in an AP the sum of first 10 terms is 50 and the fifth term is 3 times the second term a. find the first term b. find the common difference c. find the sum of first hundred terms

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.

answer: a.

step-by-step explanation: the answer is a. because, you need to converge the fractions to find a solution.

d

step-by-step explanation: