Question

19. In S =
[2a + (n - 1) d), make d as the subject.

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Answer 1

Answer:

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

$Let \: S_{n} \: = \: Sum \: of \: 'n' \: terms, \: then$

$S_{n} \: = \: a \: + \: (a+d) \: + \: (a+2d) \: + \: (a+3d) \: + \: ...+ \: a+(n-1)d$

Let l = last term, then,

$S_{n}=a \: + \: (a+d) \: + \: (a+2d) \: + \: (a+3d) \: +...+ \: (l-3d) \: + \: (l -2d) \: + \: (l - d) \: + \: l$

Now reversing the order of this gives:

$S_{n} = l \: + \: (l-d) \: + \: (l-2d) \: + \: (l-3d) \: +...+ \: (a+3d) \: + \: (a +2d) \: +(a + d) \: + \: a$

Adding these last two equations gives,

$2 \times S_{n} =a \: + \: l \: + \: (a+l) \: + \: (a+l) \: + \: (a+l) \: + ...+ \: (a+l) \: + \: (a+l) \: + \: (a+l) \: + \: ( a+l )$

$2 \times S_{n} =(a+l) \: to \: n \: terms.$

$2 \times S_{n} =(a + l) \times n$

$S_{n} = \frac{n}{2} (a + l)$

but l = a+(n-1)d, so substituting this gives:

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