Question

which tearm of the AP 2/3, 3/4,5/6,...........is 20/3
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Answer 1

$\mathfrak{\huge \underline{Question:}}$

which tearm of the AP 2/3, 3/4,5/6,...........is 20/3

$\mathfrak{\huge \underline{Given:}}$

First term= $\frac{2}{3}$

$Common ~difference = \frac{3}{4}- \frac{2}{3} \\\\\sf = \frac{9-8} {12} = \frac{1}{12}$

$\mathfrak{\huge \underline{Answer :}}$

Let $\frac{20}{3}$be the nth term of the AP....

So,

$T_n= a+(n-1)d \\\\\sf \frac{20}{3}= \frac{2}{3} + (n-1)\frac{1}{12} \\\\\sf \frac{20}{3}=\frac{2}{3} + \frac{n-1}{12} \\\\\sf \frac{20}{3} =\frac{7+n}{12} \\\\\sf 20x12=3(7+n) \\\\\sf 240= 21+3n \\\\\sf 3n=240-21 =219 \\\\\sf n=\frac{219} {3} \\\\\sf n=73$

Hence, 20/3 is the 73rd term of the AP