For any three positive real numbers a,b and c,n 9( 25a^2 + b^2 ) + 25( c^2 - 3ac ) = 15b(3a + c).n Then

n ( a): 9(25 a^2+b^2)+25(c^2-3 a c)=15 b(3 a+c) n => (15 a)^2+(3 b)^2+(5 c)^2-(15 a)(5 c)-(15 a)(3 b)-(3 b) n (5 c)=0 n => (1 / 2)[(15 a-3 b)^2+(3 b-5 c)^2+(5 c-15 a)^2]=0 n => (15 a-3 b)^2=0,(3 b-5 c)^2=0,(5 c-15 a)^2=0 n => 15 a=3 b=5 c => (a / 1)=(b / 5)=(c / 3) => b, c, a are in A.P. n\

For any three positive real numbers a,b and c, 9( 25a... | Sequence & Series (Mathematics)

For any three positive real numbers

a,b

and

c

,\n

9\left( 25a^{2} + b^{2} \right) + 25\left( c^{2} - 3ac \right) = 15b(3a + c)

.\n Then

b,c

and

a

are in A.P.

a,b

and

c

are in A.P.

a,b

and

c

are in G.P.

b,c

and

a

are in G.P.

Prev Next

Correct Answer:

b,c

and

a

are in A.P.

Explanation

\begin{aligned}\n& (\mathrm{a}): 9\left(25 a^2+b^2\right)+25\left(c^2-3 a c\right)=15 b(3 a+c) \\\n& \Rightarrow(15 a)^2+(3 b)^2+(5 c)^2-(15 a)(5 c)-(15 a)(3 b)-(3 b) \\\n& (5 c)=0 \\\n& \Rightarrow \frac{1}{2}\left[(15 a-3 b)^2+(3 b-5 c)^2+(5 c-15 a)^2\right]=0 \\\n& \Rightarrow(15 a-3 b)^2=0,(3 b-5 c)^2=0,(5 c-15 a)^2=0 \\\n& \Rightarrow 15 a=3 b=5 c \Rightarrow \frac{a}{1}=\frac{b}{5}=\frac{c}{3} \Rightarrow b, c, a \text { are in A.P. }\n\end{aligned}

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