The $7^{t h}$ term of an AP is $- 4$ and its $13^{t h}$ term is $- 16$ , find the AP?

1, 2, 3, 4, . . . .

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2, 4, 6, 8, . . .

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8, 6, 4, 2, 0, - 2, . . .

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0, - 2, - 4, - 6, . . .

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Solution

The correct option is C $8, 6, 4, 2, 0, - 2, . . .$
$n t h$ term of an AP is ,
$a_{n} = a + (n - 1) d$
where,
$a =$ first term
$d =$ common difference

now, the two terms given are $a_{7} = - 4$ and $a_{13} = - 16$
so,
$a + 6 d = - 4$ ..........(1)
$a + 12 d = - 16$ .......(2)

Substract equation (1) from (2),
$6 d = - 12$
$d = - 2$

From equation (1),
$a + 6 d = - 4$
$a + 6 (- 2) = - 4$
$a - 12 = - 4$
$a = 8$

so,
$a_{2} = a + d = 8 - 2 = 6$
$a_{3} = a + 2 d = 8 - 4 = 4$

Therefore, the AP is,
$8, 6, 4, 2, 0, . . . . .$

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