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Show that $$a_1, a_2 … , a_n , …$$ form an AP where an is defined as below (i) $$a_n = 3+4_n$$ (ii) $$a_n = 9 - 5_n$$ Also find the sum of the first 15 terms in each case.

(i) $$\because a_n = 3+4_n$$
$$a_1 = 3+4(1) = 7$$
$$a_2 = 3+4(2) = 3+8 = 11$$
$$a_3 = 3+4(3) = 3+12 = 15$$
$$a_4 = 3+4(4) = 3+16 = 19$$

We can see here, the common difference between the terms are;

$$a_2 - a_1 = 11-7 = 4$$
$$a_3 - a_2 = 15-11 = 4$$
$$a_4 - a_3 = 19-15 = 4$$

Hence, $$a_k + 1 - a_k$$ is the same value every time. Therefore, this is an AP with common difference as 4 and first term as 7.

Now, we know, the sum of $$n^{th}$$ term is;
$$S_n = {{n} \over {2}}[2a+(n -1)d]$$
$$\therefore S_{15} = {{15} \over {2}}[2(7)+(15-1)×4]$$
= $${{15} \over {2}}[(14)+56]$$
= $${{15} \over {2}}(70)$$
= 15×35
= 525

(ii) $$a_n = 9 - 5n$$
$$a_1 = 9-5×1 = 9-5 = 4$$
$$a_2 = 9-5×2 = 9-10 = -1$$
$$a_3 = 9-5×3 = 9-15 = -6$$
$$a_4 = 9-5×4 = 9-20 = -11$$

We can see here, the common difference between the terms are;

$$a_2 - a_1 = -1-4 = -5$$
$$a_3 - a_2 = -6-(-1) = -5$$
$$a_4 - a_3 = -11-(-6) = -5$$

Hence, $$a_k + 1 - a_k$$ is same every time. Therefore, this is an A.P. with common difference as -5 and first term as 4.

Now, we know, the sum of nth term is;
$$S_n = {{n} \over {2}} [2a +(n-1)d]$$
$$S_{15} = {{15} \over {2}}[2(4) +(15 -1)(-5)]$$
= $${{15} \over {2}}$$[8 + 14(-5)]
= $${{15} \over {2}}$$(8 - 70)
= $${{15} \over {2}}$$(-62)
= 15(-31)
= -465