If the sum of $n$ terms of an $A.P.$ is $3 n^{2}+5 n$ and its $m^{\text {th }}$ term is $164,$ find the value of $m$
Let $a$ and $b$ be the first term and the common difference of the $A.P.$ respectively.
$a_{m}=a+(m-1) d=164$ ............$(1)$
Sum of $n$ terms: $S_{n}=\frac{n}{2}[2 a+(n-1) d]$
Here,
$\frac{n}{2}[2 a+n d-d]=3 n^{2}+5 n$
$\Rightarrow n a+n^{2} \cdot \frac{d}{2}-\frac{n d}{2}=3 n^{2}+5 n$
Comparing the coefficient of $n^{2}$ on both sides, we obtain
$\frac{d}{2}=3$
$\Rightarrow d=6$
Comparing the coefficient of $n$ on both sides, we obtain
$a-\frac{d}{2}=5$
$\Rightarrow a-3=5$
$\Rightarrow a=8$
Therefore, from $(1),$ we obtain
$8+(m-1) 6=164$
$\Rightarrow(m-1) 6=164-8=156$
$\Rightarrow m-1=26$
$\Rightarrow m=27$
Thus, the value of $m$ is $27 .$
Let the digits $a, b, c$ be in $A.P.$ Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in $A.P.$ at least once. How many such numbers can be formed?
If $\frac{1}{{b - c}},\;\frac{1}{{c - a}},\;\frac{1}{{a - b}}$ be consecutive terms of an $A.P.$, then ${(b - c)^2},\;{(c - a)^2},\;{(a - b)^2}$ will be in
The sum of $n$ terms of two arithmetic progressions are in the ratio $(3 n+8):(7 n+15) .$ Find the ratio of their $12^{\text {th }}$ terms.
Find the $25^{th}$ common term of the following $A.P.'s$
$S_1 = 1, 6, 11, .....$
$S_2 = 3, 7, 11, .....$
Find the sum of all numbers between $200$ and $400$ which are divisible by $7.$