Question

in an AP the sum of first 5th term is 55 and fourth term is 5 more than the sum of next two term find AP​

Answer 1

Answer:

Let the 5 terms be (a - 2d), (a - d), a , (a + d), (a + 2d)

Sum of all terms = 5a = 55

⇒ a = 11

4ᵗʰ term = 5 + (1ˢᵗ term + 2ⁿᵈ term)

⇒ a + d = 5 + a - 2d + a - d

⇒ 4d = 5 + a

⇒ d = 16/4 = 4

On substituting the values of a and d we get the terms as 3, 7, 11, 15, 19.

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Answer 2

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

Let assume that

• First term of an AP = a

• Common difference of an AP = d

Now, Given that sum of first 5 terms is 55.

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

$\rm \: \dfrac{5}{2} \bigg(2a + (5 - 1)d\bigg) = 55$

$\rm \: \dfrac{1}{2} \bigg(2a +4d\bigg) = 11$

$\bf\implies \:a + 2d = 11 - - - (1)$

Now, Further given that fourth term is 5 more than the sum of next two term.

$\rm \: a_4 = 5 + a_5 + a_6$

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.

So, using this, we get

$\rm \: a + 3d = 5 + a + 4d + a + 5d$

$\rm \: 3d = 5 + 4d + a + 5d$

$\bf\implies \:a + 6d = - 5 - - - (2)$

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

$\rm \: 4d = - 16$

$\bf\implies \:d \: = \: - \: 4$

On substituting d = - 4 in equation (1), we get

$\rm \: a + 2( - 4) = 11$

$\rm \: a - 8 = 11$

$\rm \: a = 11 + 8$

$\bf\implies \:a \: = \: 19$

Hence, Required AP series is $\rm \: 19, 15, 11, 7, 3, -1, ...$