Question

10. Find the sum of all natural numbers between 300 and 500 which are divisible by 11

Answer 1

General term or nth term of A.P
The general term or nth term of A.P is given by an or tn = a + (n – 1)d, where a = a1 is the first term, d is the common difference and n is the number of term.

Sum of n terms of an AP
The sum of first n terms of an AP with first term 'a' and common difference 'd' is given by
Sn = n /2 [ 2a + ( n - 1) d] or
Sn=n /2 [ a + l ] (l = last term)

SOLUTION IS IN THE ATTACHMENT

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Answer 2

First term which is divisible by 11 between 300 and 500 is 308
Hence, a = 308

Last term which is divisible by 11 between 300 and 500 is 495
Hence, l = 495


$a_{n} = a + (n - 1)d \\ \\ \\ 495 = 308 + (n - 1)11 \\ \\ \\ 495 - 308 = (n - 1)11 \\ \\ \\ 187 = (n - 1)11 \\ \\ \\ \\ \frac{187}{11} = n - 1 \\ \\ \\ 17 = n - 1 \\ \\ \\ 18 = n$


Then,


$s_{n} = \frac{n}{2} (a + l) \\ \\ \\ s_{n} = \frac{18}{2} (308 + 495) \\ \\ \\ s_{n} = 9(803) \\ \\ \\ s_{n} = 7227$





Sum of numbers which are divisible by 11 between 300 and 500 is 7227