Let 75 ldots 57 denote the (r+2) digit number | vedclass

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(C) Let the $k$-th term be $T_k = 7 	imes 10^k + 5 	imes \frac{10^{k-1}-1}{9} + 7$. Alternatively,observe the pattern: $T_1 = 77$,$T_2 = 757$,$T_3 =

Vedclass · Accepted Answer

$1219$

Vedclass · Answer

(C) Let the $k$-th term be $T_k = 7 	imes 10^k + 5 	imes \frac{10^{k-1}-1}{9} + 7$. Alternatively,observe the pattern: $T_1 = 77$,$T_2 = 757$,$T_3 = 7557$. $T_k = 75 \ldots 57$ ($k-1$ fives). $T_k = 7 	imes 10^k + 5 	imes (10^{k-1} + 10^{k-2} + \ldots + 10^1) + 7 = 7 	imes 10^k + 5 	imes \frac{10(10^{k-1}-1)}{9} + 7$. Sum $S = \sum_{k=1}^{98} T_k = \sum_{k=1}^{98} (7 	imes 10^k + \frac{50}{9} 	imes 10^{k-1} - \frac{50}{9} + 7)$. Using the property of the sum,$9S = 10S - S$. $10S = 770 + 7570 + 75570 + \ldots + 75 \ldots 570$. $S = 77 + 757 + 7557 + \ldots + 75 \ldots 57$. $9S = 75 \ldots 57 + (75570 - 7557) + \ldots + (770 - 77) - 77$. $9S = 75 \ldots 57 + 98 	imes 13 - 77 = 75 \ldots 57 + 1274 - 77 = 75 \ldots 57 + 1210$. Thus,$S = \frac{75 \ldots 57 + 1210}{9}$. Here $m = 1210$ and $n = 9$. $m + n = 1210 + 9 = 1219$.

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