Question
Show that $a_1, a_2, ........., a_n,$ form an AP where $a_n = 3 + 4n.$
✓
Answer
Put $n = 1, 2, 3, 4, .....$ in succession, we get
$a_1 = 3 + 4(1) = 3 + 4 = 7$
$a_2 = 3 + (2) = 3 + 8 = 11$
$a_3 = 3 + 4(3) = 3 + 12 = 15$
$a_4 = 3 + 4(4) = 3 + 16 = 19$
$\therefore $$ a_2 - a1 = 11 - 7 = 4$
$a_3 - a_2 = 15 - 11 = 4$
$a_4 - a_3 = 19 - 15 = 4$
i.e.$ a_{k+1} - a_k$_ is the same every time.
So,$ a_1, a_2, .....,$ an, .... from an AP.
Here, $a = a_1 = 7$
$d = a_2 - a_1 = 4$
$\therefore $ Sum of the first 15 terms $= S_{15}$_
$ = \frac{{15}}{2}\left[ {2a + (15 - 1)d} \right]$
$\because {S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ = \frac{{15}}{2}\left[ {2a + 14d} \right]$
$= 15(a + 7d)$
$ = (15)(7 + 7 \times 4)$
$= (15) (7 + 28)$
$= (15) (35)$
$= 525$
$a_1 = 3 + 4(1) = 3 + 4 = 7$
$a_2 = 3 + (2) = 3 + 8 = 11$
$a_3 = 3 + 4(3) = 3 + 12 = 15$
$a_4 = 3 + 4(4) = 3 + 16 = 19$
$\therefore $$ a_2 - a1 = 11 - 7 = 4$
$a_3 - a_2 = 15 - 11 = 4$
$a_4 - a_3 = 19 - 15 = 4$
i.e.$ a_{k+1} - a_k$_ is the same every time.
So,$ a_1, a_2, .....,$ an, .... from an AP.
Here, $a = a_1 = 7$
$d = a_2 - a_1 = 4$
$\therefore $ Sum of the first 15 terms $= S_{15}$_
$ = \frac{{15}}{2}\left[ {2a + (15 - 1)d} \right]$
$\because {S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ = \frac{{15}}{2}\left[ {2a + 14d} \right]$
$= 15(a + 7d)$
$ = (15)(7 + 7 \times 4)$
$= (15) (7 + 28)$
$= (15) (35)$
$= 525$
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