Show that a_1, a_2, ........, a_n, form an AP where a_n = 9 - 5n. - Vidyadip

Question

Show that $a_1, a_2, ........, a_n$, form an AP where $a_n = 9 - 5n.$

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Answer

We have $a_n = 9 - 5n$
Put n = 1, 2, 3, 4,.... in succession, we get
$a_1 = 9 - 5(1) = 9 - 5 = 4$
$a_2 = 9 - 5(2) = 9 - 10 = - 1$
$a_3 = 9 - 5(3) = 9 - 15 = -6$
$a_4 = 9 - 5(4) = 9 - 20 = -11$
$\therefore $ $a_2 - a_1 = - 1 - 4 = - 5$
$a_3 - a_2 = -6 - (-1) = -6 + 1 = -5$
$a_4 - a_5 = -11 - (-6) = -11 + 6 = - 5$
$i.e. a_{k+1} - a_k$ is the same everytime
So, $a_1, a_2$, ..., an, ...... form an AP
Here, $a = a_1 = 4$
$a = a_2 - a_1 = -5$
$\therefore $ Sum of the first 15 terms $= S_{15}$
$ = \frac{{15}}{2}\left[ {2a + (n - 1)d} \right]$ $\because {S_n} = \frac{n}{2}[2a + (n - 1)d]$
$ = \frac{{15}}{2}\left[ {2 \times a + (15 - 1)d} \right]$
$ = \frac{{15}}{2}\left[ {2a + 14d} \right]$
$= 15(a + 7d)$
$= (15) [4 + 7(-5)]$
$= (15) (4 - 35)$
$= (15) (-31)$
$= -465$

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