For an A.P. T6=47 and T10=10 then find T30.

$\begin{array}{l}\text{Given that,}\\ \text{For an A.P }T_6=47\ \&\ T_{10}=10\ \\ \text{Let, first term of is }\ a\ \text{and common diffrence is d}.\\ T_6=a+5d=47\ \Rightarrow a=47-5d\\ \text{Now substuite the value,}\\ \ T_{10}=a+9d=47-5d+9d=10\ \Rightarrow d=-9.25\ \\ a=47-5d=47-5\times9.25=0.75\\ \text{so,}\\ T_{30}=a+29d=0.75+\left(29\times-9.25\ \right)=-267.5\ \end{array}$

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