Question

the last term of the series x², x ,1,......to 31 term is solve it with GP​

Answer 1

first term of GP a = x²

common ratio,

r = x / x²

r = 1 / x

$an = ar {}^{n - 1}$

$a31 = x {}^{2} ( \frac{1}{x} ) {}^{31 - 1}$

$a31 = x {}^{2}( \frac{1}{x} ) {}^{30}$

$a31 = x {}^{2} \frac{1}{x {}^{30} }$

$a31 = \frac{1}{x {}^{28} }$

$a31 = x {}^{ - 28}$

Hope the solution is clear to you!!

Answer 2

Step-by-step explanation:

r= x/x² = 1/x

N = 31

a= x²

t31=ar*n^-1

t31 = x²*1/x^31-1

t31 = 1/x^28

ans : 1/x ^28