Question

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

Answer 1

Answer:-

Given:

4th term of an AP = 10

11th term exceeds 4th term by 1.

⟹ 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.

Answer 2

$\sf{\huge{\underline{\underline{Answer:-}}}}$

$\rightarrow \sf{We \: know,} \\ \\ </p><p></p><p> \rightarrow \sf{an = a1 + (n - 1)d} \\ \\ </p><p></p><p> \rightarrow \sf{Using \: this \: formula \: in (1) and (2),} \\ </p><p>$

$\rightarrow \sf{a + 3d = 10 (3)} \\ </p><p></p><p> \\ \rightarrow\sf{a + 10d = 3(a + 3d) + 1 + d } \\ \\ \rightarrow \sf{= 2a +1→ (4)}$

$\rightarrow \sf{Putting (4) in (3)} \\ \\ \rightarrow\sf{, a + 3(2a + 1) 10 a = 1} \\ </p><p></p><p> \\ \rightarrow\sf{Thus, d=3} \\ </p><p></p><p> \\ \rightarrow \sf{Now, S, = (2a + (n - 1)d)}$

$\rightarrow\sf{S20 (2 x (1) + (20 1) 3) 2} \\ </p><p></p><p> \\ \rightarrow\sf{S20 = 590}$