IPMAT Indore 2024 · Quantitative Ability · Question 38

Step 1 — write the two conditions. With first term a and ratio r:

a{r}^{\,p-1}=q, a{r}^{\,q-1}=p

Step 2 — take logarithms. This turns the GP into an AP in \log terms.

\log a+\left( p-1 \right)\log r=\log q

\log a+\left( q-1 \right)\log r=\log p

Step 3 — subtract (2) from (1).

\left( p-q \right)\log r=\log q-\log p \Rightarrow \log r=\frac{\log q-\log p}{p-q}

Step 4 — substitute back into (1).

\log a=\log q-\left( p-1 \right)\cdot \frac{\log q-\log p}{p-q}

=\frac{\left( p-q \right)\log q-\left( p-1 \right)\log q+\left( p-1 \right)\log p}{p-q}

=\frac{\left( p-q-p+1 \right)\log q+\left( p-1 \right)\log p}{p-q}=\frac{\left( 1-q \right)\log q+\left( p-1 \right)\log p}{p-q}

Step 5 — match the printed form. Since \left( p-1 \right)=-\left( 1-p \right):

\log a=\frac{\left( 1-q \right)\log q-\left( 1-p \right)\log p}{p-q}

Answer: Option 4

Shortcut. Logs turn this into the standard AP result. Setting A=\log a, R=\log r, conditions (1) and (2) are exactly the classic “pth term =q, qth term =p” linear system.

Then eliminate by inspection of the options. All four share the denominator \left( p-q \right); they differ only in which log pairs with which bracket, and in one sign. Two checks settle it:

The numerator must pair \log q with \left( 1-q \right) — because equation (1), which contains \log q, is the one weighted by p. That removes (a) and (b).

A quick test with p=1: then a{r}^{0}=q, so \log a=\log q. Substituting p=1 into (c) gives \frac{\left( 1-q \right)\log q}{1-q}=\log q ✔ and into (d) gives the same ✔ — so use q=1 instead, where \log a must equal \log p: (c) gives \frac{\left( 1-p \right)\log p}{p-1}=-\log p ✘, while (d) gives +\log p ✔.

Plugging p=1 or q=1 into the options is faster than deriving whenever the answer choices are algebraic rearrangements of one another.