Question
Determine the $n^{th}$ term of the $AP$ whose $7^{th}$ term is $-1$ and $16^{th}$ term is $17.$
✓
Answer
The general term of an $AP$ is given by
$a_n= a + (n - 1)d$
Given that $a_7= -1$
$⇒ a + 6d = -1 ...(i)$
Now,
$a_{16}= 17$
$⇒ a + 15d = 17...(ii)$
Subtract from $(i)$ from $(ii).$
$9d = 18$
$⇒ d = 2$
Substituting in $(i),$ we get $a = -13.$
$\text { Thus, the } \mathrm{n}^{\text {th }} \text { term will be, }$
$a_n=a+(n-1) d$
$\Rightarrow a_n=-13+(n-1)(2)$
$\Rightarrow a_n=-13+2 n-2$
$\Rightarrow a_n=2 n-15 .$
$a_n= a + (n - 1)d$
Given that $a_7= -1$
$⇒ a + 6d = -1 ...(i)$
Now,
$a_{16}= 17$
$⇒ a + 15d = 17...(ii)$
Subtract from $(i)$ from $(ii).$
$9d = 18$
$⇒ d = 2$
Substituting in $(i),$ we get $a = -13.$
$\text { Thus, the } \mathrm{n}^{\text {th }} \text { term will be, }$
$a_n=a+(n-1) d$
$\Rightarrow a_n=-13+(n-1)(2)$
$\Rightarrow a_n=-13+2 n-2$
$\Rightarrow a_n=2 n-15 .$
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