Given that, the students of a school decided to beautify the school on the annual day by fixing colorful flags on the straight passage of the school. Given that, the number of flags = 27 and distance between each flag = 2m. Also, the flags are stored at the position of the middle most flag i.e., 14th flag and Ruchi was given the responsibility of placing the flags. Ruchi kept her books, where the flags were stored i.e., 14th flag and she could carry only one flag at a time. Let she placed 13 flags into her left position from middle most flag i.e., 14th flag. For placing second flag and return his initial position distance travelled = 2 + 2 = 4m. Similarly, for placing third flag and return his initial position, distance travelled = 4 + 4 = 8 m. For placing fourth flag and return his initial position, distance travelled = 6 + 6 = 12 m. For placing fourteenth flag and return his initial position, distance travelled = 26 + 26 = 52 m So this becomes a series, 4, 8, 12, 16, – – – , 52 Proceed same manner into her right position from middle most flag i.e., 14 flag. (Also, when Ruchi placed the last flag in her rightmost, she return his middle most position and collect her books. This distance also included in placed the last flag.) We get the same series 4, 8, 12, 16, – – – , 52 Clearly this series is an AP with First term, a = 4 , common difference, d = 4 and no of terms. n = 13 Total distance covered by Ruchi for placing these flags = 2Sn = 13(56) = 13(56) = 728 m Hence, the required is 728 m in which she did cover in completing this job and returning back to collect her books. Now, the maximum distance she travelled carrying a flag = Distance travelled by Ruchi during placing the 14th flag in her left position or 27th flag in her right position = 13(2) = 26 m

Hence, the maximum distance she travelled carrying a flag is 26 m.