Question
Find the fourth term from the end in an A.P. $– 11, – 8, – 5, . . ., 49.$
✓
Answer
First term from end $a = 49$
$t_n = – 11$
$t_{n – 1} = – 8$
Common difference $d = t_n – t_{n – 1} = – 11 – ( – 8) = – 11 + 8 = – 3$
Now, By using $n^{th}$ term of an A.P. formula
$t_n = a + (n – 1)d$
where n = no. of terms
$a$ = first term
$d$ = common difference
$t_n$ = $n^{th}$ terms
no. of terms $n = 4$
$\Rightarrow t_4 = 49 + (4 – 1) \times ( – 3)$
$\Rightarrow t_4 = 49 + 3 \times ( – 3)$
$\Rightarrow t_4 = 49 – 9 = 40$
$t_n = – 11$
$t_{n – 1} = – 8$
Common difference $d = t_n – t_{n – 1} = – 11 – ( – 8) = – 11 + 8 = – 3$
Now, By using $n^{th}$ term of an A.P. formula
$t_n = a + (n – 1)d$
where n = no. of terms
$a$ = first term
$d$ = common difference
$t_n$ = $n^{th}$ terms
no. of terms $n = 4$
$\Rightarrow t_4 = 49 + (4 – 1) \times ( – 3)$
$\Rightarrow t_4 = 49 + 3 \times ( – 3)$
$\Rightarrow t_4 = 49 – 9 = 40$
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