UPSEE Solved Model Paper 1

Let two numbers are x any y then x, a, b,c,y are in A.P. Let common difference of this A.P. = d Then we can take x = b-2d a = b-d b = b y = b + 2d now, a + b + c = 15 ⇒ (b-d) + b + (b + d) = 15 ⇒ b = 5 x + y = (b - 2d) + (b + 2d) x + y = 2b x+ y = 10 ......(i) Now, x, p, q, r, y are in H.P. Then $\frac{1}{x},\frac{1}{p},\frac{1}{q},\frac{1}{r},\frac{1}{y}$ will be in A.P. Let common difference of this A.P. = d' Then we can take $\frac{1}{x}$ = $\frac{1}{q}-2d'$ $\frac{1}{p}$ = $\frac{1}{q}-d'$ $\frac{1}{q}$ = $\frac{1}{q}$ $\frac{1}{r}$ = $\frac{1}{q}+d'$ $\frac{1}{y}$ = $\frac{1}{q}+2d'$ Now, $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}$ = $\frac{5}{3}$ ⇒ $\frac{3}{q}$ = $\frac{5}{3}$ ⇒ $\frac{1}{q}$ = $\frac{5}{9}$ and $\frac{1}{x}+\frac{1}{y}$ = $\frac{2}{q}$ and $\frac{1}{x}+\frac{1}{y}$ = $\frac{10}{9}$ $\frac{x+y}{xy}$ = $\frac{10}{9}$ ⇒ $\frac{10}{xy}$ = $\frac{10}{9} (x+y=10)$ xy = 9 ... (ii) x - y = $\sqrt{(x^2+y^2)-4xy}$ = $\sqrt{(10)^2-4.9}$ 𝑥 − 𝑦 = 8 . . . . . . (𝑖𝑖𝑖) from (i) and (iii) x = 9 y = 1