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The sum of eight consecutive odd numbers is $656$. The average of four consecutive even numbers is $87$. What is the sum of the smallest odd number and second largest even number?
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Let the eight consecutive odd numbers be $n,n+2,n+4,n+6,n+8,n+10,n+12\;\text{and}\;n+14$
Sum of these numbers is $656, \implies 8n+56=656 , \; \text{so,}\; n= 75.$

Let the four consecutive even numbers be $m,m+2,m+4,\;\text{and}\;m+6$

Average of these numbers is $87, \implies \dfrac{(4m+12)}{4}=87 , \; \text{so,}\; m= 84$

Sum of  smallest odd number and second largest even number is $n + (m+4)= 75+88=163$
edited by
+4 votes

firstly do 656/8 by which u got 82 then with the help of it u decide where we start taking odd number. so bcz they said 8 consecutive odd no so take 4 consusuctive no before 82 and takes 4 odd consecutive no after 82 like this

75,77,79,81   83,85,87,89,  if u add them u get 656 .so in it smallest odd number is 75

then they given avg of 4 consecutive even no is 87  so take in same way so 84,86,88,90  here you got second largest even no is 88

then they ask sum of smallest oddd no and second largest even no which is 88+75=163 ans


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