Question

In an A.P., if Sn-3 = 5n2 - 28n + 39, then its 9th term is​

Answer 1

Appreciate Question :-

In an AP, the 9th term is ______, if

$\rm \: S_{(n - 3)} = 5 {n}^{2} - 28n + 39$

$\large\underline{\sf{Solution-}}$

Given that,

In an AP series,

$\rm \: S_{(n - 3)} = 5 {n}^{2} - 28n + 39$

Replace n by k + 3, we get

$\rm \: S_{(k + 3 - 3)} = 5 {(k + 3)}^{2} - 28(k + 3) + 39$

$\rm \: S_{k} = 5( {k}^{2} + 9 + 6k) - 28k - 84 + 39$

$\rm \: S_{k} = 5{k}^{2} + 45 + 30k - 28k - 45$

$\rm \: S_{k} = 5{k}^{2} + 2k$

Now,

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n  terms of an arithmetic progression is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the progression.

n is the no. of terms.

d is the common difference.

So, using this

$\rm \: \dfrac{k}{2} \bigg(2a + (k - 1)d \bigg) = {5k}^{2} + 2k$

$\rm \: \dfrac{k}{2} \bigg(2a + (k - 1)d \bigg) = k(5k + 2)$

$\rm \: \dfrac{1}{2} \bigg(2a + (k - 1)d \bigg) = 5k + 2$

$\rm \: 2a + (k - 1)d = 2(5k + 2)$

$\rm \: 2a + kd - d = 10k + 4$

$\rm \: (2a - d)+ kd = 10k + 4$

So, on comparing we get

$\rm \: d = 10$

and

$\rm \: 2a - d = 4$

$\rm \: 2a - 10 = 4$

$\rm \: 2a =10 + 4$

$\rm \: 2a =14$

$\rm\implies \:a = 7$

Now, we know that

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

Tʜᴜs,

$\rm \: a_{9}$

$\rm \: =  \: a + (9 - 1)d$

$\rm \: =  \: a + 8d$

$\rm \: =  \: 7 + 8 \times 4$

$\rm \: =  \: 7 + 32$

$\rm \: =  \: 39$

$\rm\implies \:\boxed{\sf{  \:\rm \: a_{9} = 39 \: \: }}$