Since, the first 9 multiple of 4 starts from 0 are,
0, 4, 8, 12, 16, 20, ................
Which is an AP,
Having the first term, [tex]a_1 = 0[/tex]
And, the successive difference, d = 4,
Since, the sum of the n term of an AP,
[tex]S_n = \frac{n}{2}[2a + (n-1)d][/tex]
Hence the sum of 9 term of the above AP,
[tex]S_9 = \frac{9}{2}[2\times 0 + (9-1)\times 4][/tex]
[tex]S_9 = \frac{9}{2}(0 + 8\times 4)[/tex]
[tex]S_9 = \frac{9}{2}(32)[/tex]
[tex]S_9 = \frac{288}{2}=144[/tex]
