Find the nth term of the following sequence : 5 + 55 + 555 + . . . . Tn
A. 5(10n - 1)
B. 5n(10n - 1)
C. $$\frac{5}{9} \times \left( {{{10}^n} - 1} \right)$$
D. $${\left( {\frac{5}{9}} \right)^n} \times \left( {{{10}^n} - 1} \right)$$
Answer: Option C
Solution (By Examveda Team)
We will it through option checking method:$$\eqalign{ & {\frac{5}{9}} \times \left( {{{10}^n} - 1} \right) \cr & {\text{We}}{\kern 1pt} {\kern 1pt} {\text{put}}{\kern 1pt} {\kern 1pt} n = 1, \cr & {\frac{5}{9}} \times \left( {{{10}^1} - 1} \right) = 5 \cr & n = 2\left( {\frac{5}{9}} \right) \times \left( {{{10}^2} - 1} \right) = 55 \cr & n = 3\left( {\frac{5}{9}} \right) \times \left( {{{10}^3} - 1} \right) = 555 \cr} $$
It means Option C is satisfying the sequence so the nth term would be
$${\kern 1pt} {\frac{5}{9}} \times \left( {{{10}^n} - 1} \right)$$
This Question Belongs to Arithmetic Ability >> Progressions
Sum = 5 + 55 + 555 + ....
= 5(1) + 5(11) +5(111) + ....
= 5[1 + 11 + 111 + ...]
= 5/9 [9(1 + 11 + 111 +...)]
= 5/9 [9 + 99 + 999 +..]
= 5/9 [(10-1) + (10^2 -1) + (10^3 -1) +...]
= 5/9(10^n - 1)
Grt solution
Can u give me the formula