Question

For an AP given below. Find t20 and S20.
$\frac{1}{6} \: \: \frac{1}{4} \: \: \frac{1}{3}$
comma(,) in between those fraction. 1/6 is t1, 1/4 is t3, 1/3 is t3. Pls give correct answer and explanation if possible. If u don't know how to solve then don't even attempt. Pls leave it to others. If u give irrelevant answer u will be reported if u give correct answer and suitable explanation u will be marked brainliest.
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Answer 1

Answer:  $t_{20} = \frac{21}{12}$,  $S_{20} = \frac{115}{6}$

Step-by-step explanation:

$\frac{1}{6} , \frac{1}{4} , \frac{1}{3}, ....$

Here, $a = \frac{1}{6}$   and $d = \frac{1}{4} - \frac{1}{6} = \frac{3-2}{12} = \frac{1}{12}$

=>  $a_{n} = a + (n - 1)d$

=>  $a_{20} = a + (20-1)d$

=>  $a_{20} = a + 19d$

=>  $a_{20} = \frac{1}{6} + 19(\frac{1}{12} )$

=>  $a_{20} = \frac{2 + 19}{12}$

=>  $a_{20} = \frac{21}{12}$

Therefore, $a_{20} = t_{20} = \frac{21}{12}$

Now, $S_{n} = \frac{n}{2} [2a + (n-1)d]$

=>  $S_{20} = \frac{20}{2} [2(\frac{1}{6}) + (20-1)\frac{1}{12}]$

=>  $S_{20} = 10 ( \frac{1}{3} + \frac{19}{12} )$

=>  $S_{20} = 10 ( \frac{4+19}{12} )$

=>  $S_{20} = 5 ( \frac{23}{6} ) = \frac{115}{6}$

Therefore, $S_{20} = \frac{115}{6}$

Hope it helps...

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