Step 1: Recall formula of the $n$th term of an A.P.
The $n$th term of an arithmetic progression is given by:
\[
a_n = a + (n-1)d
\]
where $a =$ first term, $d =$ common difference.
Step 2: Identify values
First term $a = 6$
Common difference $d = 10 - 6 = 4$
We need the $25$th term $\,(n=25)$.
Step 3: Substitute values
\[
a_{25} = 6 + (25-1)\times 4
\]
\[
= 6 + 24 \times 4
\]
\[
= 6 + 96
\]
\[
= 102
\]
Step 4: Conclusion
The $25$th term of the A.P. is $102$.
The correct answer is option (B).
The common difference of the A.P.: $3,\,3+\sqrt{2},\,3+2\sqrt{2},\,3+3\sqrt{2},\,\ldots$ will be:
Let $a_1, a_2, a_3, \ldots$ be an AP If $a_7=3$, the product $a_1 a_4$ is minimum and the sum of its first $n$ terms is zero, then $n !-4 a_{n(n+2)}$ is equal to :
Let $a_1, a_2, \ldots, a_n$ be in AP If $a_5=2 a_7$ and $a_{11}=18$, then $12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots+\frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$ is equal to
$PQ$ is a chord of length $4\ \text{cm}$ of a circle of radius $2.5\ \text{cm}$. The tangents at $P$ and $Q$ intersect at a point $T$. Find the length of $TP$.