Question

The ${{7}^{th}}$ term of an A.P is $32$ and its ${{13}^{th}}$ term is $62$. Find the A.P.

Answer 1

Hint: Use the ${{n}^{th}}$ term of A.P and solve it. You will get two equations. Subtract the equations you will get the value of $d$ and then find the value of $a$ by substituting $d$. You will get a series which is an A.P.

Arithmetic Progression (A.P) is a sequence of numbers in a particular order. If we observe in our regular lives, we come across progression quite often. For example, roll numbers of a class, days in a week, or months in a year This pattern of series and sequences has been generalized in Mathematics as progressions. Let us learn here AP definition, important terms such as common difference, the first term of the series, nth term and sum of nth term formulas along with solved questions based on them.
It is a mathematical sequence in which the difference between two consecutive terms is always a constant and it is abbreviated as A.P.
The fixed number that must be added to any term of an A.P to get the next term is known as the common difference of the A.P.
An arithmetic sequence or progression is defined as a sequence of numbers in which for every pair of consecutive terms, the second number is obtained by adding a fixed number to the first one.
${{n}^{th}}$term of A.P, ${{a}_{n}}=a+(n-1)d$
Where,
$a=$ First-term
$d=$ Common difference
$n=$ number of terms
${{a}_{n}}={{n}^{th}}$ term
So we have given in the question that ${{7}^{th}}$ term of an A.P is $32$.
So we get, ${{a}_{7}}=32$
${{a}_{7}}=a+(7-1)d$
$32=a+6d$ .…… (1)
Now we have been given that the ${{13}^{th}}$ term is $62$.
${{a}_{13}}=62$,
${{a}_{13}}=a+(13-1)d$
$62=a+12d$ …… (2)
So subtracting (1) from (2), we get,
$62-32=(a+12d)-(a+6d)$
So simplifying in a simple manner we get,
$30=d6$
So we get $d=5$.
Now substituting $d=5$ in (1).
$32=a+6(5)$
$32=a+30$
$a=2$
So we get,
First-term $=a=2$ and common difference $=d=5$.
So the series forming an A.P is $2,7,12,17,.......$

Note: Carefully read the question. Properly use the term of A.P sometimes mistakes occur while solving or substituting the values. Be careful while subtracting no term should be missing.