$\begin{array}{1 1}112 \\ -112 \\ 116 \\ -116 \end{array} $

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- Sum of first $n$ terms of an $A.P.$=$S_n=\large\frac{n}{2}$$[2a+(n-1)d]$
- $t_n=a+(n-1)d$

Given: In the A.P. First term$=a=2$ and

Sum of first five terms =one fourth of sum of the next five terms.

Sum of next 5 terms=$t_6+t_7+.....t_{10}=S_{10}-S_5$

$\Rightarrow\:S_5=\large\frac{1}{4}$$(S_{10}-S_5)$...........(i)

We know that $S_n=\large\frac{n}{2}$$[2a+(n-1)d]$

$\Rightarrow\:S_5=\large\frac{5}{2}$$[2\times 2+(5-1)d]$

$\therefore\:S_5=10+10d$ and

$S_{10}=\large\frac{10}{2}$$[2\times 2+(10-1)d]$

$S_{10}=5(4+9d)$

$\therefore\:$ By substituting the values of $S_5\:and\:S_{10}$ in (i) we get

$10+10d=\large\frac{1}{4}$$(20+45d-10-10d)$

$\Rightarrow\:40+40d=10+35d$

$\Rightarrow\:d=-6$

We know that $t_n=a+(n-1)d$

$\therefore\:t_{20}=2+(20-1)(-6)=-112$

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