Question

The 4th term of an a.p is 10 and its 11th term is one more than 3 times its 4th term find the sum of First 209 Terms

Answer 1

Answer:given below

Step-by-step explanation: observe the steps of the sum

Evaluate the rest.....

Answer 2

Answer:

65417

Step-by-step explanation:

The 4th term of an a.p is 10

11th term is one more than 3 times its 4th term

11th term = 1+3(10)=31

Formula of nth term = $a_n=a+(n-1)d$

substitute n = 4

$a_4=a+(4-1)d$

$10=a+3d$  -- 1

Substitute n = 11

$a_{11}=a+(11-1)d$

$31=a+10d$ --2

Subtract 1 from 2

$31-10=a+10d-a-3d$

$21=7d$

$d=3$

Substitute the value of d in 1

$10=a+3(3)$

$a=1$

Sum of n terms =$S_n=\frac{n}{2}(2a+(n-1)d)$

Substitute n = 209

$S_{209}=\frac{209}{2}(2(1)+(209-1)3)$

$S_{209}=\frac{209}{2}(2+624)$

$S_{209}=65417$

Hence the sum of 209 terms is 65417