The ratio of the sum of first 'a' odd natural numbers to the sum of first 'b' even natural numbers is 32 : 51. If a and b are in ratio 4 : 5, then find the sum of all 'b' even numbers and all 'a' odd number.
Answers
Answer:
Sum of first "a" odd numbers = 1600
Sum of first "b" even numbers = 2550
where
a = 40
b = 50
Step-by-step explanation:
Let S(a) denote the sum of the first "a" odd natural numbers
Now, this is an AP with first term = 1, common difference = 2
=> S(a) = (a/2)[2 + (a-1)*2]
(Note: "a" is the number of terms in ths case!)
=> S(a) = (a/2)[2 + 2a - 2]
= (a/2)(2a)
= a²
Let S(b) denote the sum of the first "b" even natural numbers
Now, this is an AP with first term = 2, common difference = 2
=> S(b) = (b/2)[4 + (b-1)*2]
= (b/2)[4 + 2b - 2}
= (b/2)(2b + 2)
= (b/2)*(b+1)*2
= b(b+1)
Note: We have used
Sum of first "n" terms of an AP
= (n/2) { 2*first term + (n-1)*common difference]
We are given that:
S(a) : S(b) = 32:51
=> a²/[b(b+1)] = 32/51 .............Eqn 1
We are also given that
a:b = 4:5
=> a/b = 4/5
=> 5a = 4b
=> a = 4b/5 ............Eqn 2
Substituting the value of "a" from Eqn 2 in Eqn 1, we get:
16b²/[25(b²+b)] = 32/51
16b²/[25b(b+1)] = 32/51
16b/[25(b+1)] = 32/51
on cross multiplying,
16b * 51 = 32*25(b+1)
816b = 800(b + 1)
816b = 800b + 800
16b = 800
b = 50
Using Eqn 2, we get:
a = (4*50)/5
a = 40
In the second part, we need:
Sum of first b (= 50) even numbers, denoted as SUME
This is an AP with
first term = 2; common difference = 2; no. of terms = 50
SUME = (50/2)[2*2 + (50 - 1)*2]
= 25 [ 4 + 49*2]
= 25 [ 102]
= 2550
Sum of first a (= 40) odd numbers, denoted as SUMO
This is an AP with
first term = 1; common difference = 2; no. of terms = 40
SUMO = (40/2)[2*1 + (40-1)*2]
= 20(2 + 78]
= 1600