need help in seriess :( please (1 Viewer)

[joannee--]

how do i do this question ??


how many terms of the series 45 + 47 + 49 + ... give a sum of 1365


help would be greatly appreciated thanks

 

[watatank]

a = 45

d = 2

Sn = 1365

(n/2) * [ 2a + (n-1)d] = 1365

n [ 2a + (n-1)d] = 2730

n [ 2(45) + 2(n-1)] = 2730

n [ 90 + 2n- 2)] = 2730

n [ 88 + 2n] = 2730

88n + 2n^2 = 2730

2n^2 + 88n - 2730 = 0

n^2 + 44n - 1365 = 0



EDIT: if you use a bit of trial and error you can factorise this to

(n+65)(n-21) = 0

n = -65, 21

ignore the negative answer since you cant have a negative number of terms.

n = 21, .'. there are 21 terms in the series.

alternatively solve for n using the quadratic formula, ignore negative answer.

 

[joannee--]

thankss so much .. that helped

 

[samih91]

a = 45

d = 2

Sn = 1365

(n/2) * [ 2a + (n-1)d] = 1365

n [ 2a + (n-1)d] = 2730

n [ 2(45) + 2(n-1)] = 2730

n [ 90 + 2n- 2)] = 2730

n [ 88 + 2n] = 2730

88n + 2n^2 = 2730

2n^2 + 88n - 2730 = 0

n^2 + 44n - 1365 = 0



EDIT: if you use a bit of trial and error you can factorise this to

(n+65)(n-21) = 0

n = -65, 21

ignore the negative answer since you cant have a negative number of terms.

n = 21, .'. there are 21 terms in the series.

alternatively solve for n using the quadratic formula, ignore negative answer.

Is there a formula to solve a question like this?

 

[jannny]

Formula for Sum of all terms for Arithmetic Progression.

Sn = n/2 ( 2a + (n-1)d )

Nth term equation

Tn = A + (n - 1)d

For Geometric

if r > 0

Sn = a( r^n - 1) / r - 1

r < 0

Sn = a( 1 - r^n) / 1 - r

Nth term equation

Tn = a x r ^(n -1)