Question
Find the arithmetic progression whose third term is $16$ and seventh term exceeds its fifth term by $12.$
✓
Answer
Let $a, a + d, a + 2d, a + 3d$, ………. be the A.P.
$a_n = a + (n – 1)d$
But $a_3 = 16$
$a_7 - a_5 = 12$
Now $a_3 = a + (3 - 1)d = a + 2d$
$a_5 = a + (5 - 1)d = a + 4d$
and $a_7 = a + (7 - 1)d = a + 6d$
$\therefore$
$a + 2d = 16$
$\Rightarrow a = 16 - 2d$
and $a_7 - a_5 = a + 6d - a - 4d$
$\Rightarrow\ 12 = 2\text{d}\Rightarrow \text{d}=\frac{12}{2}=6$
$\therefore$
$a = 16 - 2d = 16 - 2 \times 6$
$\Rightarrow a = 16 - 12 = 4$
$\therefore$ Sequencing (A.P.) will be
$4, 10, 16, 22, .....$
$a_n = a + (n – 1)d$
But $a_3 = 16$
$a_7 - a_5 = 12$
Now $a_3 = a + (3 - 1)d = a + 2d$
$a_5 = a + (5 - 1)d = a + 4d$
and $a_7 = a + (7 - 1)d = a + 6d$
$\therefore$
$a + 2d = 16$
$\Rightarrow a = 16 - 2d$
and $a_7 - a_5 = a + 6d - a - 4d$
$\Rightarrow\ 12 = 2\text{d}\Rightarrow \text{d}=\frac{12}{2}=6$
$\therefore$
$a = 16 - 2d = 16 - 2 \times 6$
$\Rightarrow a = 16 - 12 = 4$
$\therefore$ Sequencing (A.P.) will be
$4, 10, 16, 22, .....$
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