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Question: In an AP, given \(d = 5,{S_9} = 75\) , find \(a\) , and \({a_9}\) ....

In an AP, given d=5,S9=75d = 5,{S_9} = 75 , find aa , and a9{a_9} .

Explanation

Solution

For solving this question we are going to use the sum formula and number of terms in an AP formula. The sum formula is given by Sn=n2[2a+(n−1)d]{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] and the formula for the number of terms in an AP is given by an=a+(n−1)d{a_n} = a + \left( {n - 1} \right)d. By using these two we will get the required values.

Formula used:
Sum of the terms of AP,
Sn=n2[2a+(n−1)d]{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]
Number of terms in an AP,
an=a+(n−1)d{a_n} = a + \left( {n - 1} \right)d
Here,
Sn{S_n} , will be the sum of an AP
nn , will be the number of terms
aa , will be the first term
dd , will be the common difference

Complete Step by Step Solution:
So we have the term given as d=5,S9=75d = 5,{S_9} = 75 . Firstly we will find out the first term and for this, we will use the formula of Sn{S_n} .
So it is given by Sn=n2[2a+(n−1)d]{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]
On substituting the values, in the above formula, we get the equations
⇒75=92[2a+(9−1)5]\Rightarrow 75 = \dfrac{9}{2}\left[ {2a + \left( {9 - 1} \right)5} \right]
Now on solving the above equation, we get the equation as
⇒75=92[2a+40]\Rightarrow 75 = \dfrac{9}{2}\left[ {2a + 40} \right]
Again solving the above solution, we get
⇒25=3[a+20]\Rightarrow 25 = 3\left[ {a + 20} \right]
So on multiplying and removing the braces, we get
⇒3a=25−60\Rightarrow 3a = 25 - 60
And on dividing and taking the constant term to the right side, we get
⇒a=−353\Rightarrow a = \dfrac{{ - 35}}{3}
So, now by using the formula of the number of terms in AP, and substituting the values we have known so far, we get
⇒a9=−353+(9−1)5\Rightarrow {a_9} = \dfrac{{ - 35}}{3} + \left( {9 - 1} \right)5
And on solving the braces, we get the equation as
⇒a9=−353+40\Rightarrow {a_9} = \dfrac{{ - 35}}{3} + 40
Now taking the LCM and solving it, we get
⇒a9=−35+1203\Rightarrow {a_9} = \dfrac{{ - 35 + 120}}{3}
And on adding we get
⇒a9=853\Rightarrow {a_9} = \dfrac{{85}}{3}

Hence, the value of aa will be −353\dfrac{{ - 35}}{3} and a9{a_9} will be 853\dfrac{{85}}{3}.

Note:
As we all know the number series is very high scoring in terms of competition and to be the best in this department we have to know the commonly used formulas and so on. IF we get an expert in this type of series problem, then it also helps in improving the skills and knowledge for sure.