Question

If the sum of a certain number of terms of the AP 25, 22, 19, …… is 116. Find the last term.

Answer 1

Hint: Use formula of finding sum to find n, Sum = \[\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]. Here, n is terms, a is first term and d is common difference then use the formula, \[{{T}_{n}}=a+\left( n-1 \right)d\], to find the last term.

Complete step-by-step answer:
The sequence given is in the form of AP.
At first we briefly understood what is arithmetic progression.
In mathematics, an arithmetic progression (AP) or an arithmetic sequence of numbers such that the difference between the consecutive terms is constant. Here, difference means the second minus first. For instance, the sequence 5, 7, 9, 11, 13, 15 …….. is an arithmetic progression with a common difference of 2.
If the initial term of an arithmetic is considered as a and common difference in successive terms is d, then the \[{{n}^{th}}\] term of a sequence \[\left( {{T}_{n}} \right)\] is denoted by :
\[{{T}_{n}}=a+\left( n-1 \right)d\]
And in general, \[{{T}_{n}}={{T}_{m}}+\left( n-m \right)d\].
A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes called an arithmetic progression. The sum of finite arithmetic progression is called an arithmetic series.
Let’s first analyze the sequence, its common difference is (22 - 25) which is (- 3), its first term is 25 and let’s suppose it number of terms be n. It’s sum is also given 116.
So, at first we will write formula which is to find sum if a is first term d is difference and n is number of terms,
Sum = \[\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]
Here, n will remain n, d will be (- 3), a will be 25 and sum will be 116. So, on substituting we get,
\[116=\dfrac{n}{2}\left[ 50+\left( n-1 \right)\left( -3 \right) \right]\]
Here, on cross multiplication we get,
\[232=n\left[ 50-3n+3 \right]\]
Or
\[n\left( 53-3n \right)=232\]
Now, on simplification and taking all terms on one side we get,
\[53n-3{{n}^{2}}=232\]
Or, \[3{{n}^{2}}-53n+232=0\]
Or, \[3{{n}^{2}}-24n-29n+232=0\]
Hence, on factorizing we get,
\[3n\left( n-8 \right)-29\left( n-8 \right)=0\]
So, \[\left( 3n-29 \right)\left( n-8 \right)=0\]
Hence, \[n=\dfrac{29}{3}\] or 8
As n represents the number of terms so it can’t be a fraction hence 8 is the number of terms.
So, the \[{{8}^{th}}\] term will be the last term.
\[{{T}_{8}}=a+\left( 8-1 \right)d\]
Here, a = 25 and d = - 3
So, \[{{T}_{8}}=25+7\left( -3 \right)\]
\[{{T}_{8}}=25-21=4\]
Hence, last term is 4.

Note: We can also do it in another way. Let the last term be \[{{n}^{th}}\] term, then use formula for \[{{T}_{n}}\] which is \[a+\left( n-1 \right)d\], where a is first term and d is common difference and n represents number of terms. After that use the formula for sum which is \[\dfrac{n}{2}\left( a+l \right)\], where n is number of terms, a is first term and l is the \[{{T}_{n}}\] value and then equate it with 116.