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If a is constant then a + 2a + 3a + ..... + na is

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If a is constant then $a + 2 a + 3 a + ..... + na$ is

\frac{an ( n + 1 )}{4}

\frac{an ( n + 1 )}{2}

\frac{n ( n + 1 )}{2}

\frac{an ( n + 1 )}{6}

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If a is constant then

a + 2 a + 3 a + ..... + na

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Text solutionVerified

a_{2} - a_{1} = 2 a - a

∴ a_{2} - a_{1} = a

a_{3} - a_{2} = 3 a - 2 a

∴ a_{3} - a_{2} = a

Thus, difference between consecutive terms of A.P. is same. Thus, given series is in A.P.

Sum of first n terms of A.P. is given by,

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

∴ S_{n} = \frac{n}{2} [2 a + (n - 1) a]

∴ S_{n} = \frac{n}{2} [2 a + na - a]

∴ S_{n} = \frac{n}{2} [a (n + 2) - a]

∴ S_{n} = \frac{n}{2} [a (n + 2 - 1)]

∴ S_{n} = \frac{n}{2} [a (n + 1)]

∴ S_{n} = \frac{an ( n + 1 )}{2}

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