Question

find the sum of all two-digit numbers which are multiple of 11​

Answer 1

Answer:

297

Step-by-step explanation:

The question is related to AP.

11, 22, 33, ....... 99

a = 11, d = 11, n = 9.

Sum = n/2 [ 2a + ( n-1 )d ]

= 9/2[ 2 × 11 + 10×11 ]

= 9/2[ 11 × 6 ]

= 297.

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Answer 2

SOLUTION

TO DETERMINE

The sum of all two-digit numbers which are multiple of 11

FORMULA TO BE IMPLEMENTED

The sum of first n natural numbers

$\displaystyle \sf{ = \frac{n(n + 1)}{2} }$

EVALUATION

The two-digit numbers which are multiple of 11 are 11 , 22 , 33 , 44 , 55 , 66 , 77 , 88 , 99

Hence the required sum

= 11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99

= 11 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 )

$\displaystyle \sf{ = 11 \times \frac{9(9 + 1)}{2} }$

$\displaystyle \sf{ = 11 \times \frac{9 \times 10}{2} }$

$\displaystyle \sf{ = 11 \times 9 \times 5 }$

$\displaystyle \sf{ = 495 }$

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