1. Using the formula for nth term ( = a1 * r^(n-1)
the 5th term = (-2)*(1/4)^4
= -2/256 = -1/128 (answer)
2. The 4 terms between 1 and 32 are found as follows:-
a1 = 1 and a6 = 32 so 32/1 = a6/a1
therefore 32/1 = a1r^5 / a1
so r^5 = 32 and r = 2
so 4 terms are a1r, a1r^2, a1^r^3 and a1r^4
= 2, 4, 8, 16 (answer)
3. You need the sum of 10 terms where a1 = -2 and r = 3
S10 = -2 * (3^10 - 1) / (3 - 1)
= -59058
4. Sum to infinity = a1 / (1 - r)
a1 = 1 and r = 1/2
so the answer is 1 / (1 - 1/2)
= 1 / 1/2
= 2 (answer)
