CATAlgebra > Hard191919161616171717181818✅ Correct Option: 1Related questions:CAT 2024 Slot 3Consider the sequence t1=1t_1 = 1t1=1, t2=−1t_2 = -1t2=−1 and tn=(n−3n−1)tn−2t_n = \left(\frac{n-3}{n-1}\right) t_{n-2}tn=(n−1n−3)tn−2 for n≥3n \ge 3n≥3. The, the value of the sum 1t2+1t4+1t6+⋯+1t2022+1t2024\frac{1}{t_2} + \frac{1}{t_4} + \frac{1}{t_6} + \dots + \frac{1}{t_{2022}} + \frac{1}{t_{2024}}t21+t41+t61+⋯+t20221+t20241 isCAT 2017 Slot 2An infinite geometric progression a1,a2,a3,...a_1, a_2, a_3,...a1,a2,a3,... has the property that an=3(an+1+an+2+....)a_n = 3(a_{n+1} + a_{n+2} +....)an=3(an+1+an+2+....) for every n≥1n \ge 1n≥1. If the sum a1+a2+a3+.....=32a_1 + a_2 + a_3 +..... = 32a1+a2+a3+.....=32, then a5a_5a5 isCAT 2019 Slot 2Let a1,a2,...a_1, a_2, ...a1,a2,... be integers such that a1−a2+a3−a4+...+(−1)n−1an=na_1 - a_2 + a_3 - a_4 + ... + (-1)^{n - 1} a_n = na1−a2+a3−a4+...+(−1)n−1an=n, for all n≥1n \ge 1n≥1. Then a51+a52+...+a1023a_{51} + a_{52} + ... + a_{1023}a51+a52+...+a1023 equals