**Slot -2 – Quantitative Aptitude – Modern Maths – Sequence and Series – Let t1, t2,… be real numbers**

**Let t _{1}, t_{2},… be real numbers such that t_{1}+t_{2}+…+t_{n} = 2n^{2}+9n+13, for every positive integer n ≥ 2. If t_{k}=103, then k equals?**

**Answer:** 24

**Solution:**

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

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