Question

In an AP the sum of the first four terms is 26 and the sum of the first term and square of the fourth term is 125, then find first 4terms of an AP.




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Answer 1

Appropriate Question :-

In an AP the sum of the first four terms is 26 and the sum of the square of first term and square of the fourth term is 125, then find first 4terms of an AP.

Given :-

In an AP series,

• The sum of the first four terms is 26

• The sum of the square of first term and square of the fourth term is 125.

To Find :-

• First 4 terms of an AP.

Formula Used :-

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

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

And

↝ 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,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

$\red{\large\underline{\bf{Solution-}}}$

Let assume that,

• First term of an AP = a

• Common difference of an AP = d

Given that,

• The sum of the first four terms is 26

Using the formula of sum of n terms,

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

$\rm :\longmapsto\:\dfrac{4}{2} \bigg(2 \:a\:+\:(4\:-\:1)\:d \bigg) = 26$

$\bf\implies \:2a + 3d = 13 - - - (1)$

Again,

Given that

• The sum of the square of first term and square of the fourth term is 125

$\rm :\longmapsto\: {(a_1)}^{2} + {(a_4)}^{2} = 125$

$\rm :\longmapsto\: {a}^{2} + {(a + 3d)}^{2} = 125$

$\rm :\longmapsto\: {a}^{2} + {(13 - a)}^{2} = 125$

$\boxed{\because \sf \: 2a + 3d = 13 \implies \: a + 3d = 13 - a}$

$\rm :\longmapsto\: {a}^{2} + {a}^{2} + 169 - 26a = 125$

$\rm :\longmapsto\: 2{a}^{2} - 26a + 44 = 0$

$\rm :\longmapsto\: {a}^{2} - 13a + 22 = 0$

$\rm :\longmapsto\: {a}^{2} - 11a - 2a + 22 = 0$

$\rm :\longmapsto\:a(a - 11) - 2(a - 11) = 0$

$\rm :\longmapsto\:(a - 2)(a - 11) = 0$

$\bf\implies \:a = 11 \: \: \: \: or \: \: \: \: a = 2$

Hence,

On substituting the values of a in equation (1), we get

$\begin{gathered}\boxed{\begin{array}{c|c} \bf a & \bf d \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 2 & \sf 3 \\ \\ \sf 11 & \sf - 3 \end{array}} \\ \end{gathered}$

Hence,

Two cases arises,

When a = 2 and d = 3,

The required 4 terms of AP are

$\rm :\longmapsto\:2,5,8,11$

Or

When a = 11 and d = - 3

$\rm :\longmapsto\:11,8,5,2$