Offline paper
Mathematics · 2017
JEE Main 2017 · 2017-04-02 Offline · Q68
For any three positive real numbers a, b and c, 9(25 a^2+b^2)+25(c^2-3 a c)=15 b(3 a+c). Then:
For any three positive real numbers a, b and c,
$$9\left(25 a^2+b^2\right)+25\left(c^2-3 a c\right)=15 b(3 a+c) .
$$Then :
Official answer
(1)
b, c and a are in A.P.
More from Progressions
- If m arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between 3 and 243 such that 4^th A.M. is equal to 2^nd G.M., then m is…2020
- 1/(3^2-1)+1/(5^2-1)+1/(7^2-1)+… …+1/((201)^2-1) is equal to:2021
- In an increasing geometric progression of positive terms, the sum of the second and sixth terms is 70/3 and the product of the third and fifth terms…2024
- Let a, b, c and d be positive real numbers such that a+b+c+d=11. If the maximum value of a^5 b^3 c^2 d is 3750 β, then the value of β is2023
- If the sum of the series (1/2-1/3)+(1/2^2-1/(2 · 3)+1/3^2)+(1/2^3-1/(2^2 · 3)+1/(2 · 3^2)-1/3^3)+ (1/2^4-1/(2^3 · 3)+1/(2^2 · 3^2)-1/(2 ·…2023
- The sum of all 3-digit numbers less than or equal to 500, that are formed without using the digit "1" and they all are multiple of 11, is \ \ \ \.2021
- If (1/(α+1)+1/(α+2)+… …+1/(α+1012))-(1/(2 · 1)+1/(4 · 3)+1/(6 · 5)+… …+1/(2024 · 2023))=1/2024, then α is equal to ____.2024
- lim_n → ∞ ((1^2-1)(n-1)+(2^2-2)(n-2)+⋯ ⋯+((n-1)^2-(n-1)) · 1)/((1^3+2^3+⋯ ⋯+n^3)-(1^2+2^2+⋯+n^2)) is equal to:2024
JEE Main 2017 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.