Question

How many terms of the A.P: 9, 17, 25………………….. must be taken to give a sum of 636?

Answer 1

Hint: Numbers of a series are said to be in A.P if and only if the common difference that is the difference between the consecutive terms remains constant throughout the series. Use the direct formula for the sum of n terms of an AP, to find out the value of n for the given sum.

Complete step by step solution:

The given A.P is
$9,17,25,...................$
So the first term (a) of the A.P is = 9.
And the common difference (d) of the A.P is
$ \Rightarrow d = \left( {17 - 9} \right) = \left( {25 - 17} \right) = 8$
Now we have to find out the number of terms in the A.P if the sum of an A.P is 636.
As we know that the sum (Sn) of an A.P is given as
$ \Rightarrow {S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
Where symbols have their usual meanings.
So substitute the values in the above equation we have,
$ \Rightarrow 636 = \dfrac{n}{2}\left( {2 \times 9 + \left( {n - 1} \right)8} \right)$
Now simplify the above equation we have,
$
   \Rightarrow 636 = n\left( {9 + \left( {n - 1} \right)4} \right) \\
   \Rightarrow 636 = n\left( {4n + 5} \right) \\
   \Rightarrow 4{n^2} + 5n - 636 = 0 \\
 $
Now this is a quadratic equation so apply quadratic formula we have,
$ \Rightarrow n = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where (a = 4, b = 5, c = -636)
So, $ \Rightarrow n = \dfrac{{ - 5 \pm \sqrt {{5^2} - 4\left( 4 \right)\left( { - 636} \right)} }}{{2\left( 4 \right)}} = \dfrac{{ - 5 \pm \sqrt {25 + 10176} }}{8}$
$ \Rightarrow n = \dfrac{{ - 5 \pm \sqrt {25 + 10176} }}{8} = \dfrac{{ - 5 \pm \sqrt {10201} }}{8} = \dfrac{{ - 5 \pm \sqrt {{{\left( {101} \right)}^2}} }}{8} = \dfrac{{ - 5 \pm 101}}{8}$
$ \Rightarrow n = \dfrac{{ - 5 + 101}}{8},\dfrac{{ - 5 - 101}}{8}$
$ \Rightarrow n = \dfrac{{96}}{8},\dfrac{{ - 106}}{8}$
Now as we know negative value of n is not possible
$ \Rightarrow n = \dfrac{{96}}{8} = 12$
So we have to take 12 terms in an A.P to give a sum of 636.
So n = 12 is the required answer.

Note: Whenever we face such types of problems it is advised to have a good grasp over the general formula related to that specific series which is being asked of in the question, just like arithmetic progression in this case. This helps to save a lot of time and to get on the right track to reach the answer.