Question

first term of an ap is 5 and the 29th term is 106 then find the 35th tetm

Answer 1

First term = a = 5
a29 = 106
=> a+28d=106
=> 5+28d = 106
=>28d=101
=>d=101/28

a35=a+34d = 5+34 * 101/28 = 5+ 17 * 101/14 = (70+1717)/14 = 1777/28

Answer 2

Answer:

$a_{35}=\frac{1787}{14}$

Step-by-step explanation:

First term =a = 5

Formula of nth term in A.P. = $a_n=a+(n-1)d$

Where a = first term

d= common difference

Substitute n = 29 and a = 5  in the formula

$a_{29}=5+(29-1)d$

Sice we are given that 29th term is 106

$106=5+(28)d$

$106-5=28d$

$101=28d$

$\frac{101}{28}=d$

So, common difference =  $\frac{101}{28}$

Now to find 35th term :

$a_{35}=5+(35-1)\frac{101}{28}$

$a_{35}=5+(34)\frac{101}{28}$

$a_{35}=\frac{1787}{14}$

Hence the 35th term of A.P. is $\frac{1787}{14}$