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# In an $A.P.$ the first term is $2$ and the sum of first $5$ terms is one-fourth of the next five terms. Find its $20^{th}$ term.

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

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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 $t_n=a+(n-1)d$
$\therefore\:t_{20}=2+(20-1)(-6)=-112$
answered Feb 19, 2014

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