Question

$\large\mathbf{HEY\: QUESTION❤}$



Find the sum of all three digit natural numbers, which are multiples of 11.



★ REQUIRED QUALITY ANSWER​

Answer 1

Answer:

110 and 990

Step-by-step explanation:

hope you understand well

Answer 2

Answer:

We know that the sum of terms for an A.P is given by

Sn = n/2[2a + (n − 1)d]

Where; a = first term for the given A.P. d

= common difference of the given A.P. n

= number of terms All 3 digit natural number which are multiples of 11.

So, we know that the first 3 digit multiple of 11 is 110 and the last 3 digit multiple of 13 is 990.

And, these terms form an A.P. with the common difference of 11.

Here, first term (a) = 110 and the last term (l) = 990 Common difference (d) = 11

Finding the number of terms in the A.P. by, an = a + (n − 1)d

We get,

990 = 110 + (n – 1)11

⟹ 990 = 110 + 11n -11

⟹ 990 = 99 + 11n

⟹ 11n = 891

⟹ n = 81

Now, using the formula for the sum of n terms,

we get

S81 = 81/2[2(110) + (81 − 1)11]

= 81/2[220 + 880]

= 81/2[1100]

= 81(550)

= 44550

Hence, the sum of all 3 digit multiples of 11 is 44550.

Step-by-step explanation:

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