Question

the sum of first 14 terms of Ap 1050 and fourth terms is 40 than find the 20th term​

Answer 1

Answer:

The 20th term in this AP is 200

Step-by-step explanation:

We know that,

a(nth) = a + (n - 1)d

also,

Sn = (n/2)[2a + (n - 1)d]

Now,

We are given,

S14 = 1050

a(4th) = 40

Now, let's start with a(40th)

n = 4

a = a

d = d

a(4th) = a + (4 - 1)d

40 = a + 3d

a = 40 - 3d ----- 1

Now,

S14 = (14/2)[2a + (14 - 1)d]

1050 = 7 × (2a + 13d)

1050 = 14a + 91d ----- 2

Now, putting eq.1 in eq.2, we get,

14(40 - 3d) + 91d = 1050

560 - 42d + 91d = 1050

49d = 1050 - 560

49d = 490

d = 490/49

d = 10

Now,

a = 40 - 3d

a = 40 - 3(10)

a = 40 - 30

a = 10

Hence, AP = 10, 20, 30,.......

Thus,

a(20th) = a + (20 - 1)d

= 10 + (19 × 10)

= 10 + 190

= 200

Thus, the 20th term in this AP is 200

Hope it helped and you understood it........All the best