Question

The 4th term from the end of the AP -11,-8,-5,…..49 is​

Answer 1

Given:

AP  = -11, -8 , -5, ........ 49

To be found:

The 4th term from the end of the given AP.

$\bf{\underline{To\:\:get\:\:n^{th}\:\:term\:\:from\:\:the\:\:end\:\:of\:\:AP }}$

= $\red{\boxed{l-(n-1)d}}$

[ ∵ '$\red{l}$' is the last term of AP, '$\red{n}$' is nth term, '$\red{d}$' is the common difference.]

So,

$\red{l}$ of the AP = 49

$\red{n}$ of the AP = 4

$\red{d}$ of the AP = (-8) - (-11) = -8+11 = 3

The 4th term from the end of the given AP -

$= \red{l-(n-1)d} \\ \\ = 49 - (4-1)3 \\ \\ = 49 - 3(3) \\ \\ = 49 - 9 \\ \\ = \boxed{\boxed{40}}$

Hence,

The 4th term from the end of the given AP - 40

Answer 2

Given ,

The AP is -11 , -8 , -5 , ... , 49

• First term (a) = -11
• Common difference (d) = 3
• Last term (l) = 49

We know that , the nth term of an AP from end is given by

$\sf \fbox{ a_{n} = l - (n - 1)d }$

Thus ,

The fourth term from end of an AP is

$\sf \Rightarrow a_{4} = 49 - (4 - 1)3 \\ \\\sf \Rightarrow a_{4} =49 - 9 \\ \\ \sf \Rightarrow a_{4} =40$

$\therefore \sf \bold{ \underline{The \: 4th \: term \: from \: end \: of \: an \: AP \: is \: 40}}$