Find the fifth term of the sequence if a15=29 and a21=47

We have a15 = a1 + 14 x r and a21 = a1 + 20 x r <=> 29 = a1 + 14 x r and 47 = a1 + 20 x r;

Then, 47 - 29 = (20 x r + a1) - (14 x r + a1);

18 = 6 x r;

r = 3;

We have 29 = a1 + 42;

a1 = 29 - 42;

a1 = - 13;

Finally, a5 = a1 + 4 x r = -13 + 4 x 3 = -13 + 12 = -1.

1260=3*6*7*10=3^2*2^2*7*5 perfect cude = 3^3*2^

step-by-step explanation: