Let the angles of a triangle ABC be in A.P. and letn b:c = sqrt(3):sqrt(2). Find the angle A.

n n A+B+C=180^°, 2 B=A+C(Because A, B, C are in A . P) n Therefore 3 B=180^° => B=60^°n nnnGiven fracbsqrt(3)=fraccsqrt(2)=k Therefore b=k sqrt(3), c=k sqrt(2)nBy the sine rule, (b / sin B)=(c / sin C)nn n frack sqrt(3)sin 60^°=frack sqrt(2)sin C Therefore sin C n =frac1sqrt(2) n C=45^° n C=135^° is impossible n A=180^°-(B+C)=180^°-(60^°+45^°)=75^°n n

Let the angles of a triangle ABC be in A.P. and let b... | Sequence & Series (Mathematics)

\n\begin{aligned}\n& A+B+C=180^{\circ}, 2 B=A+C(\because A, B, C \text { are in } A . P) \\\n& \therefore 3 B=180^{\circ} \Rightarrow B=60^{\circ}\n\end{aligned}\n

\n\nGiven

\frac{b}{\sqrt{3}}=\frac{c}{\sqrt{2}}=k \quad \therefore b=k \sqrt{3}, c=k \sqrt{2}

\nBy the sine rule,

\frac{b}{\sin B}=\frac{c}{\sin C}

\n\begin{aligned}\n& \frac{k \sqrt{3}}{\sin 60^{\circ}}=\frac{k \sqrt{2}}{\sin C} \therefore \sin C \\\n& =\frac{1}{\sqrt{2}} \\\n& C=45^{\circ} \\\n& C=135^{\circ} \text { is impossible } \\\n& A=180^{\circ}-(B+C)=180^{\circ}-\left(60^{\circ}+45^{\circ}\right)=75^{\circ}\n\end{aligned}\n

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

Let a_{1},a_{2},a_{3},. be a G.P. of increasing positive numbers. Let the sum of its 6^{{th}{~}} and 8^{{th}{~}} terms be 2 and the product of its 3^{{rd}{~}} and 5^{{th}{~}} terms be 1/9. Then 6( a_{2} + a_{4} )( a_{4} + a_{6} ) is equal to

For any odd integer n ? 1,n^{3} - (n - 1)^{3} + + ( - 1)^{n - 1}1^{3} =

The sum of the first n terms of the series (1^{2} + 2.2^{2} + 3^{2} + 2.4^{2} + 5^{2} + 2 . 6^{2} + ..) is (\frac{n(n + 1)^{2}}{2}, when n is even. when n is odd, the sum is

Let a,b be two non-zero real numbers. If p and r are the roots of the equation x^{2} - 8ax + 2a = 0 and q and s are the roots of the equation \end{enumerate} x^{2} + 12bx + 6b = 0, such that \frac{1}{p},\frac{1}{q},\frac{1}{r},\frac{1}{s} are in A.P., then a^{- 1} - b^{- 1} is equal to

If a_{1},a_{2},a_{3}, are in A.P. such that a_{1} + a_{7} + a_{16} = 40, then the sum of the first 15 terms of this A.P. is

Let \frac{1}{x_{1}},\frac{1}{x_{2}},,\frac{1}{x_{n}}( x_{i} eq 0 .\ for . \ i = 1,2,,n ) be in A.P. such that x_{1} = 4 and x_{21} = 20. If n is the least positive integer for which x_{n} > 50, then \sum_{i = 1}^{n}\mspace{2mu}( \frac{1}{x_{i}} ) is equal to

Let 3,a,b,c be in A.P. and 3,a - 1,b + 1,c + 9 be in G.P. Then, the arithmetic mean of a,b and c is

Number of 4 -digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11 , is equal to

Download Divine JEE App

Quick Links

Exam Categories

Contact Info

Divine JEE