Question

sum of (n+1) terms of arithmetic sequence is Pn²+Qn+R prove that P+R=Q ?​

Answer 1

I think the ans is right..i proved it

Answer 2

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

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of first n terms of an arithmetic sequence 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 first n terms of an AP

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

Tʜᴜs,

↝ Sum of (n + 1 ) terms is,

$\rm :\longmapsto\:S_{n + 1}\:=\dfrac{n + 1}{2} \bigg(2 \:a\:+\:(n + 1\:-\:1)\:d \bigg)$

$\rm :\longmapsto\:S_{n + 1}\:=\dfrac{n + 1}{2} \bigg(2 \:a\:+\:n\:d \bigg)$

$\rm :\longmapsto\:S_{n + 1}\:=\dfrac{1}{2} \bigg(2an + {dn}^{2} + 2a + nd \bigg)$

$\rm :\longmapsto\:S_{n + 1}\:=\dfrac{1}{2} \bigg({dn}^{2} + nd + 2an + 2a \bigg)$

$\rm :\longmapsto\:S_{n + 1}\:=\dfrac{1}{2} \bigg({dn}^{2} + n(d + 2a) + 2a \bigg)$

$\rm :\longmapsto\:S_{n + 1}\:= \bigg(\dfrac{1}{2}{dn}^{2} + \dfrac{1}{2}n(d + 2a) + a \bigg)$

But it is given that,

$\rm :\longmapsto\:S_{n + 1}\: = P {n}^{2} + Qn + R$

So, on comparing we get,

$\rm :\longmapsto\:\red{ \bf \: P = \dfrac{d}{2}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \rm :\longmapsto\:\red{ \bf \: Q = \dfrac{1}{2}(2a + d)} \\ \rm :\longmapsto\:\red{ \bf \: R = a} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Now,

Consider,

$\rm :\longmapsto\:P + R$

$\sf \: = \: \dfrac{d}{2} + a$

$\sf \: = \: \dfrac{d + 2a}{2}$

$\sf \: = \: \dfrac{1}{2}(d + 2a)$

$\sf \: = \: Q$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ 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.