Question

Find sum of first 10 terms of an AP whose first term is 20 and 4th term is 5

Answer 1

Since it is given that 4th term is 5 and first term is 20 so
A=20 , d=? , N=4
So from formula a+(n-1)*d=Tn
So 20+(4-1)d=5 cause we have the 4th term
So from above we get d=(-5)
And then from sum formula I.e S=n/2(2*a+(n-1)*d)
S=10/2(2*20+(10-1)*(-5))
S=5*(40+9*(-5))
S=5*(40-45)
S=5*(-5)
S=(-25)
So the ans to your question is -25

Answer 2

Given : First term of an AP is 20 and it's 4th term is 5.

To find : Sum of first 10 terms.

Solution :

nth term of an AP is given by,

• an = a + (n - 1)d

where a = first term, d = common difference

Therefore,

a = 20 ...(i)

a4 = a + (4 - 1)d = 5

a + 3d = 5

Substitute value of a from (i)

20 + 3d = 5

3d = -15

d = -5

Now, sum of n terms of an AP is given by,

• Sn = n/2*[2a + (n - 1)d]

Therefore,

S10 = 10/2*[2(20) + (10 - 1)(-5)]

S10 = 5*[40 + 9(-5)]

S10 = 5*[40 - 45]

S10 = 5*[-5]

S10 = -25

Therefore the sum of 10 terms of AP is -25.