Question

Q. If the sum of nine consecutive natural number is 117,then the greatest of these natural number is.​

Answer 1

$\bigstar\huge\bf\orange{Answer}$

$The \: consecutive \: integers \: are \: \\1st \: number \:= n, \\2nd \: number= n + 1, \\ 3rd \: number=n + 2, \\ 4th \: number=n + 3, \\5th \: number= n + 4, \\ 6th \: number=n + 5, \\7th \: number= n + 6, \\8th \: number= n + 7 ,and \\9th \: number= n + 8$

By problem,

n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)+(n+6)+(n+7)+(n+8)=117

$9n + 36 = 117 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 9n = 117 - 36 \\ 9n = 81 \\ n = \frac{81}{9 } \\ n = 9$

$there \: fore, \\ 1st \: integer=n \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =9 \\ 2nd \: number=n+1=9+1 \\ =10 \\ 3rd \: number=n+2=9+2 \\ =11 \\ 4th \: number=n+3=9+3 \\ =12 \\ 5th \: number=n+4=9+4 \\ =13 \\ 6th \: number=n+5=9+5 \\ =14 \\ 7th \: number=n+6=9+6 \\ =15 \\ 8th \: number=n+7=9+7 \\ =16 \\ 9th \: number=n+8=9+8 \\ </p><p>=17$

The greatest of these natural numbers is

$\huge\bf\pink{17}$

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