Question

The sum of terms of the series \[2,5,8,......\] is 950.find \[n\]

Answer 1

Hint: Here,first find out which series it is then use the sum of n terms of that series. Given series is in the form of A.P.

Complete step-by-step answer:
Given 2,5,8….
As we can see the difference between two consecutive terms are constant i.e 3. Hence the given series is in AP.
So, \[a = 2,d = 3,{S_n} = 950\]
We know that the formula for \[{S_n}\] is given by \[{S_n} = \dfrac{n}{2}\left( {2a + (n - 1)d} \right) = 950\]
\[ \Rightarrow \dfrac{n}{2}\left( {2a + (n - 1)d} \right) = 950\]
\[ \Rightarrow \dfrac{n}{2}\left( {2 \times 2 + (n - 1)3} \right) = 950\]
\[ \Rightarrow n(4 + 3n - 3) = 950 \times 2\]
\[ \Rightarrow n(1 + 3n) = 1900\]
\[ \Rightarrow n + 3{n^2} = 1900\]
\[ \Rightarrow 3{n^2} + n - 1900 = 0\]
Using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]we get
\[ \Rightarrow n = \dfrac{{ - 1 \pm \sqrt {1 - 4 \times 3 \times ( - 1900)} }}{{2 \times 3}}\]
\[ \Rightarrow n = \dfrac{{ - 1 \pm \sqrt {1 + 22800} }}{6}\]
\[ \Rightarrow n = \dfrac{{ - 1 \pm \sqrt {22801} }}{6}\]
\[ \Rightarrow n = \dfrac{{ - 1 \pm 151}}{6}\]
\[ \Rightarrow n = \dfrac{{150}}{6},\dfrac{{ - 152}}{6}\]
Now n cannot be negative hence \[n = 25\]

Note: (i) The negative value n needs to be discarded since, n is a positive integer and terms cannot be negative.
(ii) Three things determine an AP First term, common difference and number of terms knowing these you can find the AP. Depending on data given apply the appropriate formula and find the unknowns. Be careful with calculations.