The sum of second term and the seventh term of an A.P. is 30. If its fifteenth term is 1 less than twice its eighth term, find the A.P.
Answer
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Hint: We will first write the formula for ${n^{th}}$ term of an A.P. Then apply the given condition which will help us form two different equations and we have two variables as well a and d. So, we can solve the equations for a and d and get the answer.
Complete step-by-step answer:
Let us first of all write the formula for ${n^{th}}$ term of an A.P.
$ \Rightarrow {a_n} = a + (n - 1)d$, where a is the first term of the A.P, d is the common difference and ${a_n}$ is the ${n^{th}}$ term of A.P.
We are given that: sum of second term and the seventh term of an A.P. is 30.
We can write this as:- ${a_2} + {a_7} = 30$.
Applying the formula for ${a_n}$ that is ${a_n} = a + (n - 1)d$ in ${a_2}$ and ${a_7}$. So, we will have:-
$ \Rightarrow (a + d) + (a + 6d) = 30$
Simplifying this by combining the like terms, we will get:-
$ \Rightarrow 2a + 7d = 30$ …………….(1)
We also are given that: fifteenth term is 1 less than twice its eighth term.
We can rewrite it as: ${a_{15}} = 2{a_8} - 1$.
Applying the formula for ${a_n}$ that is ${a_n} = a + (n - 1)d$ in ${a_8}$ and ${a_{15}}$. So, we will have:-
$ \Rightarrow a + 14d = 2(a + 7d) - 1$
Simplifying this by opening the brackets, we will get:-
$ \Rightarrow a + 14d = 2a + 14d - 1$
Simplifying it by clubbing the like terms, we will get:-
$ \Rightarrow - a = - 1$
Multiplying by -1 on both the sides, we will get:-
$ \Rightarrow a = 1$.
Hence, the first term of the A.P. is 1.
Putting this value of a in (1), we will get:-
$ \Rightarrow 2 + 7d = 30$
Taking the 2 from LHS to RHS, we will get:-
$ \Rightarrow 7d = 28$
Taking the 7 from LHS to RHS, we will get:-
$ \Rightarrow d = \dfrac{{28}}{7} = 4$
Hence, the common difference is 4.
Since an AP is given by a, a + d, a + 2d, a + 3d, a + 4d, ………..
Hence, the A.P. will be:- 1, 5, 9, 13, 17 ……..
Note: The students must remember that they need to have as much equation as the number of unknown variables to find their value.
Fun Fact:- If a, b, c are in AP, then 2b = a + c.
If each term of an AP is increased, decreased, multiplied, or divided by a constant non-zero number, then the resulting sequence is also in AP.
Complete step-by-step answer:
Let us first of all write the formula for ${n^{th}}$ term of an A.P.
$ \Rightarrow {a_n} = a + (n - 1)d$, where a is the first term of the A.P, d is the common difference and ${a_n}$ is the ${n^{th}}$ term of A.P.
We are given that: sum of second term and the seventh term of an A.P. is 30.
We can write this as:- ${a_2} + {a_7} = 30$.
Applying the formula for ${a_n}$ that is ${a_n} = a + (n - 1)d$ in ${a_2}$ and ${a_7}$. So, we will have:-
$ \Rightarrow (a + d) + (a + 6d) = 30$
Simplifying this by combining the like terms, we will get:-
$ \Rightarrow 2a + 7d = 30$ …………….(1)
We also are given that: fifteenth term is 1 less than twice its eighth term.
We can rewrite it as: ${a_{15}} = 2{a_8} - 1$.
Applying the formula for ${a_n}$ that is ${a_n} = a + (n - 1)d$ in ${a_8}$ and ${a_{15}}$. So, we will have:-
$ \Rightarrow a + 14d = 2(a + 7d) - 1$
Simplifying this by opening the brackets, we will get:-
$ \Rightarrow a + 14d = 2a + 14d - 1$
Simplifying it by clubbing the like terms, we will get:-
$ \Rightarrow - a = - 1$
Multiplying by -1 on both the sides, we will get:-
$ \Rightarrow a = 1$.
Hence, the first term of the A.P. is 1.
Putting this value of a in (1), we will get:-
$ \Rightarrow 2 + 7d = 30$
Taking the 2 from LHS to RHS, we will get:-
$ \Rightarrow 7d = 28$
Taking the 7 from LHS to RHS, we will get:-
$ \Rightarrow d = \dfrac{{28}}{7} = 4$
Hence, the common difference is 4.
Since an AP is given by a, a + d, a + 2d, a + 3d, a + 4d, ………..
Hence, the A.P. will be:- 1, 5, 9, 13, 17 ……..
Note: The students must remember that they need to have as much equation as the number of unknown variables to find their value.
Fun Fact:- If a, b, c are in AP, then 2b = a + c.
If each term of an AP is increased, decreased, multiplied, or divided by a constant non-zero number, then the resulting sequence is also in AP.
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