SIMPLE INTEREST.
5, a the first term,
>ve series we have
expressed by the
le annuity,
f £32 5 forborne 12
i == .035
3 the amount,
oeing given,
l.i
, z)
ve
the number of years, and by the sum divide twice the amount of the
annuity.
Example. What annuity forborne 12 years will amount to
¿£4650 15 0 at 3^ per cent simple interest ?
z
s = 4650. 75
zz = 12
5
n
- 1 = 11
n.(n
— 1) = 132
z = .035
annuity of ¿£a in
660
396
mber of years less
n (n —
■ 1) z= 4.620
2zz = 24
half of this product
2 n + n (zz —
1) z = 28.62 ;
i = .035
4650.75
2
9301.50
8586
7155~
5724
14310
14310
16. To find (n) the number of years, the rest being given,
(Art. 15.) 2 s = a (2 n + n (zz — 1) z)
divide each side by a, we have
2s
—= 2 zz -j- zz (zz — 1) z = zzz 2 + 2 n_— in = zzz 2 + n (2 — z)
dividing by z, zz 2 + -—r— n
2 — ¿V
ai
adding ^ ^ l J to each side to complete the square (Arithmetic and
Algebra, 206).
, 2 — i /2 — A 2 2 s . (2 — z) 2
z V 21 J ai 4 z 2
8 z ~ + (2 — z) 2
— 4? ’
extracting the square root of each side;
2 -J. _ \/ 8 ^ + (2- ¿)
zz +
by transposition,
2 z
2 i
\/ 87 (2 — z) 2 . -(2 — i)
2 z
: number of years less
this product add twice
n =