Find 5th term in the series 5, 15, 45, ...
350
405
Explanation:
a = 5, r - $\dfrac{15}{5}$ = 3, n=5
5th term, t5
=$ar^{n-1} = 5 \times 3^{5-1}$
= $5 \times 3^4 = 5 \times 81$ = 405
a = 5, r - $\dfrac{15}{5}$ = 3, n=5
5th term, t5
=$ar^{n-1} = 5 \times 3^{5-1}$
= $5 \times 3^4 = 5 \times 81$ = 405