Solveeit Logo

Question

Question: The \(26{\text{th}}\), \(11{\text{th}}\) and the last term of an A.P. are \(0\), \[3\] and \(\dfrac{...

The 26th26{\text{th}}, 11th11{\text{th}} and the last term of an A.P. are 00, 33 and 15\dfrac{{ - 1}}{5} respectively. Find the common difference and number of terms.

Explanation

Solution

In this question, we will use the nthn{\text{th}} term formula to find the first term and common difference of A.P. then substitute the obtained values in the equation formed for the last term to find the numbers of terms.

Formula Used: We know that the formula for nthn{\text{th}} term formula of an A.P. series is an=a+(n1)d{a_n} = a + \left( {n - 1} \right)d. Here, an{a_n} is the nthn{\text{th}} term, aa is the first term, dd is the common difference and nn is the number of terms.

Complete step-by-step answer:
We know that 26th26{\text{th}} term of the given A.P. is 00. Substitute the values in the nthn{\text{th}} term formula for the 26th26{\text{th}} term.
\Rightarrow 0=a+(261)d0 = a + \left( {26 - 1} \right)d
\Rightarrow 0=a+25d    ...(1)0 = a + 25d\;\;...\left( 1 \right)
We know that 11th11{\text{th}} term of the given A.P. is 33. Substitute the values in the nthn{\text{th}} term formula for the 11th11{\text{th}} term.
\Rightarrow 3=a+(111)d3 = a + \left( {11 - 1} \right)d
\Rightarrow 3=a+10d    ...(2)3 = a + 10d\;\;...\left( 2 \right)
Subtract equation (2)\left( 2 \right) from equation (1)\left( 1 \right) and solve.
\Rightarrow 3=a+25da10d - 3 = a + 25d - a - 10d
\Rightarrow 3=15d - 3 = 15d
\Rightarrow 315=d\dfrac{{ - 3}}{{15}} = d
\Rightarrow 15=d\dfrac{{ - 1}}{5} = d
Now, substitute the common difference in equation (1)\left( 1 \right).
\Rightarrow 0=a+25(15)0 = a + 25\left( {\dfrac{{ - 1}}{5}} \right)
\Rightarrow 0=a50 = a - 5
\Rightarrow a=5a = 5
We know that the last term of A.P. is 15\dfrac{{ - 1}}{5}. Substitute all the values in nthn{\text{th}} term formula of the A.P. series and calculate the number of terms.
15=5+(n1)(15)\dfrac{{ - 1}}{5} = 5 + \left( {n - 1} \right)\left( {\dfrac{{ - 1}}{5}} \right)
1=25+(n1)(1)\Rightarrow - 1 = 25 + \left( {n - 1} \right)\left( { - 1} \right)
1=25n+1\Rightarrow - 1 = 25 - n + 1
n=27\Rightarrow n = 27
Therefore, the common difference is 15\dfrac{{ - 1}}{5} and the number of terms of A.P. is 2727.

Note: In this question, we need to find the values of first term and common difference. Put these values in the equation obtained by the nthn{\text{th}} term formula of A.P for the last term of A.P. to find the number of terms.