Question

The sum of first 30 terms
of
an A. P. 2/5,1,8/5,11/5......is
​

Answer 1

$Given \:A.P : \frac{2}{5} , 1 , \frac{8}{5}, \frac{11}{5} ,\cdot\cdot\cdot$

$First \:term (a) = \frac{2}{5}$

$Common \: differnce (d) = a_{2} - a_{1} \\= 1 - \frac{2}{5} \\= \frac{5-2}{5} \\= \frac{3}{5}$

$\boxed { \pink {Sum \:of \: n \:terms (S_{n}) = \frac{n}{2} [2a+(n-1)d] }}$

$\implies Sum \:of \: 30 \:terms (S_{30})\\= \frac{30}{2}[ 2\times \frac{2}{5} + ( 30 - 1 ) \times \frac{3}{5} ] \\= 15[ \frac{4}{5} + 29 \times \frac{3}{5}] \\= 15 [ \frac{4}{5} + \frac{87}{5}]</p><p>\\= 15 \times \frac{4+87}{5} \\= 15 \times \frac{91}{5} \\= 3 \times 91 \\= 273$

Therefore.,

$\red { Sum \:of \: 30 \:terms (S_{30})} \green { = 273}$

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