Question

how many terms of AP 54,51,48,45,... be taken so that their sum is 513?

Answer 1

so common difference(d) = 51 - 54 = -3 using formula of sum of terms  s = n/2(2a +(n - 1)d) here s = sum  a = first term n = no . of terms d = common difference so 513 = n/2(54 ×2 +(n -1)(-3)) so solving for n we get n = 19 terms

Answer 2

Answer:

18 or 19

Step-by-step explanation:

Given : AP 54,51,48,45,...

To Find:how many terms of AP 54,51,48,45,... be taken so that their sum is 513?

Solution:

a = first term = 54

d = common difference = 51-54=48-51=-3

Sum of first n terns = $S_n=\frac{n}{2}(2a+(n-1)d)$

We are given that sum of n terms is 513

So, To calculate n

$513=\frac{n}{2}(2(54)+(n-1)(-3))$

$1026=n(108-3n+3)$

$1026=108n-3n^2+3n$

$3n^2-111n+1026=0$

$n^2-37n+342=0$

$n^2-19n-18n+342=0$

$n(n-19)-18(n-19)=0$

$(n-19)(n-18)=0$

$n=19,18$

So, n can be 19 or 18

To verify substitute values in Formula:

At n = 19

$S_n=\frac{19}{2}(2(54)+(19-1)(-3))$

$S_n=513$

At n = 18

$S_n=\frac{18}{2}(2(54)+(18-1)(-3))$

$S_n=513$

Hence 18 or 19 terms can be taken so that their sum is 513.