If $p, q and r$ are in A.P, then which of the following is/are true?

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

p^{th}, q^{th} and r^{th}

terms of A.P., are in A.P.

p^{th}, q^{th} and r^{th}

terms of H.P., are in H.P.

None of these

p^{th}, q^{th} and r^{th}

terms of G.P., are in G.P

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

q - p = r - q \to (I)

(i) $t_{p} = a + (p - 1) d, t_{q} = a + (q - 1) d, t_{r} = a + (r - 1) d$

$\Rightarrow \frac{t_{p} - a}{d} + 1 = p, \frac{t_{q} - a}{d} + 1 = q, \frac{t_{r} - a}{d} + 1 = r$

From (I) $\frac{t_{q} - a}{d} + 1 - \frac{t_{p} - a}{d} - 1 = \frac{t_{r} - a}{d} + 1 - \frac{t_{q} - a}{d} - 1$

$\Rightarrow t_{q} - t_{p} = t_{r} - t_{q}$

$∴ t_{p}, t_{q}, t_{r}$ are in A.P

(ii) $t_{p} = a r^{p - 1}, t_{q} = a r^{q - 1}, t_{r} = a r^{r - 1}$

$\Rightarrow \frac{t_{p}}{a} = r^{p - 1} \Rightarrow p - 1 = \log_{r} (\frac{t_{p}}{a}) \Rightarrow p = \log (\frac{t_{p}}{a}) + 1$

$q = \log_{r} (\frac{t_{q}}{a}) + 1, r = \log_{r} (\frac{t_{r}}{a}) + 1$

$(I) \Rightarrow \log_{r} (\frac{t_{q}}{a}) + 1 - \log_{r} (\frac{t_{p}}{a}) - 1 = \log_{r} (\frac{t_{r}}{a}) + 1 - \log_{r} (\frac{t_{q}}{a}) - 1$

$\Rightarrow \log_{r} (\frac{t_{q}}{t_{p}}) = \log_{r} (\frac{t_{r}}{t_{q}}) \Rightarrow t_{q}^{2} = t_{p} t_{r} \Rightarrow t_{p}, t_{q}, t_{r}$ are in G.P.

(iii) $t_{p} = \frac{1}{a + (p - 1) d}, t_{q} = \frac{1}{a + (q - 1) d}, t_{r} = \frac{1}{a + (r - 1) d}$

$\Rightarrow a + (p - 1) d = \frac{1}{t_{p}} \Rightarrow (p - 1) d = \frac{1}{t_{p}} - a = \frac{1 - a t_{p}}{t_{p}} \Rightarrow p = \frac{1 - a t_{p}}{d t_{p}} + 1$

Similarly $q = \frac{1 - a t_{q}}{d t_{q}} + 1, r = \frac{1 - a t_{r}}{d t_{r}} + 1$

$(I) \Rightarrow \frac{1 - a t_{q}}{d t_{q}} + 1 - \frac{1 - a t_{p}}{d t_{p}} - 1 = \frac{1 - a t_{r}}{d t_{r}} + 1 - \frac{1 - a t_{q}}{d t_{q}} - 1$

$\frac{t_{p} - a t_{p} t_{q} - t_{q} + a t_{p} t_{q}}{t_{p} . t_{q}} = \frac{t_{q} - a t_{r} t_{q} - t_{r} + a t_{r} t_{q}}{t_{r} . t_{q}}$

$1 - \frac{t_{q}}{t_{p}} = \frac{t_{q}}{t_{r}} - 1 \Rightarrow 2 = \frac{t_{q}}{t_{r}} + \frac{t_{q}}{t_{p}} = t_{q} (\frac{1}{t_{r}} + \frac{1}{t_{p}})$