# Sum of n terms of a certain series is given by S_n=2n+3n^2, what is the type of the series and what is its 20^(th) term?

Aug 9, 2017

It is an arithmetic progression with first term as $5$ and common difference as $6$ and ${20}^{t h}$ term is $119$

#### Explanation:

As sum of $n$ terms of a certain series is given by ${S}_{n} = 2 n + 3 {n}^{2}$,

Sum of $20$ terms is 2×20+3×20^2=40+1200=1240.

Further, sum of $19$ terms is 2×19+3×19^2=38+1083=1121,.
Hence ${20}^{t h}$ term is $1240 - 1121 = 119$.

As sum of $1$ term is 2×1+3×1^2=5, sum of first two terms is 2×2+3×2^2=4+12=16, second term is $16 - 5 = 11$ and common difference is $11 - 5 = 6$. If it is arithmetic progression the third term should be $11 + 6 = 17$.

As sum of first three terms is 2×3+3×3^2=6+27=33, third term is $33 - 16 = 17$ hence it is an arithmetic progression.

Aug 9, 2017

$d = 6 , {a}_{20} = 119 , {S}_{20} = 1240$

#### Explanation:

$\text{calculate the first 'few' terms of the sequence}$

$\text{using } {S}_{n} = 2 n + 3 {n}^{2}$

${S}_{1} = 2 + 3 = 5 \Rightarrow {a}_{1} = 5$

${S}_{2} = 4 + 12 = 16$

$\Rightarrow {a}_{2} = {S}_{2} - {S}_{1} = 16 - 5 = 11$

${S}_{3} = 6 + 27 = 33$

$\Rightarrow {a}_{3} = {S}_{3} - {S}_{2} = 33 - 16 = 17$

${S}_{4} = 8 + 48 = 56$

$\Rightarrow {a}_{4} = {S}_{4} - {S}_{3} = 56 - 33 = 23$

$\text{the first 4 terms are } 5 , 11 , 17 , 23$

$\text{common difference ( d)}$

$d = 23 - 17 = 17 - 11 = 11 - 5 = 6$

$\Rightarrow \text{ these terms are an arithmetic sequence with } d = 6$

$\text{the sum to n terms of an arithmetic sequence is}$

•color(white)(x)a_n=a_1+(n-1)d

$\Rightarrow {a}_{20} = 5 + \left(19 \times 6\right) = 119$

$\Rightarrow {S}_{20} = \left(2 \times 20\right) + \left(3 \times {20}^{2}\right) = 1240$