Question

Which term of A.P. -3, -1/2, 2, 9/2, 7,…… is 47​

Answer 1

Answer:

21st term

Step-by-step explanation:

Here First term (a) = -3

Common difference (d) = third term - second term= 2 -(-1/2) = 5/2  [Difference of any two consecutive terms]

$a_n = 47$

We know that

$a+(n-1)d=a_n$

$\therefore \ -3+(n-1)\times \frac{5}{2}=47$

$(n-1)\times \frac{5}{2}=47+3$

$(n-1)=50\times \frac{2}{5}$

$n-1=20$

$n=21$

Hence 21st term of given A.P. is 47

Answer 2

Hint:

If $a_{1}$ be the first term and $d$ be the common ratio of an A.P., then its $n-th$ term be

$\quad a_{n}=2a+(n-1)d$

Step-by-step explanation:

The given A.P. is

$\quad -3,-\dfrac{1}{2},2,\dfrac{9}{2},7,...$

The first term, $a_{1}=-3$ and the common difference, $d=-\dfrac{1}{2}-(-3)=\dfrac{5}{2}$

If we consider $47$ to be the $n-th$ term of the A.P., then

$\quad a_{n}=47$

$\Rightarrow a_{1}+(n-1)d=47$

$\Rightarrow -3+(n-1)\times\dfrac{5}{2}=47$

$\Rightarrow (n-1)\times\dfrac{5}{2}=50$

$\Rightarrow n-1=\dfrac{100}{5}$

$\Rightarrow n=20+1$

$\Rightarrow n=21$

Final answer:

The $21st$ term of the given A.P. is $47$.

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