Sums IV
What is the sum of all even number between 10 and 500 inclusive?
The answer is 62730.
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2 solutions
500=10+(n-1)xd so n=246 Put this in sum equation i.e S of 246 terms=10 +(246-1)x2 d=2 because the series is of even numbers
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please help me to give this solution
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What help do u want? Which step u couldn't understand?
S n = 2 n ( 2 a + ( n − 1 ) b )
To define the n , we can divide the last term by 2. There are 2 5 0 term ( 2 5 0 0 ) , but because it starts in 1 0 so 2 5 0 is subtracted with 4 ( 2 , 4 , 6 , 8 ) .
S 2 4 6 = 2 2 4 6 ( 2 ( 1 0 ) + ( 2 4 6 − 1 ) 2 )
S 2 4 6 = 1 2 3 ( 2 0 + 4 9 0 )
S 2 4 6 = 6 2 7 3 0
here, we can use the formula S = n/2 (a+l) where a = 10, l (last term) = 500, d(common difference)=2 . So, the AP formed will be - 10,12,14,16......488,500 & l = a + (n-1)d. on solving, we get n = 246. So, on putting the values a=10 , l =500 , & n= 246 in the formula S = n/2 (a+l), we get S = 62730.