Answer:

[tex]\boxed{\bf\:a_{20} \: = \: - 112 \: }\\ [/tex]

Step-by-step explanation:

Let assume that first term and common difference of an AP be a and d respectively.

Given that, first term of an AP, a = 2

Further given that, sum of first five terms is equal to one fourth of the sum of next five terms.

[tex]\sf \: S_5 = \dfrac{1}{4}(S_{10} - S_5) \\ [/tex]

[tex]\sf \: 4S_5 = S_{10} - S_5 \\ [/tex]

[tex]\sf \: 4S_5 + S_5 = S_{10} \\ [/tex]

[tex]\sf \: 5S_5 = S_{10} \\ [/tex]

[tex]\sf \: 5 \times \dfrac{5}{2} \bigg( 2a + (5 - 1)d\bigg) = \dfrac{10}{2} \bigg(2a + (10 - 1)d \bigg) \\ [/tex]

[tex]\sf \: 5( 2a + 4d) = 2(2a + 9d) \\ [/tex]

[tex]\sf \: 10a + 20d = 4a + 18d \\ [/tex]

[tex]\sf \: 20d - 18d = 4a - 10a \\ [/tex]

[tex]\sf \: 2d = - 6a \\ [/tex]

[tex]\sf \: d = - 3a \\ [/tex]

[tex]\sf \: d = - 3 \times 2 \\ [/tex]

[tex]\implies \sf \:d = - \: 6 \\ [/tex]

Now, Consider

[tex]\sf \: a_{20} \\ [/tex]

[tex]\sf \: = \: a + ({20} - 1)d \\ [/tex]

[tex]\sf \: = \: a +19d \\ [/tex]

[tex]\sf \: = \: 2 +19 \times ( - 6) \\ [/tex]

[tex]\sf \: = \: 2 - 114 \\ [/tex]

[tex]\sf \: = \: - 112\\ [/tex]

Hence,

[tex]\implies \sf \:\boxed{\bf\:a_{20} \: = \: - 112 \: }\\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Formulae Used

↝ nᵗʰ term of an arithmetic progression is,

[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}[/tex]

↝ Sum of n  terms of an arithmetic progression is,

[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}[/tex]

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the progression.

n is the no. of terms.

d is the common difference.

aₙ is the nᵗʰ term.

Verified answer

EXPLANATION.

First term of an ap is 2.

Sum of first five terms is equal to one fourth of next five term.

As we know that,

Formula of :

Sum of first n terms of an ap.

Sₙ = n/2[2a + (n - 1)d].

Using this formula in this question, we get.

First term : a = 2.

Sum of first five terms is equal to one fourth of next five term.

⇒ S₅ = 1/4 x (S₁₀ - S₅).

⇒ 4S₅ = S₁₀ - S₅.

⇒ 4S₅ + S₅ = S₁₀.

⇒ 5S₅ = S₁₀.

⇒ 5 x [5/2[2a + (5 - 1)d] = 10/2[2a + (10 - 1)d].

⇒ 25/2[2a + 4d] = 5[2a + 9d].

⇒ 25/2[2(a + 2d)] = 5[2a + 9d].

⇒ 25(a + 2d) = 5(2a + 9d).

⇒ 25a + 50d = 10a + 45d.

⇒ 25a - 10a = 45d - 50d.

⇒ 15a = - 5d.

⇒ 3a = - d.

Put the value of a = 2 in the expression, we get.

⇒ 3(2) = - d.

⇒ d = - 6.

To find : The value of 20th term.

As we know that,

General term of an ap.

⇒ Tₙ = a + (n - 1)d.

Using this formula in this question, we get.

⇒ T₂₀ = a + 19d.

⇒ T₂₀ = 2 + 19 x (-6).

⇒ T₂₀ = 2 - 114.

⇒ T₂₀ = - 112.

∴ The value of 20th term of an ap T₂₀ = - 112.