Question

In an A.P t11=35 and t31=115 then find t50. Pleeease solvee​

Answer 1

$\bold{\boxed{Answer}}$

$\: \: \: given\\ \\ t(11) = 35 \\ = > a + 10d = 35 - - - - - - (1) \\ \: \: \: \: \: \: \: and \: \: \: t(31) = 115 \\ = > a + 30d = 115 - - - - - - (2)$

$(2) - (1) = > 20d = 80 \\ = > d = 4$

putting the value of d in eq. 1 ,

a=-5

$now \\ \: \: \: t(50) = a + 49d \\ = - 5 + 49 \times 4 \\ = - 5 + 196 \\ = 191$

so this is your answer mate

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