If the sum of a certain number of terms of the $A.P.$ $25,22,19, \ldots \ldots .$ is $116$ Find the last term
If the sum of a certain number of terms of the $A.P.$ $25,22,19, \ldots \ldots .$ is $116$ Find the last term
Let the sum of $n$ terms of the given $A.P.$ be $116$
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
Here, $a=25$ and $d=22-25=-3$
$\therefore S_{n}=\frac{n}{2}[2 \times 25+(n-1)(-3)]$
$\Rightarrow 116=\frac{n}{2}[50-3 n+3]$
$\Rightarrow 232=n(53-3 n)=53 n-3 n^{2}$
$\Rightarrow 3 n^{2}-53 n+232=0$
$\Rightarrow 3 n^{2}-24 n-29 n+232=0$
$\Rightarrow 3 n(n-8)-29(n-8)=0$
$\Rightarrow(n-8)(3 n-29)=0$
$\Rightarrow n=8$ or $n=\frac{29}{3}$
Howerer, $n$ cannot be equal to $\frac{29}{3}$ therefore, $n=8$
$\therefore a_{8}=$ Last term $=a+(n-1) d=25+(8-1)(-3)$
$=25+(7)(-3)=25-21$
$=4$
Thus, the last term of the $A.P.$ is $4.$
Similar Questions
Consider a sequence whose sum of first $n$ -terms is given by $S_n = 4n^2 + 6n, n \in N$, then $T_{15}$ of this sequence is -
Consider a sequence whose sum of first $n$ -terms is given by $S_n = 4n^2 + 6n, n \in N$, then $T_{15}$ of this sequence is -
The sum of the first four terms of an $A.P.$ is $56$. The sum of the last four terms is $112$. If its first term is $11$, the number of terms is
The sum of the first four terms of an $A.P.$ is $56$. The sum of the last four terms is $112$. If its first term is $11$, the number of terms is
For any three positive real numbers $a,b,c$ ; $9\left( {25{a^2} + {b^2}} \right) + 25\left( {{c^2} - 3ac} \right) = 15b\left( {3a + c} \right)$ then
For any three positive real numbers $a,b,c$ ; $9\left( {25{a^2} + {b^2}} \right) + 25\left( {{c^2} - 3ac} \right) = 15b\left( {3a + c} \right)$ then
- [JEE MAIN 2017]
Three numbers are in $A.P.$ whose sum is $33$ and product is $792$, then the smallest number from these numbers is
Three numbers are in $A.P.$ whose sum is $33$ and product is $792$, then the smallest number from these numbers is
Given sum of the first $n$ terms of an $A.P.$ is $2n + 3n^2.$ Another $A.P.$ is formed with the same first term and double of the common difference, the sum of $n$ terms of the new $A.P.$ is
Given sum of the first $n$ terms of an $A.P.$ is $2n + 3n^2.$ Another $A.P.$ is formed with the same first term and double of the common difference, the sum of $n$ terms of the new $A.P.$ is
- [JEE MAIN 2013]