-1) / (10-1)  = 10(10^n - 1) / 9

so we get :

S(n) = $\frac{10(10^{n} - 1)}{9} - 2^{n}$