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The question below is from previous year CAT question from CAT 2018 exam comes from CAT Modern Maths – Sequence and Series – Let t1, t2,… be real numbers… Find out by answering this question which tests an aspirant’s Quantitative Ability Skills:
Q. 4: Let t1, t2,… be real numbers such that t1+t2+…+tn = 2n2+9n+13, for every positive integer n ≥ 2. If tk=103, then k equals?
Given, t(base1)+t(base2)+…+t(base n) = 2n^2+9n+13
So t(base1) + t(base2) = 2×2^2+9×2+13=39
t(base1) + t(base2) + t(base3) = 2×3^2+9×3+13=58 means t(base 3) = 58 – 39 = 19
t(base1) + t(base2) + t(base3)+t(base4) = 2×4^2+9×4+13=81 means t(base4) = 81 – 58 = 23
t(base1) + t(base2) + t(base3)+t(base4) + t(base5) = 2×5^2+9×5+13=108 means t(base4) = 108 – 81 = 27
t(base1) + t(base2) + t(base3)+t(base4) + t(base5) + t(base6) = 2×6^2+9×6+13=139 means t(base4) = 139 – 108 = 31
So we can see that t(base3), t(base4) , t(base5) , t(base6) form an A.P. 19, 23, 27, 31
t(basek) = 4(k+1) + 3
Given , t(basek) = 103
4(k+1) + 3 = 103
k = 24