Question

In an arithmetic sequence the sum of 1 st 19 terms is 209 and sum of first 22 terms is 473. Then its 21 st term is​

Answer 1

Answer:

The 21st term of the A.P. is 88.

Step-by-step explanation:

Given: Sum of first 19 terms of A.P. is 209 and sum of first 22 terms of A.P. is 473.

Let the first term of the given A.P. be a and its common difference is d.

w.k.t., Sum of n terms of an A.P. with first term a and common difference d is $\tt{ \dfrac{n}{2}\{2a+(n-1)d\} }$.

$\tt{ \implies \dfrac{19}{2}\{ 2a + (19-1)d \} = 209 }$ and $\tt{ \dfrac{22}{2}\{ 2a + (22-1)d \} = 473 }$

$\tt{ \implies 2a + 18d = 209 \times \dfrac{2}{19} }$ and $\tt{ 2a + (22-1)d = 473 \times \dfrac{2}{22} }$

$\tt{ \implies 2a + 18d = 22 }$ and $\tt{ 2a + 21d = 43 }$

On subtracting above two equations we get,

2a + 18d - (2a+21d) = 22 - 43

⇒ 2a + 18d - 2a - 21d = -21

⇒ -3d = -21

⇒ d = 21/3 = 7

On putting d=7 in one of the above equations we get,

2a + 18×7 = 22

⇒ 2a + 126 = 22

⇒ 2a = 22 - 126

⇒ 2a = -104

⇒ a = -104/2 = -52

Also, general term of an A.P. with first term a and common difference d is $\tt{ a_{n} = a + (n-1)d }$.

⇒ a₂₁ = a + (21-1)d

⇒ a₂₁ = -52 + 20×7

⇒ a₂₁ = -52 + 140

⇒ a₂₁ = 88