if x,y a and b are real numbers and if x+y = a/b+cosA+SinA.Show that (b^2-1)(x^2+y^2)+a^2=2abx?

Question

2.find the sum of the term from the (n+1)th to the mth term inclusive of a arithemetrical progression whose first term is a and whose second term is b, if m=13,n=3 and the sum is 12a find the ratio b:a

Answer 1

1.
Suppose b = 2, a = 0, A = 0, x = 0, y = 1. Obviously x+y = a/b + cos A + sin A (and A = a in case those are not meant to be different), but (b^2 - 1)(x^2+y^2) + a^2 = 3 while 2abx = 0. So the statement is false.

2.
kth term is a + (k-1)(b-a)
Sum of first r terms is ra + r(r-1)(b-a)/2
So sum of (n+1)th to mth term is (sum of first m terms) - (sum of first n terms)
= (ma + m(m-1)(b-a)/2) - (na + n(n-1)(b-a)/2)
= (m-n) a + (b-a)/2 [m(m-1) - n(n-1)]
If m=13, n=3 we have
10a + (b-a)/2 [(13(12) - 3(2)]
= 10a + 75(b-a)
= 12a => 75(b-a) = 2a
=> b - a = (2/75) a
=> b = (77/75) a
so b:a :: 77:75.