Question:
Let \(a, b, c\) be positive real numbers such that \(a, 7, c, b\) are in an arithmetic progression. If \(ab=3(a+b)\), then the largest value of \(b\) is :
Let \(a, b, c\) be positive real numbers such that \(a, 7, c, b\) are in an arithmetic progression. If \(ab=3(a+b)\), then the largest value of \(b\) is :
Show Hint
Let the common difference be d and write a and b in terms of d using the second term 7.
Updated On: Nov 12, 2026
- \(9-\sqrt{2}\)
- \(9-3 \sqrt{2}\)
- \(9+3 \sqrt{2}\)
- \(9+6 \sqrt{2}\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Let the common difference be \(d\). Then \(a=7-d\) and \(b=7+2d\) (the terms are \(7-d,\ 7,\ 7+d,\ 7+2d\)).
\(ab=3(a+b)\Rightarrow(7-d)(7+2d)=3(14+d)\).
\(49+7d-2d^2=42+3d\Rightarrow 2d^2-4d-7=0\Rightarrow d=\dfrac{4\pm\sqrt{16+56}}{4}=1\pm\dfrac{3\sqrt2}{2}\).
\(b=7+2d=9\pm3\sqrt2\). The larger value is \(9+3\sqrt2\) (here \(d\approx3.12\), \(a\approx3.88>0\), \(c\approx10.12>0\), so all terms are positive).
Answer: (3) \(9+3 \sqrt{2}\)
\(ab=3(a+b)\Rightarrow(7-d)(7+2d)=3(14+d)\).
\(49+7d-2d^2=42+3d\Rightarrow 2d^2-4d-7=0\Rightarrow d=\dfrac{4\pm\sqrt{16+56}}{4}=1\pm\dfrac{3\sqrt2}{2}\).
\(b=7+2d=9\pm3\sqrt2\). The larger value is \(9+3\sqrt2\) (here \(d\approx3.12\), \(a\approx3.88>0\), \(c\approx10.12>0\), so all terms are positive).
Answer: (3) \(9+3 \sqrt{2}\)
Was this answer helpful?
0
0
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
View More Questions
Top JEE Main Arithmetic Progression Questions
- \(3, 7, 1,......., 404\) and \(4, 7, 10,......, 403\). Find sum of common terms.
- Let m and n be the coefficient of \(7^{th}\) and \(13^{th}\) term in expansion of \(\bigg(\frac{1}{3x^{\frac{1}{3}}} +\frac{1}{2x^{\frac{2}{5}}}\bigg)^{18}\), then \(\bigg(\frac{m}{n}\bigg)^{\frac{1}{3}}\) is:
- If ln a, ln b, ln c are in AP and ln a – ln 2b, ln 2b – ln 3c, ln 3c – ln a are in AP then a : b : c is
- If 2nd, 8th, 44th terms of A.P. are 1st, 2nd and 3rd terms respectively of G.P. and first term of A.P. is 1 then the sum of first 20 terms of A.P. is
- In an A.P., the sixth term a6=2. If the product a1a4a5 is the greatest, then the common difference of the A.P. is equal to:
View More Questions
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
View More Questions