Question

how many terms of the AP -6, -11/2, -5 are needed to give the sum -25

Answer 1

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

Given AP series is

$\rm :\longmapsto\: - 6, \: - \dfrac{11}{2}, \: - 5, \: - - - -$

So,

First term of an AP series, a = - 6

Common difference, d is

$\rm \: = \: \: a_2 - a_1$

$\rm \: = \: \: - \dfrac{11}{2} - ( - 6)$

$\rm \: = \: \: - \dfrac{11}{2} + 6$

$\rm \: = \: \: \dfrac{ - 11 + 12}{2}$

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

Further, given that

$\rm :\longmapsto\:S_n = - \: 25$

Let number of terms required to get the sum be 'n'.

Now,

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

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

$\rm :\longmapsto\: - 25 = \dfrac{n}{2}\bigg( - 12 + (n - 1)\dfrac{1}{2} \bigg)$

$\rm :\longmapsto\: - 25 = \dfrac{n}{2}\bigg(\dfrac{ - 24 + n - 1}{2} \bigg)$

$\rm :\longmapsto\: - 25 = \dfrac{n}{2}\bigg(\dfrac{ n - 25}{2} \bigg)$

$\rm :\longmapsto\: {n}^{2} - 25n = - 100$

$\rm :\longmapsto\: {n}^{2} - 25n + 100 = 0$

$\rm :\longmapsto\: {n}^{2} - 20n - 5n+ 100 = 0$

$\rm :\longmapsto\:n(n - 20) - 5(n - 20) = 0$

$\rm :\longmapsto\:(n - 20)(n - 5) = 0$

$\bf\implies \:n = 20 \: \: \: or \: \: \: n = 5$

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.