Question

for a GP..............​

Answer 1

Answer:

t^3=63 and t^6=1701 tn=1638

Answer 2

Answer:

t3=63

t6=1701

$a {r}^{2} = 63..........(1) \\ ar {}^{5} = 1701.......(2) \\ (2) \div (1) \\ \frac{ar {}^{5} }{ar {}^{2} } = \frac{1701}{63} \\ {r}^{3} = 27 \\ r {}^{3} = {3}^{3} \\ therefore = \: r = 3 \\ put \: r \: value \: in \: (1) \\ a(3) {}^{2} = 63 \\ a9 = 63 \\ a = \frac{63}{9} = 7 \\ \\ tn = ar {}^{n - 1} \\ 7 \times 3 {}^{n - 1}$

$a {r}^{2} = 63..........(1) \\ ar {}^{5} = 1701.......(2) \\ (2) \div (1) \\ \frac{ar {}^{5} }{ar {}^{2} } = \frac{1701}{63} \\ {r}^{3} = 27 \\ r {}^{3} = {3}^{3} \\ therefore \: r = 3$