ABCD is a square of length a, `a in N`, a > 1. Let `L_1, L_2 , L_3...` be points on BC such that `BL_1 = L_1 L_2 = L_2 L_3 = .... 1` and `M_1,M_2 , M_3,....`be points on CD such that `CM_1 = M_1M_2= M_2 M_3=... = 1`. Then `sum_(n = 1)^(a-1) ((AL_n)^2 + (L_n M_n)^2)` is equal to :

A B C is a right-angled triangle in which /_B=90^0 and B C=adot If n points L_1, L_2, ,L_nonA B is divided in n+1 equal parts and L_1M_1, L_2M_2, ,L_n M_n are line segments parallel to B Ca n dM_1, M_2, ,M_n are on A C , then the sum of the lengths of L_1M_1, L_2M_2, ,L_n M_n is (a(n+1))/2 b. (a(n-1))/2 c. (a n)/2 d. none of these

How many electrons in an atom can have n = 3, l = 1, m =-1 and s = + 1//2

If l : m = 2 (1)/(2) : 1(2)/(3) and m : n = 1 (1)/(4) : 3 (1)/(2) , find l : m : n

If L_1 = 2.02 m pm 0.01 m , L_2 = 1.02 m pm 0.01 m , determine L_1 + 2L_2

If a != 1 and l n a^(2) + (l n a^(2))^(2) + (l n a^(2))^(3) + ... = 3 (l n a + (l n a)^(2) + ( ln a)^(3) + (l n a)^(4) + ....) then 'a' is equal to

ABC is a right-angled triangle in which angleB=90^(@) and BC=a. If n points L_(1),L_(2),…,L_(n) on AB is divided in n+1 equal parts and L_(1)M_(1), L_(2)M_(2),…,L_(n)M_(n) are line segments paralllel to BC and M_(1), M_(2),….,M_(n) are on AC, then the sum of the lengths of L_(1)M_(1), L_(2)M_(2),...,L_(n)M_(n) is

The line l_1 x+ m_1 y+n=0 and l_2 x + m_2 y+n=0 will cut the coordinate axes at concyclic points if (A) l_1 m_1 = l_2 m_2 (B) l_1 m_2 = l_2 m_1 (C) l_1 l_2 = m_1 m_2 (D) l_1 l_2 m_1 m_2 = n^2

If l, m, n denote the side of a pedal triangle, then (l)/(a ^(2))+(m)/(b^(2))+(n)/(c ^(2)) is equal to

If vec aa n d vec b are two non-collinear vectors, show that points l_1 vec a+m_1 vec b ,l_2 vec a+m_2 vec b and l_3 vec a+m_3 vec b are collinear if |l_1 l_2 l_3 m_1m_2m_3 1 1 1|=0.

The direction cosines of the lines bisecting the angle between the line whose direction cosines are l_1, m_1, n_1 and l_2, m_2, n_2 and the angle between these lines is theta , are