Question

Find the number of terms of the AP 6,9,12,...............so that their sum is 627

Answer 1

Answer:

n=19

Step-by-step explanation:

Formula used:

The sum of n terms of an A.P a, a+d, a+2d,.... is

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

Given A.P is

6, 9, 12...............

Here, a=6, d=3

Given:

$S_n=627$

$\frac{n}{2}[2a+(n-1)d]=627$

$\frac{n}{2}[2(6)+(n-1)3]=627$

$n[12+3n-3]=1254$

$n[9+3n]=1254$

$3n[3+n]=1254$

$n[3+n]=418$

$n^2+3n-418=0$

$(n+22)(n-19)=0$

$n=-22,19$

But n cannot be negaative

$\therefore\:n=19$

Answer 2

Step-by-step explanation:

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