Question

The sum of the first 25 terms of an A.P. is 1700, while its common difference is 6. What is the 31st

Answer 1

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

Given that

• The sum of the first 25 terms of an A.P. is 1700.

• It's common difference is 6.

Let assume that

• First term of an AP is a

• Common difference of an AP is d, so d = 6

• Number of terms is n, so n = 25

• Sₙ is the sum of n terms of AP, so Sₙ = 1700

We know that

↝ Sum of 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 n terms of AP.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

So, on substituting the values, we get

$\rm :\longmapsto\:1700 = \dfrac{25}{2} \bigg(2a + (25 - 1) \times 6\bigg)$

$\rm :\longmapsto\:68 = \dfrac{1}{2} \bigg(2a + (24) \times 6\bigg)$

$\rm :\longmapsto\:68 = a + 72$

$\rm \implies\:\boxed{\tt{ a \: = \: - \: 4}}$

Now,

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.

Tʜᴜs,

$\rm :\longmapsto\: a_{31} = a + (31 - 1)d$

$\rm :\longmapsto\: a_{31} = a + 30d$

$\rm :\longmapsto\: a_{31} = - 4 + 30 \times 6$

$\rm :\longmapsto\: a_{31} = - 4 + 180$

$\rm \implies\:\boxed{\tt{ \: a_{31} \: = \: 176 \: }}$

Answer 2

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

Given that

The sum of the first 25 terms of an A.P. is 1700.

It's common difference is 6.

Let assume that

First term of an AP is a

Common difference of an AP is d, so d = 6

Number of terms is n, so n = 25

Sₙ is the sum of n terms of AP, so Sₙ = 1700

We know that

↝ Sum of 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 n terms of AP.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

So, on substituting the values, we get

$\rm :\longmapsto\:1700 = \dfrac{25}{2} \bigg(2a + (25 - 1) \times 6\bigg)$

$\rm :\longmapsto\:68 = \dfrac{1}{2} \bigg(2a + (24) \times 6\bigg)$

$\rm :\longmapsto\:68 = a + 72$

$\rm \implies\:\boxed{\tt{ a \: = \: - \: 4}}$

Now,

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.

Tʜᴜs,

$\rm :\longmapsto\: a_{31} = a + (31 - 1)d$

$\rm :\longmapsto\: a_{31} = a + 30d$

$\rm :\longmapsto\: a_{31} = - 4 + 30 \times 6$

$\rm :\longmapsto\: a_{31} = - 4 + 180$

$\rm \implies\:\boxed{\tt{ \: a_{31} \: = \: 176 \: }}$