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

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

Explanation

Solution

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 = a = 25
And Common difference = d = second term – first term =2225=3 = 22 - 25 = - 3
Given, Sum of n terms =Sn=116 = {S_n} = 116
We need to find last term ana{}_n
First, we have to find,nn
We know that, sum of first n terms of an AP is given by
Sn=n2[2a+(n1)d]{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]
Putting values of a, d and Sn
116=n2[2×25+(n1)(3)]116 = \dfrac{n}{2}\left[ {2 \times 25 + \left( {n - 1} \right)\left( { - 3} \right)} \right]
116=n2[50+(n1)(3)]\Rightarrow 116 = \dfrac{n}{2}\left[ {50 + \left( {n - 1} \right)\left( { - 3} \right)} \right]
116×2=n[50+(n1)(3)]\Rightarrow 116 \times 2 = n\left[ {50 + \left( {n - 1} \right)\left( { - 3} \right)} \right]
232=n[503n+3]\Rightarrow 232 = n\left[ {50 - 3n + 3} \right]
232=n[533n]\Rightarrow 232 = n\left[ {53 - 3n} \right]
232=53n3n2\Rightarrow 232 = 53n - 3{n^2}
3n253n+232=0\Rightarrow 3{n^2} - 53n + 232 = 0
Solving quadratic equation using middle term splitting method, we have
3n224n29n+232=03{n^2} - 24n - 29n + 232 = 0
3n(n8)29(n8)=0\Rightarrow 3n\left( {n - 8} \right) - 29\left( {n - 8} \right) = 0
(3n29)(n8)=0\Rightarrow \left( {3n - 29} \right)\left( {n - 8} \right) = 0
Either
(3n29)=03n=29n=293\left( {3n - 29} \right) = 0 \Rightarrow 3n = 29 \Rightarrow n = \dfrac{{29}}{3}
Or
(n8)=0n8=0n=8\left( {n - 8} \right) = 0 \Rightarrow n - 8 = 0 \Rightarrow n = 8
n=293n = \dfrac{{29}}{3} is not possible as nn 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, an={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 44.

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.