Question

what is the value of (81+82+....+130)

Answer 1

this is required answer

Answer 2

$Given \: sequence \: 81+82+\cdot\cdot\cdot+130\:is \:an \:A.P$

$First \:term (a) = 81$

$Common \: difference (d) = a_{2} - a_{1} \\= 82 - 81 \\= 1$

$\boxed { \pink { n^{th} \:term (a_{n}) = a+(n-1)d }}$

$a_{n} = 130 \: (given)$

$\implies a + (n-1)d = 130$

$\implies 81 + (n-1) \times 1 = 130$

$\implies 81 + n-1= 130$

$\implies 80 + n= 130$

$\implies n= 130 - 80$

$\implies n= 50$

$\boxed { \pink { Sum\:of \:n \:terms (S_{n}) = \frac{n}{2}\Big(a+a_{n}\Big) }}$

$</p><p>[tex]Now , Sum \:of \:50\:terms (S_{50}) \\= \frac{50}{2} \Big( 81 + 130\Big) \\= 25 \times 211\\= 5275$

Therefore.,

$\red{ Sum \:of \:50\:terms (S_{50})}\green {= 5275 }$

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