Question

The sum of 5th and 9th terms of an A.P. is 72 and the sum of 7th and 12th terms is 97. Find the
(a)25th term of an AP
(b)Sum of first 50 terms of the AP

Answer 1

Here we will first assume the assume of the first term and the common difference between the terms to be any variable and then we will find the 5th and 9th term of an AP using its properties and then we will add both the terms and we will equate it with the given sum. Similarly, we will find the 7th and 12th term of an AP using its properties and then we will add both the terms and we will equate it with the given sum. From there, we will get two equations and after simplifying these two equations, we will get the value of the first term and the common difference. Using these values, we will find the 25th term and sum of the first fifty terms of an AP.

Complete step-by-step answer:
Let the first term of an AP be $a$ and common difference be $d$.
We know that ${{n}^{th}}$ term of an AP is given by:-
${{T}_{n}}=a+(n-1)d$ ……… $\left( 1 \right)$
We will find the 5th term of an AP by substituting the value of $n$ as 5 in equation 1.
Therefore,
$\Rightarrow {{T}_{5}}=a+(5-1)d$
On further simplification, we get
$\Rightarrow {{T}_{5}}=a+4d$ …….. $\left( 2 \right)$
We will find the 9th term of an AP by substituting the value of $n$ as 9 in equation 1.
Therefore,
$\Rightarrow {{T}_{9}}=a+(9-1)d$
On further simplification, we get
$\Rightarrow {{T}_{9}}=a+8d$ …….. $\left( 3 \right)$
Now, we will add equation 2 and equation 3.
$\Rightarrow {{T}_{5}}+{{T}_{9}}=a+4d+a+8d$
On adding the like terms, we get
$\Rightarrow {{T}_{5}}+{{T}_{9}}=2a+12d$
It is given that the sum of 5th and 9th term of an AP is equal to 72.
Therefore,
$\Rightarrow 72=2a+12d$ ……… $\left( 4 \right)$
We will find the 7th term of an AP by substituting the value of $n$ as 7 in equation 1.
Therefore,
$\Rightarrow {{T}_{7}}=a+(7-1)d$
On further simplification, we get
$\Rightarrow {{T}_{7}}=a+6d$ …….. $\left( 5 \right)$
We will find the 12th term of an AP by substituting the value of $n$ as 12 in equation 1.
Therefore,
$\Rightarrow {{T}_{12}}=a+(12-1)d$
On further simplification, we get
$\Rightarrow {{T}_{12}}=a+11d$ …….. $\left( 6 \right)$
Now, we will add equation 5 and equation 6.
$\Rightarrow {{T}_{7}}+{{T}_{12}}=a+6d+a+11d$
On adding the like terms, we get
$\Rightarrow {{T}_{7}}+{{T}_{12}}=2a+17d$
It is given that the sum of 7th and 12th term of an AP is equal to 97.
Therefore,
$\Rightarrow 97=2a+17d$ ……… $\left( 7 \right)$
Now, we will subtract equation 4 from equation 7.
$\Rightarrow 97-72=2a+17d-2a-12d$
On subtracting the like terms, we get
$\Rightarrow 25=5d$
Dividing both sides by 5, we get

& \Rightarrow \dfrac{25}{5}=\dfrac{5d}{5} \\\ & \Rightarrow d=5 \\\ \end{aligned}$$ Substituting value of $d$in equation 7, we get $$\Rightarrow 97=2a+17\times 5$$ On multiplying the terms, we get $$\Rightarrow 97=2a+85$$ Subtracting 85 on both sides, we get $$\begin{aligned} & \Rightarrow 97-85=2a+85-85 \\\ & \Rightarrow 12=2a \\\ \end{aligned}$$ Dividing on both sides by 2, we get $$\begin{aligned} & \Rightarrow \dfrac{12}{2}=\dfrac{2a}{2} \\\ & \Rightarrow a=6 \\\ \end{aligned}$$ Now, we will find 25th term of an AP by putting value of $a$, $d$ and $n$ in equation 1. $$\Rightarrow {{T}_{25}}=6+(25-1)\times 5$$ On simplifying the terms, we get $$\begin{aligned} & \Rightarrow {{T}_{25}}=6+24\times 5 \\\ & \Rightarrow {{T}_{25}}=6+120=126 \\\ \end{aligned}$$ Thus, 25th term of an AP is 126. We know the formula of sum of $n$ terms of an AP is $${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$$ We will sum of first 50 terms of an AP by putting the value of $a$, $d$ and $n$ in formula of sum. $$\Rightarrow {{S}_{50}}=\dfrac{50}{2}\left[ 2\times 6+(50-1)\times 5 \right]$$ On simplifying the terms, we get $$\Rightarrow {{S}_{50}}=25\left( 12+49\times 5 \right)$$ On further simplification, we get $$\Rightarrow {{S}_{50}}=25\left( 12+245 \right)=25\times 257=6425$$ **Therefore, the sum of the first 50 terms of an AP is equal to 6425.** **Note:** We have obtained the sum of an arithmetic progression. An arithmetic progression is defined as a sequence in which the difference between the term and the preceding term is constant or in other words, we can say that an arithmetic progression is a sequence such that every element after the first is obtained by adding a constant term to the preceding element.