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