Question

Find the 9th term from the end of an A.P 5, 9 ,13 ,.........185

Answer 1

$\bf{\underline{QUESTION:-}}$

Find the 9th term from the end of an A.P 5, 9 ,13.......185

$\bf{\underline{Formula:-}}$

$\bf\large a_n = a + (n-1) d$

$\bf → Last\:term\: ( a_n )= 185$

$\bf → common \: difference (d) = 4$

$\bf → First \: term ( a ) = 5$

___________________

$\bf{\underline{SOLUTION:-}}$

$\bf → 185 = 5 + (n-1) 4$

$\bf → 185 - 5 = 4n - 4$

$\bf → 185 - 5 + 4 = 4n$

$\bf → 184 = 4n$

$\bf → n =\frac{184}{4}$

$\bf → n = 46$

______________

We have to find from end

• $\bf it\:means \:46\: - \:9\: + \:1$
• $\bf 46\: -\: 8$
• $\bf 38$

$\bf → a_{38}= 5 + ( 38 -1 ) 4$

$\bf → a_{38} = 5 + (37) 4$

$\bf → a_{38} = 5 + 148$

$\bf → a_{38}= 153$

or $\bf→ a_9 = 153$

_______________

Answer 2

Answer:

The given AP is

5, 9 ,13 ,.........185

First term = a = 5

Common Difference = d = 9 - 5 = 4

Let 185 be the nth term

So

a + ( n-1)d = 185

➙ 5 + (n-1)×4 = 185

➙ (n-1)×4 = 180

➙ (n-1) = 45

➙ n = 46

Now the 9th term from end

= ( 46 - 9 + 1) th term from First

= 38 th th term from First

= a + ( 38 - 1) × d

= 5 + 37×4

= 5 + 148

= 153