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 =