When the 9^{th} term of an A.P. is divided by | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the first term be $a$ and the common difference be $d$. The $n^{th}$ term is $T_n = a + (n-1)d$.Given $T_9 = 5T_2$ $\Rightarrow a + 8d = 5(a +

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let the first term be $a$ and the common difference be $d$. The $n^{th}$ term is $T_n = a + (n-1)d$.Given $T_9 = 5T_2$ $\Rightarrow a + 8d = 5(a + d)$ $\Rightarrow a + 8d = 5a + 5d$ $\Rightarrow 4a - 3d = 0$ $\Rightarrow d = \frac{4a}{3}$.Given $T_{13} = 2T_6 + 5$ $\Rightarrow a + 12d = 2(a + 5d) + 5$ $\Rightarrow a + 12d = 2a + 10d + 5$ $\Rightarrow 2d - a = 5$.Substitute $d = \frac{4a}{3}$ into the second equation: $2(\frac{4a}{3}) - a = 5$.$\frac{8a}{3} - a = 5$ $\Rightarrow \frac{5a}{3} = 5$ $\Rightarrow a = 3$.

When the $9^{th}$ term of an $A.P.$ is divided by its $2^{nd}$ term,the quotient is $5$. When the $13^{th}$ term is divided by the $6^{th}$ term,the quotient is $2$ and the remainder is $5$. Find the first term of the $A.P.$

Similar Questions

If the sum of $n$ terms of an $A.P.$ is $S_n = nP + \frac{1}{2}n(n-1)Q$,where $P$ and $Q$ are constants,find the common difference.

If the sum of $n$ terms of an arithmetic progression is $2n^2 + 5n$,then its $n^{th}$ term is.........

The sum of the first $p, q,$ and $r$ terms of an $A.P.$ are $a, b,$ and $c,$ respectively. Prove that $\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$.

Let $x, y > 0$. If $x^{3} y^{2} = 2^{15}$,then the least value of $3x + 2y$ is

Let $a_1, a_2, a_3, \ldots$ be in an $A.P.$ such that $\sum_{k=1}^{12} a_{2k-1} = -\frac{72}{5} a_1$,where $a_1 \neq 0$. If $\sum_{k=1}^{n} a_k = 0$,then $n$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module