Question

The 4th term from the end of the AP: -11,-8,-5,...,49 is (a) 37,(b) 40,(c) 43,(d) 58​

Answer 1

$\huge\red{Answer}$

(b) 40

$\huge\red{Explanation :-}$

Let a be the first term of this AP

and d be the common difference

a = -11

d = -8-(-11)

= -8+11

= 3

l = 49

4th term from last l-(n-1)d

= 49-(4-1)3

= 49-3×3

= 49-9

= 40

HOPE IT HELPS YOU !!

Answer 2

$\mathfrak{ANSWER}$

To find the nth term from the end, we will take the last term to be the first term and the additive inverse of the common difference,

a = 49

d = -8-(-11) = 3

=> d = -3

n = 4

To find ${a_{n}}$

$\boxed{a_{n}= a+(n-1)d}$

Using the formula we get

${a_{4}}$ = 49 + 3×-3

${a_{4}}$ = 49 - 4

${a_{4}}$ = 40

Answer - Option b