Question

Find in 31st term of an ap whose 10th term is 38 and 16th term is 9.​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

The $31^{st}$ term of the arithmetic progression is  -63.5

Step-by-step explanation:

The $n^{th}$ term of an arithmetic progression having the first term $a$ and the common difference $d$ is given by,

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

Given $10^{th}$ term of the arithmetic progression is 38. That is

$a_{10}=a+(10-1)d=38\\\\\implies a+9d=38$

Also, the $16^{th}$ term is 9. That is,

$a_{16}=a+(16-1)d=9\\\\\implies a+15d=9$

Now we have two equations with two unknowns $a$ and $d$ . Subtract the second equation from the first equation, we get

$a+9d-(a+15d)=38-9\\\\\implies 9d-15d=29\\\\\implies -6d=29\\\\\implies d=\dfrac{-29}{6}$

Substitute the above value of $d$ in the first equation, we obtain,

$a+9(\dfrac{-29}{6})=38\\\\\implies a =38+\dfrac{29\times 3}{2} =\dfrac{76 + 87}{2} = \frac{163}{2} = 81.5$

So the first  term is 81.5 and the common difference is $\dfrac{-29}{6}$  .

Thus the $31^{st}$ term is,

$a_{31}=a+(31-1)d=a+30d = 81.5+30(\dfrac{-29}{6})=81.5-145=-63.5$