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8. Sequences and Series
easy
If the sum of $n$ terms of an $A.P$. is $2{n^2} + 5n$, then the ${n^{th}}$ term will be
A
$4n + 3$
B
$4n + 5$
C
$4n + 6$
D
$4n + 7$
Solution
(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$.
Standard 11
Mathematics