Question

How many terms of the A.P. 25,22, 19,… are needed to give the sum 116? Also find the last term.

Answer 1

Hint: Here, first we find the common difference of the given AP. Then, apply the formula of sum of n terms of an AP, as we are given a sum of AP. From there we will get the number of terms, we can find the last term using formula.

Complete step by step answer:
As we know, Arithmetic Progression (AP) is a sequence of numbers in order in which the difference of any two consecutive terms is a constant, which is called a common difference.
Here, the given AP is 25, 22, 19, …
First term \[ = a = 25\]
And Common difference = d = second term – first term \[ = 22 - 25 = - 3\]
Given, Sum of n terms \[ = {S_n} = 116\]
We need to find last term \[a{}_n\]
First, we have to find,\[n\]
We know that, sum of first n terms of an AP is given by
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
Putting values of a, d and Sn
\[116 = \dfrac{n}{2}\left[ {2 \times 25 + \left( {n - 1} \right)\left( { - 3} \right)} \right]\]
\[ \Rightarrow 116 = \dfrac{n}{2}\left[ {50 + \left( {n - 1} \right)\left( { - 3} \right)} \right]\]
\[ \Rightarrow 116 \times 2 = n\left[ {50 + \left( {n - 1} \right)\left( { - 3} \right)} \right]\]
\[ \Rightarrow 232 = n\left[ {50 - 3n + 3} \right]\]
\[ \Rightarrow 232 = n\left[ {53 - 3n} \right]\]
\[ \Rightarrow 232 = 53n - 3{n^2}\]
\[ \Rightarrow 3{n^2} - 53n + 232 = 0\]
Solving quadratic equation using middle term splitting method, we have
\[3{n^2} - 24n - 29n + 232 = 0\]
\[ \Rightarrow 3n\left( {n - 8} \right) - 29\left( {n - 8} \right) = 0\]
\[ \Rightarrow \left( {3n - 29} \right)\left( {n - 8} \right) = 0\]
Either
\[\left( {3n - 29} \right) = 0 \Rightarrow 3n = 29 \Rightarrow n = \dfrac{{29}}{3}\]
Or
\[\left( {n - 8} \right) = 0 \Rightarrow n - 8 = 0 \Rightarrow n = 8\]
\[n = \dfrac{{29}}{3}\] is not possible as \[n\] cannot be a fraction
So, the possible value of n is 8.
To find last term, we use the formula
an = a + (n – 1) × d
Here, \[{a_n} = \] last term, \[a = 25,\]\[n = 8,\]\[d = - 3\]
\[{a_n} = 25 + \left( {8 - 1} \right)\left( { - 3} \right)\]\[ = 25 + \left( 7 \right)\left( { - 3} \right)\]
\[ = 25 - 21\]\[ = 4\]
Hence, last term is \[4\].

Therefore, the number of terms needed to give the sum of 116 is 8 and the last term is 4.

Note:
 In these types of questions, always be careful while choosing the AP formula. We have many formulas of AP like sum of terms, nth term and relation between sum of n terms and nth term. First check what is given and what to find, because using one formula only one unknown thing we can get.