IIT JEE 1982 Maths - 'Adapted' - Subjective to Multi-Correct Q14

Assuming that the GP exists and that the first term is $a$ , the common ratio is $r$ and:

$\begin{cases} ar^i = 8 \\ ar^j = 12 \\ ar^k = 27 \end{cases}$ , where $i<j<k$ are natural numbers .

Then, we have:

$\begin{cases} \dfrac {ar^j}{ar^i} = \dfrac {12}8 & \implies r^{j-i} = \dfrac 32 \\ \dfrac {ar^k}{ar^j} = \dfrac {27}{12} & \implies r^{k-j} = \dfrac 94 \end{cases}$

$\begin{aligned} \implies r^{k-j} & = r^{2(j-i)} \\ \implies k - j & = 2j -2i \\ k & = 3j - 2i \end{aligned}$ .

Statement A: Let us check if $i = 1$ , $j=2$ , then $k = 3(2)-2(1) = 4$ . From $r^{j-i} = \frac 32$ $\implies r = \frac 32$ . From $ar=8$ $\implies a=\frac {16}3$ . $ar^2 = \frac {16}3 \times \left(\frac 32\right)^2 = 12$ (correct) and $ar^4 = \frac {16}3 \times \left(\frac 32\right)^4 = 27$ (correct). Therefore, statement A is false .

Statement B: From $k=3j-2i$ , for any $j>i$ , there is a $k$ , therefore, there are infinite such GP. Therefore, statement A is false .

Statement C: If $i$ and $j$ are consecutive, then $j=i+1$ . From $k=3j-2i$ , we have $k=3i+3-2i=i+3\ne i+2$ . Therefore, they cannot be consecutive terms and statement C is false .

Statement D: Let $i=1$ (2nd term) and $j=3$ (4th term). From $k=3j-2i$ , we have $k=3(3)-2(1)=7$ (8th term). Therefore, statement D is true .

Therefore, the answer is $\boxed{9991}$ .