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
$${\kern 1pt} {\frac{5}{9}} \times \left( {{{10}^n} - 1} \right)$$
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Related Questions on Progressions
Find the first term of an AP whose 8th and 12th terms are respectively 39 and 59.
A. 5
B. 6
C. 4
D. 3
E. 7
The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is
A. 600
B. 765
C. 640
D. 680
E. 690
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