Question

nth term of an A.P. is 4n -1 then find sum
of first 18 terms.​

Answer 1

Answer:

666

Step-by-step explanation:

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Answer 2

Answer:

The sum of first 18 terms is 9 (142 - 17d).

Step-by-step explanation:

The nth term of an AP is:

$T_{n}=a+(n-1)d$

a = first term

d = common difference

It is provided that the nth term is (4n - 1).

Then the relation between a and d is:

$T_{n}=a+(n-1)d\\4n-1=a+(n-1)d\\4n-1=a+nd-d\\a=4n-nd-1+d\\=n(4-d)+(d-1)$

The sum of first n terms is:

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

For n = 18 compute the sum of first 18 terms as follows:

$S_{18}=\frac{18}{2}[2a+(18-1)d]\\=9[2(n(4-d)+d-1)+17d]\\=9[2(18(4-d)+d-1)+17d]\\=9[144-36d+2d-2+17d]\\=9(142-17d)$

Thus, the sum of first 18 terms is 9 (142 - 17d).