in an AP, U1=3 and the common difference is 0.8. if Sn=231, how do i find the value of n?

Answer 1

Use the formula: Sn = n/2(2u1 + (n-1)d)

so 231 = n/2 (2x3 + (n-1)x0.8)

This gives you a (quadratic) equation in n. It should have rational solutions and one of them will be invalid (see if you can figure why!).

Good luck!

Answer 2

The sum of an arithmetic series is given by:

Sn = n/2(2u1 + (n-1)d)

So we have:

231 = n/2(6+0.8n - 0.8)
231 = n/2(5.2 + 0.8n)
462 = 5.2n + 0.8n^2
0.8n^2 + 5.2n - 462 = 0

Multiply through by 2.5:

2n^2 + 13n - 1155 = 0

Usig the quadratic formula, we find that the positive solution to this equation is n=21.

Therefore the value of n is 21.

Answer 3

the formula for the sum of the nth
term is
Sn=n/2{2a+(n-1)d}......(1)
where
a=1st term
d=the common difference
Sn=the sum of n terms

n is found by substituting the values
into equation(1) and solving the
resulting quadratic equation

the series is
a,a+d,a+2d,a+3d,....a+(n-1)d
n=1,2,3,.......

substitute into equation(1)
231=n/2{2*3+(n-1)0.8}
231=n{3+0.4(n-1)}
231=3n+0.4n(n-1)
multiply through by 10
2310=30n+4n^2-4n
this gives
4n^2+26n-2310=0....(2)
using the quadratic formula
n=
{-26+or-sqrt(26^2-16*2310)}/8
={-26+or-194}/8=21 or -27.5
since n has to be +ve,
n=21
{disregard the -ve answer}

i hope that this helps

Answer 4

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

is the formula to use .

Sn = 231
n to find
a = 3
d = 0.8

So 231 = n/2 (2(3) + (n -1)0.8)
462 = n (6 + 0.8n - 0.8)
462 = 6n + 0.8n^2 - 0.8n
462 = 5.2n + 0.8n^2
This is now a quadratic equation in 'n' ; rearranging

0.8n^2 + 5.2n - 462 = 0

n = - 5.2 +/- Sq Rt(5.2^2 - 4(0.8)(-462))
___________________________
2(0.8)

n = -5.2 +/- Sq Rt(27.04 + 1478.4)
________________________
1.6

n = -5.2 +/- Sq Rt (1505.44)
__________________
1.6

n = -5.2 +/- 38.8
__________
1.6
n = -5.2+38.8
________
1.6
n= 21 (ANSWER)

The other value of 'n' is impossible as it gives negative result.