Solve : If a, b, c, d are in H.P. , then prove that ab + bc + cd = 3ad

Question

My try : Since a, b, c, d are in HP, 1/a, 1/b, 1/c and 1/d are in AP.

If the common difference of AP is k, then 1/d = 1/a + 3k

==> k = (1/3)(1/d - 1/a) = (1/3){(a - d)/ad} = (a-d)/3ad

and 1/c = 1/a + 2k = 1/a + 2(a-d)/3ad = [3d + 2a - 2d]/3ad = (2a + d)/3ad ==> c = 3ad/(2a + d)

Further, a, b, c are in HP, ==> b = 2ac/(a+c) So, ab + bc = b(a+c) = 2ac/(a+c) = 2ac

ab + bc + cd = 2ac + cd = c(2a+d)

Substituting for c from step (ii) above, ab + bc + cd = {3ad/(2a+d)}*(2a+d) = 3ad [Proved] I think that this process is quite long , is there any another approach of this question.

Answer 1

Since $a$, $b$, $c$, $d$ are in HP, $c^{-1}+a^{-1}=2b^{-1}$ and $d^{-1}-b^{-1}=c^{-1}-a^{-1}$.

\begin{align} (c^{-1}+a^{-1})(d^{-1}-b^{-1})&=2b^{-1}(c^{-1}-a^{-1}) \\ (cd)^{-1}+(ad)^{-1}-(ab)^{-1}-(bc)^{-1}&=2(bc)^{-1}-2(ab)^{-1} \\ (cd)^{-1}+(ad)^{-1}+(ab)^{-1}&=3(bc)^{-1} \\ (abcd)[(cd)^{-1}+(ad)^{-1}+(ab)^{-1}]&=(abcd)[3(bc)^{-1}] \\ ab+bc+cd&=3ad \\ \end{align}

Answer 2

To organize your method: $$k=\frac13\left(\frac1d-\frac1a\right)=\frac12\left(\frac1c-\frac1a\right)=\frac1b-\frac1a,$$ where $k$ is the common difference of the AP: $\frac1a,\frac1b,\frac1c,\frac1d$.

Equating $(1,3),(2,3),(1,2)$ we get: $$ab=3ad-2bd\\ bc=2ac-ab\\ cd=3ad-2ac\\ ab+bc+cd=3ad-2bd+2ac-ab+3ad-2ac=\\ =3ad+(3ad-2bd)-ab=3ad.$$ Alternative method: $$k=\frac1b-\frac1a=\frac1c-\frac1b=\frac1d-\frac1c,$$ where $k$ is the common difference of the AP: $\frac1a,\frac1b,\frac1c,\frac1d$. It results in: $$ab=\frac{a-b}{k}; bc=\frac{b-c}{k}; cd=\frac{c-d}{k};\\ ab+bc+cd=\frac{a-b+b-c+c-d}{k}=\frac{a-d}{k}=3ad,$$ because: $$\frac{a-d}{k}=3ad \iff k=\frac13\left(\frac1d-\frac1a\right).$$

Answer 3

This is no quicker, but for what it’s worth:

If $a,b,c,d$ are in harmonic progression, then $\frac1a,\frac1b,\frac1c, \frac1d$ are in arithmetic progression, so $\frac1a+\frac1d=\frac1b+\frac1c$, or (just adding the fractions) $$\displaystyle \frac{a+d}{\color{red}{ad}}=\frac{b+c}{bc}.$$

Also, $\frac1a+\frac1c=\frac2b$ and $\frac1b+\frac1d=\frac2c$, or $\frac{a+c}{ac}=\frac2b$ and $\frac{b+d}{bd}=\frac2c$, from which $$\displaystyle\frac{2a+2c}{\color{green}{abc}}=\frac4{b^2}\mbox{ and }\displaystyle\frac{2b+2d}{\color{blue}{bcd}}=\frac4{c^2}$$.

From this, $\displaystyle \color{red}{ad}=\frac{(a+d)bc}{b+c}=\frac{abc+bcd}{b+c}$, and you therefore need to show that $\displaystyle ab+bc+cd=3\frac{abc+bcd}{b+c}$, or equivalently that $$\displaystyle (b+c)(ab+bc+cd)=3(abc+bcd).$$

Now, $3(abc+bcd)\\=(abc+bcd)+2(\color{green}{abc}+\color{blue}{bcd})\\=abc+bcd+(ab^2+b^2c)+(bc^2+c^2d)\\=ab^2+b^2c+bcd+abc+bc^2+c^2d\\=b(ab+bc+cd)+c(ab+bc+cd).$