The 4th term of ap is 10 and the 11th term of it exceeds 4th term by 1 find the sum of 20 terms

4th term of an AP = 10

11th term exceeds 4th term by 1.

the sum of 20 terms

⟹ 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.

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source