Class 10 Math All topics Arithmetic Progressions

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

(a)

$4an(n+1) $

(b)

$2an(n+1) $

(c)

$2n(n+1) $

(d)

$6an(n+1) $

Correct answer: (b)

Solution:

1) $a_{2}−a_{1}=2a−a$

$∴a_{2}−a_{1}=a$

2) $a_{3}−a_{2}=3a−2a$

$∴a_{3}−a_{2}=a$

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,

$S_{n}=2n [2a+(n−1)d]$

$∴S_{n}=2n [2a+(n−1)a]$

$∴S_{n}=2n [2a+na−a]$

$∴S_{n}=2n [a(n+2)−a]$

$∴S_{n}=2n [a(n+2−1) ]$

$∴S_{n}=2n [a(n+1) ]$

$∴S_{n}=2an(n+1) $

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