Question

In an A.P., if 4th term is 1/6 and 6 th term is 1/4 ,prove that the sum of first 24 terms is 25/2


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Answer 1

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

Let assume that

• First term of an AP = x

• Common difference of an AP = d

It is given that,

$\rm \: a_4 = \frac{1}{6}$

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,

$\rm \: a + (4 - 1)d = \frac{1}{6}$

$\rm\implies \:\bf\: a + 3d = \frac{1}{6} - - - (1)$

Now, further given that

$\rm \: a_6 = \frac{1}{4}$

$\rm \: a + (6 - 1)d = \frac{1}{4}$

$\rm\implies \:\bf \: a + 5d = \frac{1}{4} - - - (2)$

On Subtracting equation (2) from equation (1), we get

$\rm \: 2d = \dfrac{1}{4} - \dfrac{1}{6}$

$\rm \: 2d = \dfrac{3 - 2}{12}$

$\rm \: 2d = \dfrac{1}{12}$

$\rm\implies \:\bf \: d = \dfrac{1}{24}$

On substituting the value of d in equation (1), we get

$\rm \: a + 3 \times \dfrac{1}{24} = \dfrac{1}{6}$

$\rm \: a +\dfrac{1}{8} = \dfrac{1}{6}$

$\rm \: a = \dfrac{1}{6} - \dfrac{1}{8}$

$\rm \: a = \dfrac{4 - 3}{24}$

$\rm\implies \:\bf \: a = \dfrac{1}{24}$

Now, we have to find sum of first 24 terms of an AP.

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ 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

$\rm \: S_{24}\:=\dfrac{24}{2} \bigg(2 \times \dfrac{1}{24} \:+\:(24\:-\:1) \times \dfrac{1}{24} \bigg)$

$\rm \: S_{24}\:= \: 12 \bigg(\dfrac{1}{12} \:+\:\dfrac{23}{24} \bigg)$

$\rm \: S_{24}\:= \: 12 \bigg(\dfrac{2 + 23}{24} \bigg)$

$\rm\implies \:\boxed{ \rm{ \:\bf \: S_{24}\:= \:\dfrac{25}{2} \: \: }}$