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Question: Write the expression \({a_n} - {a_k}\) for the A.P. a+d ,a+2d,…… and find the common difference of t...

Write the expression an−ak{a_n} - {a_k} for the A.P. a+d ,a+2d,…… and find the common difference of the AP for which
1.11th terms is 5 and 13th terms is 79
2.20 th term is 10 more than the 18th term

Explanation

Solution

We are given a AP with the first term a+d and difference d . We need to find d value using the formula am=a1+(m−1)d{a_m} = {a_1} + (m - 1)d to find the expression an−ak{a_n} - {a_k}. And in (i) we are given the 11th term is 5 and the 13th term is 79 . Using this we get two linear equation in two variables and solving that gives the value of d. and in (ii) it is said that the 20th term is 10 more than 18th term and we get a20−a18=10{a_{20}} - {a_{18}} = 10
And using the expression an−ak=(n−k)d{a_n} - {a_k} = (n - k)d we can find the value of d.

Complete step-by-step answer:
We are given AP a+d , a+2d ,……..
Here in this AP the first term a1=a+d{a_1} = a + d and the common difference
⇒d=a+2d−(a+d) ⇒d=a+2d−a−d ⇒d=2d−d=d  \Rightarrow d = a + 2d - (a + d) \\\ \Rightarrow d = a + 2d - a - d \\\ \Rightarrow d = 2d - d = d \\\
We are asked to find the expression an−ak{a_n} - {a_k}
Firstly , lets find .an,ak{a_n},{a_k}.using the formula,
⇒am=a1+(m−1)d\Rightarrow {a_m} = {a_1} + (m - 1)d
So ,
⇒an=a+d+(n−1)d ⇒an=a+d+nd−d ⇒an=a+nd  \Rightarrow {a_n} = a + d + (n - 1)d \\\ \Rightarrow {a_n} = a + d + nd - d \\\ \Rightarrow {a_n} = a + nd \\\
And
⇒ak=a+d+(k−1)d ⇒ak=a+d+kd−d ⇒ak=a+kd  \Rightarrow {a_k} = a + d + (k - 1)d \\\ \Rightarrow {a_k} = a + d + kd - d \\\ \Rightarrow {a_k} = a + kd \\\
Now ,
⇒an−ak=a+nd−(a+kd) ⇒an−ak=a+nd−a−kd=(n−k)d  \Rightarrow {a_n} - {a_k} = a + nd - (a + kd) \\\ \Rightarrow {a_n} - {a_k} = a + nd - a - kd = (n - k)d \\\
Then we are asked to find the common difference of a AP under two conditions
11th terms is 5 and 13th terms is 79
We are given that the 11 th term is 5
⇒a11=a+d+(11−1)d ⇒5=a+d+10d ⇒5=a+11d  \Rightarrow {a_{11}} = a + d + (11 - 1)d \\\ \Rightarrow 5 = a + d + 10d \\\ \Rightarrow 5 = a + 11d \\\
Let this be equation 1
We are given that the 13th term is 79
⇒a13=a+d+(13−1)d ⇒79=a+d+12d ⇒79=a+13d  \Rightarrow {a_{13}} = a + d + (13 - 1)d \\\ \Rightarrow 79 = a + d + 12d \\\ \Rightarrow 79 = a + 13d \\\
Let this be equation 2
Solving equation 1 and 2 by elimination method
{\text{ }}a + 11d = 5 \\\ \- {\text{ }} - {\text{ }} - \\\ {\text{ }}a + 13d = 79 \\\ \\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_ \\\ 0 - 2d = - 74 \\\ \\\
The common difference of the AP under this condition is 37
20 th term is 10 more than the 18th term
We are given that 20th term is 10 more than 18th term
⇒a20=10+a18   \Rightarrow {a_{20}} = 10 + {a_{18}} \\\ \\\
From this we can write
.⇒a20−a18=10   \Rightarrow {a_{20}} - {a_{18}} = 10 \\\ \\\ .………..(3)
We already known that an−ak=(n−k)d{a_n} - {a_k} = (n - k)d
Using this we can write a20−a18=(20−18)d=2d{a_{20}} - {a_{18}} = (20 - 18)d = 2d………..(4)
Now equating the right hand side of (3) and (4)
⇒2d=10 ⇒d=102=5  \Rightarrow 2d = 10 \\\ \Rightarrow d = \dfrac{{10}}{2} = 5 \\\
Therefore the common difference of the AP under this condition is 5.

Note: A sequence of numbers is called an Arithmetic progression if the difference between any two consecutive terms is always the same. In simple terms, it means that the next number in the series is calculated by adding a fixed number to the previous number in the series.