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# In an A.P, given $d=5,\text{ }{{\text{S}}_{\text{9}}}\text{=75}$ Find $a\text{ and }{{\text{a}}_{9}}\text{.}$  Hint: We know that in an A.P if we have first term 'a', common difference ’d’ the sum up to ${{n}^{th}}$ term is given by,
${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
Also, ${{n}^{th}}$ term of the A.P is given by,
${{T}_{n}}=a+\left( n-1 \right)d$

Now, we already have some values given to us in the question. So, by using the values in these formulas we will get the required value.
We have been given in an A.P $d=5,\text{ }{{\text{S}}_{\text{9}}}\text{=75}$
Let 'a' be the first term of the A.P.
We know that in an A.P if we have first term 'a', common difference 'd' then sum up to ${{n}^{th}}$ term is given by,
\begin{align} & {{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right) \\ & \Rightarrow {{S}_{9}}=\dfrac{9}{2}\left( 2a+\left( 9-1 \right)5 \right) \\ & \Rightarrow 75=\dfrac{9}{2}\left( 2a+8\times 5 \right) \\ & \Rightarrow 75=\dfrac{9}{2}\left( 2a+40 \right) \\ \end{align}
On multiplying the equation by $\dfrac{2}{9}$ we get
\begin{align} & \Rightarrow \dfrac{75\times 2}{9}=\left( 2a+40 \right) \\ & \Rightarrow \dfrac{50}{3}=2a+40 \\ \end{align}
On subtracting 40 to both side of the equation, we get
$\dfrac{50}{3}-40=2a$
On taking $LCM$ of the terms, we get
\begin{align} & \Rightarrow \dfrac{50-120}{3}=2a \\ & \Rightarrow \dfrac{-70}{3}=2a \\ \end{align}
On dividing the equation by 2, we get
\begin{align} & \Rightarrow a=\dfrac{-70}{2\times 3}=\dfrac{-70}{6} \\ & \Rightarrow a=\dfrac{-35}{3} \\ \end{align}
Now, we have $a=\dfrac{-35}{3},d=5$
We know that ${{n}^{th}}$ term of an A.P is given by,
${{T}_{n}}=a+\left( n-1 \right)d$
On substituting the value of a, n and d, we get
\begin{align} & \Rightarrow {{T}_{9}}=\dfrac{-35}{3}+\left( 9-1 \right)\times 5 \\ & \Rightarrow {{T}_{9}}=\dfrac{-35}{3}+\left( 8\times 5 \right) \\ & \Rightarrow {{T}_{9}}=\dfrac{-35}{3}+40 \\ \end{align}
On taking $LCM$ of the terms, we get

\begin{align} & \Rightarrow {{T}_{9}}=\dfrac{-35+120}{3} \\ & \Rightarrow {{T}_{9}}=\dfrac{85}{3} \\ \end{align}
Therefore, $a=\dfrac{-35}{3}\text{ and }{{\text{a}}_{9}}\text{=}\dfrac{85}{3}$

Note: Arithmetic progression is known as A.P. It is a series of terms, in which each term after the first term is equal to the sum of the first term and a constant known as common difference. This is a very simple question, so if the students get the formula right, then he must be aware of substitution of right values and avoid calculation mistakes. If the students get the formula wrong or forgets it, then it is impossible to solve and get the results.
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