CATAlgebra > MediumEntered answer:✅ Correct Answer: 6Related questions:CAT 2020 Slot 3If x1=−1x_1 = -1x1=−1 and xm=xm+1+(m+1)x_m = x_{m+1} + (m + 1)xm=xm+1+(m+1) for every positive integer m,m,m, then x100x_{100}x100 equalsCAT 2024 Slot 2The sum of the infinite series is 15(15−17)+(15)2((15)2−(17)2)+(15)3((15)3−(17)3)+….\frac{1}{5}\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{5}\right)^{2}\left(\left(\frac{1}{5}\right)^{2}-\left(\frac{1}{7}\right)^{2}\right)+\left(\frac{1}{5}\right)^{3}\left(\left(\frac{1}{5}\right)^{3}-\left(\frac{1}{7}\right)^{3}\right)+\ldots .51(51−71)+(51)2((51)2−(71)2)+(51)3((51)3−(71)3)+….. equal toCAT 2017 Slot 1Let a1,a2,……...a3na_{1}, a_{2}, \ldots \ldots . . . a_{3 n}a1,a2,……...a3n be an arithmetic progression with a1=3a_{1}=3a1=3 and a2=7a_{2}=7a2=7. If a1+a2+….+a3n=1830a_{1}+a_{2}+\ldots .+a_{3 n}=1830a1+a2+….+a3n=1830, then what is the smallest positive integer mmm such that m(a1+a2+….+an)>1830m\left(a_{1}+a_{2}+\ldots .+a_{n}\right)>1830m(a1+a2+….+an)>1830 ?