In an $A.P.,$ the first term is $2$ and the sum of the first five terms is one-fourth of the next five terms. Show that $20^{th}$ term is $-112$
In an $A.P.,$ the first term is $2$ and the sum of the first five terms is one-fourth of the next five terms. Show that $20^{th}$ term is $-112$
First term $=2$
Let d be the common different of the $A.P.$
Therefore, the $A.P.$ is $2,2+d, 2+2 d, 2+3 d \ldots$
Sum of first five terms $=10+10 d$
Sum of next five terms $=10+35 d$
According to the given condition,
$10+10 d=\frac{1}{4}(10+35 d)$
$\Rightarrow 40+40 d=10+35 d$
$\Rightarrow 30=-5 d$
$\Rightarrow d=-6$
$\therefore a_{20}=a+(20-1) d=2+(19)(-6)=2-114=-112$
Thus, the $20^{\text {th }}$ of the $A.P.$ is $-112$
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