2 n in a d is 2 = nickels in a dime.
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The sum of an arithmetical sequence whose nth term is U(n) = a + (n-1)*d is S(n) = 1/2*n*[2a + (n-1)d] or 1/2*n(a + l) where l is the last term in the sequence.
Assuming nickles, dimes, and quarters, there are ten different ways to make change for a half dollar. Just enumerate the combinations... 10 n 8 n 1 d 6 n 2 d 4 n 3 d 2 n 4 d 5 d 5 n 1 q 3 n 1 d 1 q 1 n 2 d 1 q 2 q
1 q 1 d 1 q 2 n 2 d 3 n 3 d 1 n 1 d 5 n 7 n
This can be found out by using the formula for sum of n terms of an arithmetic progression. here, n is not known. a=2 d=2 (since it's even) nth term=98 (last even number less than 100) using formula for nth term, nth term=a+(n-1)d 98=2+(n-1)2 therefore, n=49 so, sum of n terms=n/2[2*a+(n-1)d] putting n=49, a=2, and d=2, sum=2450
Base (b) = 1 000 000 Number (n) = 2 416 304 Modular (m) =n Mod b = 416304 Integer Difference (d) = n - b = 2 if m/b is more than or equal to 0.5 then d = d+1 else d answer is (d), 2 million