(a) Given that ${S_n} = 2{n^2} + 5n$

Putting $n = 1,\;2,\;3,\;……….,\;{S_1} = 2 \times 1 + 5 \times 1 = 7\;$,

${S_2} = 2 \times 4 + 10 = 8 + 10 = 18,\;{S_3} = 18 + 15 = 33$.

So, ${T_1} = {S_1} = a = 7,\;{T_2} = {S_2} – {S_1} = 18 – 7 = 11$,

${T_3} = {S_3} – {S_2} = 33 – 18 = 15$

Therefore series is $7,\,11,\;15,\,……..$

Now, ${n^{th}}$ term $ = a + (n – 1)d = 7 + (n – 1)4 = 4n + 3$.

Aliter : As we know ${T_n} = {S_n} – {S_{n – 1}}$

$ = (2{n^2} + 5n) – \left\{ {2\,{{(n – 1)}^2} + 5\,(n – 1)} \right\}$

$ = 2{n^2} + 5n – 2{n^2} + 4n – 2 – 5n + 5 = 4n + 3$.