Question

If the sum of all terms of AP : -4 , -1,2,5 , ... x is 72 , find X. help please!!!​

Answer 1

$\large\underline\blue{\bold{Given  }}$

$\tt \:  \longrightarrow \: - 4 - 1 + 2 + 5 + ... + x = 72$

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

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.

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} (2\:a\:+\:(n\:-\:1)\:d)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of term.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

$\large\underline\purple{\bold{Solution :- }}$

☆ ʜᴇʀᴇ,

• aₙ = x

• a = - 4

• d = - 1 - (- 4) = 3

• Sₙ = 72

☆ Step :- 1

$\tt \:  \longrightarrow \: S_n = 72$

$\tt \:  \longrightarrow \: \dfrac{n}{2} (2a + (n - 1)d) = 72$

$\tt \:  \longrightarrow \: \dfrac{n}{2} (2( - 4) + (n - 1) \times 3) = 72$

$\tt \:  \longrightarrow \: \dfrac{n}{2} ( - 8 + 3n - 3) = 72$

$\tt \:  \longrightarrow \: n(3n - 11) = 144$

$\tt \:  \longrightarrow \: {3n}^{2} - 11n - 144 = 0$

$\tt \:  \longrightarrow \: {3n}^{2} - 27n + 16n - 144 = 0$

$\tt \:  \longrightarrow \: 3n(n - 9) + 16(n - 9) = 0$

$\tt \:  \longrightarrow \: (n - 9)(3n + 16) = 0$

$\tt \:  \longrightarrow \: n = 9 \: or \: n = - \dfrac{16}{3}$

$\tt\implies \: \boxed{ \purple{ \bf \: n \: = \: 9}}$

☆ Step :- 2

☆ Now,

• aₙ = x

• a = - 4

• d = - 1 - (- 4) = 3

• n = 9

$\tt \:  \longrightarrow \: a_n\:=\:a\:+\:(n\:-\:1)\:d$

$\tt \:  \longrightarrow \: x = - 4 + (9 - 1) \times 3$

$\tt \:  \longrightarrow \: x = - 4 + 8 \times 3$

$\tt \:  \longrightarrow \: x = - 4 + 24$

$\tt\implies \: \boxed{ \red{ \bf \: x \: = \: 20}}$