stolyar wrote:

we have two ariphmetical progressions

(X) the period is 10 years; x1=300 for the first year, +30 for each of the next 9 years. x10=300+9*30=300+270=570. The sum=[300+570]*10/2=4350

(Y) the period is 20 half-years; y1=200 for the first half-year; y20=200+19*15=200+285=485; the sum=[200+485]*20/2=6850

the total is 11200. no such options.

I have assumed that 200 and 300 are yearly salaries. Let's make them monthly ones.

(X) x1=3600; x10=3600+270=3870; the sum=[3600+3870]*10/2=37350

(Y) y1=2400; y20=2400+285=2685; the sum=[2400+2685]*20/2=50850

the total is 88200. no such option again.

I do not know what to do...

Actually i wanted to know if we can do this without a "formula".

Anyways...

The problem does mention that the salary is paid at the end of the month

The formula for SUM of an AP is => N/2 *[ 2*a + (N-1) *d]

X : a = 300, d = 30, N = 10;

S1 = 5(600 + 9 ├Ч (30 ├Ч 12)) = 870 ├Ч 5 ├Ч 12 = 52,200.

Y : a = 200, d = 15, N = 20;

S2 = 10(400 + 19 ├Ч (15 ├Ч 6)) = 41,100.

Total amount = 52,200 + 41,100 = 93,300

Thanks

Praetorian