Question

$Can \: the \: sum \: of \: first \\ few \: terms \: of \: Arithmetic \\ sequence \: 6 \: , \: 10 \: , \:14 ... \: be \: 500 \: ?​$
​

Answer 1

★ Answer

• FIRST TERM = 6
• COMMON DIFFERENCE = 4

We have to find that how many terms joined up to the sum of 500.

Let the number of terms be n.

As We can find the sum of any number of terms using Sⁿ = n/2[2a+(n-1)d] Formula.

→ 500 = n/2[2×6 + (n-1)4]

→ 500 = n/2[12 + 4n - 4]

→ 1000 = n[8 + 4n]

→ 1000 = 8n + 4n²

→ 250 = 2n + n²

→ n²+2n-250=0

Now, We can Factories Equation to get the values.

→ n = [ -b ± √(b²-4ac)]/2

→ n = [ -2 ± √(4+1000)]/2

→ n = [-2 ± √(1004)}/2

Hence,

It is not Possible to have sum of n terms of an AP be 500.

Answer 2

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

Given AP series is

$\rm :\longmapsto\:6, \: 10, \: 14, \: - - - -$

Here,

• First term of AP, a = 6

• Common difference, d = 10 - 6 = 4

Let if possible,

• Sum of n terms be 500.

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.

Now,

On substituting the values, we get

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

$\rm :\longmapsto\:500\:=\dfrac{n}{2} \bigg(2 \: \times 6\:+\:(n\:-\:1)\:4 \bigg)$

$\rm :\longmapsto\:500\:=\dfrac{n}{2} \bigg(12\:+\:4n\:-\:4 \bigg)$

$\rm :\longmapsto\:500\:=\dfrac{n}{2} \bigg(8\:+\:4n \bigg)$

$\rm :\longmapsto\:500\:=n(4 + 2n)$

$\rm :\longmapsto\:500\:=4n + 2 {n}^{2}$

$\rm :\longmapsto\:250\:=2n +{n}^{2}$

$\rm :\longmapsto\:{n}^{2} + 2n - 250 = 0$

So, its a quadratic equation in 'n',

So, By using quadratic formula, we have

$\rm :\longmapsto\:n = \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$

Here,

• a = 1

• b = 2

• c = - 250

$\rm :\longmapsto\:n = \dfrac{ - 2 \: \pm \: \sqrt{ {2}^{2} - 4(1)( - 250)} }{2(1)}$

$\rm :\longmapsto\:n = \dfrac{ - 2 \: \pm \: \sqrt{ 4 + 1000} }{2}$

$\rm :\longmapsto\:n = \dfrac{ - 2 \: \pm \: \sqrt{ 1004} }{2}$

$\rm :\longmapsto\:n = \dfrac{ - 2 \: \pm \: 2 \sqrt{251} }{2}$

$\rm :\longmapsto\:n = - 1 \: \pm \: \sqrt{251} \: \: \cancel \in \: \: N$

Hence, its not possible to have sum 500.

Additional Information :-

↝ 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.