Question

The $4^{th}$ term of an AP is $9$ and its $9^{th}$ term is $34.$ The sum of the first $10$ terms of the AP is ___

Answer 1

The formula to find the $n^{th}$ term of an arithmetic progression is $t_n=a+(n-1)d$.

where,

'$a$' is the first term,

${}^{\prime }{d}^{\prime }$ is the common difference,

${}^{\prime }{n}^{\prime }$ is the number of terms,

'$t_n$' is the $n^{th}$ term.

Given,

$4^{th}$ term $=9$ and the $9^{th}$ term $=34.$
$\Rightarrow a_4=a+(4-1)d=9$
$\Rightarrow a_9=a+(9-1)d=34$

On subtracting the above two equations, we get

$a+(9-1)d-[a+(4-1\left)d\right]=34-9$

$\Rightarrow 8d-3d=25$

$\therefore d=5$

On putting the value of d in any one above equation, we get

$a+3\times 5=9$

$a=9-15=-6$
From the above equations, we get $d=5$ and $a=-6.$

The formula to find out the sum of ${}^{\prime }{n}^{\prime }$ terms of an AP is

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

where $S_n$ is the sum of $n$ terms of the AP.

Now, sum of the given AP upto $10$ terms $=S_{10}$

$\Rightarrow S_{10}=\frac{10}{2}\left[2\right(-6)+(10-1\left)5\right]=5(-12+45)=165$