Question

3 The fourth term of an AP is 10 and its eleventh term exceeds three times the fourth term
by 1. Find the sum of first 30 terms of AP.​

Answer 1

Answer:

1335

Step-by-step explanation:

a₄ = a + 3d = 10......................(1)

a₁₁ - 3a₄ = 1

a + 10d - 3(10) = 1

a + 10d = 31.........................(2)

From 1 and 2,

 a + 10d =31

- a + 3d   =10

⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

       7d = 21

        d= 3

a + 3 (3) = 10

 a=10-9

 a=1

Sₙ = n/2(2a + (n-1) d)

S₃₀= 30/2 (2ₓ1 + (30-1) 3 )

     =15 (2 + 29ₓ3)

     =15 ₓ (2 +87)

     =15 ₓ89

      = 1335

Hope this helps..........