Given:
4th term of an AP = 10
11th term exceeds 4th term by 1.
Find:
the sum of 20 terms
Solution:
⟹ 11th term = 10 + 11 = 11.
We know that,
nth term of an AP (an) = a + (n - 1)d
Hence,
⟹ a₄ = a + (4 - 1)d
⟹ a + 3d = 10
⟹ a = 10 - 3d -- equation (1)
⟹ a₁₁ = a + (11 - 1)d
Substitute the value of a from equation (1).
⟹ 10 - 3d + 10d = 11
⟹ 7d = 11 - 10
⟹ 7d = 1
⟹ d = 1/7
Substitute the value of d in equation (1).
⟹ a = 10 - 3(1/7)
⟹ a = (70 - 3)/7
⟹ a = 67/7
Now,
Sum of first n terms of an AP (Sn) = n/2 * [ 2a + (n - 1)d ]
⟹ S₂₀ = 20/2 * [ 2(67/7) + (20 - 1)(1/7) ]
⟹ S₂₀ = 10 (134 + 19)/7
⟹ S₂₀ = 1530/7
∴ Sum of first 20 terms of the given AP is 1530/7.
I hope it will help you.
Regards.