(a)The common difference is 3, so the nth term has the form 3n + c. When n=1, 3(1)+c=5, so c=2.

(b)Substitute n = 10 into the formula.

(a)Substitute n=4: 4² + 1 = 17, which matches the 4th term given. For n=8: 8² + 1.

(a)Common difference is 4, so nth term = 4n + c; when n=1, 4+c=7, c=3.

(b)Substitute n=15.

(a)nth term = 110 − 10n (starts at 100 when n=1). Set 110−10n=0.

(a)These are square numbers: 1²,2²,3²,4²,5²; next are 6² and 7².

(a)Each term is ×2 the previous, so nth term = 3 × 2^(n−1).

(a)Common difference 4, nth term = 4n+c; n=1 gives c=−2.

(a)Each term adds one more than the previous addition (2,3,4,5 so far, next add 6 then 7).

(a)Each term is ×3 the previous: nth term = 2×3^(n−1).

(a)y=kx, so 15=3k, k=5; then y=5×8.