Maths Mess !?

Question

a) find the number of terms of the arithmetic progression ... 54, 51, 48 ... so that their sum is 513

b) the sum of two numbers a and b is 15, and the sum of their reciprocals 1/a and 1/b is 3/10. Find the number a and b

help :-)

Answer 1

for the first one u hav to use the sum formula i.e sn=n/2(2a+(n-1)d)
thus we get
513=n/2(2*54+(n-1)-3)
513*2=n(108-3n+3)
1026=111n-3n^2
-3n^2+111n-1026=0
(dividing by -3)
n^2-37n+342
solve by using quadratic formula we get,
(n-19)(n-18)
n=19or n=18
thus u hav to either add 19 or 18 terms to get 513.
(it comes like this b'coz the 19th term is 0)

Answer 2

a)

use the formula to find the number of series ...

b)

given.,

a+b=15 ------- (1)
1/a + 1/b = 3/10 -------- (2)

consider the equation (2), & solve it ...

a + b
------- = 3/10
ab

15 / ab = 3/10

150 / 3 = ab

ab = 50
a = 50 / b ------- (3)

now consider the equation (1),
a + b = 15

sub (3) in (1),
50 / b + b = 15


50 + b^2
----------- = 15
b

50 + b^2 = 15b

b^2 - 15b + 50 = 0
(b-10) ( b-5) = 0
b = 10, b =5

therefore ,

if b = 10, then a = 5
if b = 5 , then a = 10

Answer 3

a) use the formula sn=n/2(2a+(n-1)d)



b)

a+b=15 ------- (1)
1/a + 1/b = 3/10 -------- (2)

consider the equation (2), & solve it ...

a + b
------- = 3/10
ab

15 / ab = 3/10

150 / 3 = ab

ab = 50
a = 50 / b ------- (3)

now consider the equation (1),
a + b = 15

sub (3) in (1),
50 / b + b = 15


50 + b^2
----------- = 15
b

50 + b^2 = 15b

b^2 - 15b + 50 = 0
(b-10) ( b-5) = 0
b = 10, b =5



if b = 10, then a = 5
if b = 5 , then a = 10

Answer 4

14 numbers from 54, 51, 48, .....18, 15 = 513

a = 5
b = 10