Answer:
This question can be solved in two ways but we're gonna go with the Formula Method
For the sum of terms
Sn = n/2[2a+ (n-1)d]
Given
Sn=50
n=10terms
Substituting
50=10/2[2a + (10-1)d]
50=5[2a + 9d]
10=2a + 9d......(i)
Now the Second Statement...
""The Fifth term is 3x the second term""
What is the Fifth term(T5) of an Ap?
Using the general formula for nth term of an Ap
Tn = a+(n-1)d
T5 = a +(5-1)d
T5= a + 4d
For Second term (T2)
T2=a + (2-1)d
T2 = a + d
"It said... T5 = 3T2
ie a + 4d = 3(a + d)
a + 4d = 3a + 3d
Collecting Like terms
2a=d........(ii)
Bring down both equations
10=2a + 9d...(i)
2a = d...(ii)
Substituting d=2a into Equation (i)
10 = 2a + 9(2a)
10=2a + 18a
10=20a
a=10/20
a=1/2
Since d=2a
d=2(1/2)
d=1.
Now
We're asked to Find the Sum of the First 100terms
n=100
a=1/2
d=1
Sn =n/2[2a+(n-1)d]
Sn= 100/2[2(1/2) + (100-1).1)
Sn= 50[1 + 99]
Sn = 50(100)
Sn = 5,000.