ABCD is a square of length a, \(a \in \mathbb{N}\), \(a > 1\). Let \(L_1, L_2, L_3, \ldots\) be points on BC such that \(BL_1 = L_1L_2 = L_2L_3 = \cdots = 1\) and \(M_1, M_2, M_3, \ldots\) be points on CD such that \(CM_1 = M_1M_2 = M_2M_3 = \cdots = 1\). Then \(\sum_{n=1}^{a-1}(AL_n^2 + L_nM_n^2)\) is equal to
Let x, x, x, x be in a geometric progression. 2, 7, 9, 5 are subtracted respectively from x, x, x x then the 1 2 3 4 1 2 3 4 resulting numbers are in an arithmetic progression. Then the value of 1 24$($$x_{1}$$$x_{2}$$$x_{3}$$$$$$x_{4}$$)$is:
For Problems 19–21: Let \(A_1, A_2, A_3, \ldots, A_m\) be the arithmetic means between \(-2\) and 1027 and \(G_1, G_2, G_3, \ldots, G_n\) be the geometric means between 1 and 1024. The product of geometric means is \(2^{45}\) and sum of arithmetic means is \(1025 \times 171\).The numbers \(2A_{171},\ G_5^2 + 1,\ 2A_{172}\) are in
Let $3,7,11,15,\ldots,403$ and $2,5,8,11,\ldots,404$ be two arithmetic progressions. Then the sum of the common terms in them, is equal to
If sum of the series $1+\dfrac{\sqrt5-\sqrt3}{2\sqrt5}+\dfrac{8-2\sqrt{15}}{30}+\dfrac{14\sqrt5-18\sqrt3}{60\sqrt5}+\cdots=2+\dfrac{a+\sqrt{15}}{b}\log_b\!\left(\dfrac{a}{c}\right)$; $a,b,c\in\mathbb{N}$, $\gcd(a,b,c)=1$, then $2(a^2+b^2+c^2)$ is