Question

find the sum of the n term in the A.P root2 , 2root2, 3root2 .......................​

Answer 1

$\sqrt{2} ,\:2\sqrt{2} ,\:3\sqrt{2} ,\cdot\cdot\cdot , is \: an \: A.P$

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

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

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

$S_{n} = \frac{n}{2} [ 2\times \sqrt{2} + ( n - 1) \times \sqrt{2} ] \\= \frac{\sqrt{2}n}{2} [ 2 + n - 1] \\= \frac{\sqrt{2}n}{2} (n + 1)$

Therefore.,

$\red{ Sum \: of \: n \: terms (S_{n})}\green { = \frac{\sqrt{2}n}{2} (n + 1)}$

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