A 200 followers problem

if real numbers a,b,c,d,e satisfy

a+1=b+2=c+3=d+4=e+5=a+b+c+d+e+3

what is the value of a^2+b^2+c^2+d^2+e^2 ?


The answer is 10.

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2 solutions

Abhishek Sharma
Apr 12, 2015

Let a + 1 = b + 2 = c + 3 = d + 4 = e + 5 = a + b + c + d + e + 3 = k a+1=b+2=c+3=d+4=e+5=a+b+c+d+e+3=k .

Now a + 1 = k a+1=k .

Rearranging we get a = k 1 a=k-1 .

Similarly, b = k 2 b=k-2 , c = k 3 c=k-3 , d = k 4 d=k-4 , e = k 5 e=k-5 .

Also a + b + c + d + e + 3 = k a+b+c+d+e+3=k .

Substituting the values in we get k = 3 k=3 .

Now a = 2 a=2 , b = 1 b=1 , c = 0 c=0 , d = 1 d=-1 , e = 2 e=-2 .

Therefore a 2 + b 2 + c 2 + d 2 + e 2 = 10 \boxed{{a}^{2}+{b}^{2}+{c}^{2}+{d}^{2}+{e}^{2}=10} .

Did exactly the same way.!! 😀😀😀

Anurag Pandey - 4 years, 10 months ago
Saurav Sah
Apr 1, 2015

we can take a=2 b=1 c=0 d=-1 e=-2 . and by substituting value we get answer as 10

abhishek anand .I had seen all your problems are just copied from india pre rmo 2014 mumbai region question paper.this is also from that.

Saurav Sah - 6 years, 2 months ago

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yes, my exams are over so i was solving them so i posted them on brilliant

abhishek anand - 6 years, 2 months ago

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