Latest TCS Aptitude Question SOLUTION: The sum of 3 consecutive no of 4 no A,B,C,D are 4613,4961,5010,5099 then what is the largest no among A,B,C,d?

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The sum of 3 consecutive no of 4 no A,B,C,D are 4613,4961,5010,5099 then what is the largest no among A,B,C,d?

Asked In TCS Ashish Singh (11 years ago)

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Garima  (11 years ago)

Let, S1,S2,S3,S4 be the sums of A,B,C,D taking 3 of them at a time.

S1+S2+S3+S4 = 4613+4961+5010+5099
=>S1+S2+S3+S4 = 19683....[eqn 1]

S1= A+B+C
S2= B+C+D
S3= C+D+A
S4= D+A+B

Hence
S1+S2+S3+S4 = 3(A+B+C+D)
3(A+B+C+D)= 19683

(A+B+C+D)=6561

So,the greatest term among them =sum of all four - sum of lowest three terms
So,the greatest term among them=6561-4613 = 1948.

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