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Question: If \[{(m + 1)^{th}}\] term of an A.P is twice the \[{(n + 1)^{th}}\] term, then show that \[{(3m + 1...

If (m+1)th{(m + 1)^{th}} term of an A.P is twice the (n+1)th{(n + 1)^{th}} term, then show that (3m+1)th{(3m + 1)^{th}} term is twice the (m+n+1)th{(m + n + 1)^{th}} term.

Explanation

Solution

We assume the general arithmetic progression having first term as ‘a’ and common difference as ‘d’. Using the formula of the nth term of an A.P we write the values of all the required terms. Form equations depicting the relation between the terms.

  • An arithmetic progression is a sequence of terms having common differences between them. If ‘a’ is the first term of an AP, ‘d’ is the common difference, then the nth term of an AP can be found as an=a+(n−1)d{a_n} = a + (n - 1)d .

Complete step-by-step answer:
Let us assume an arithmetic progression having first term ‘a’ and common difference ‘d’.
We write the value of (m+1)th{(m + 1)^{th}} term
⇒am+1=a+(m+1−1)d\Rightarrow {a_{m + 1}} = a + (m + 1 - 1)d
⇒am+1=a+md\Rightarrow {a_{m + 1}} = a + md … (1)
Now we calculate the value of (n+1)th{(n + 1)^{th}}term
⇒an+1=a+(n+1−1)d\Rightarrow {a_{n + 1}} = a + (n + 1 - 1)d
⇒an+1=a+nd\Rightarrow {a_{n + 1}} = a + nd … (2)
We are given that (m+1)th{(m + 1)^{th}}term of an A.P is twice the (n+1)th{(n + 1)^{th}}term
⇒a+md=2(a+nd)\Rightarrow a + md = 2(a + nd)
⇒a+md=2a+2nd\Rightarrow a + md = 2a + 2nd
Bring all values with variable a to one side of the equation
⇒2a−a=md−2nd\Rightarrow 2a - a = md - 2nd
⇒a=(m−2n)d\Rightarrow a = (m - 2n)d … (3)
We have to show that (3m+1)th{(3m + 1)^{th}}term is twice the (m+n+1)th{(m + n + 1)^{th}}term
We calculate the value of (3m+1)th{(3m + 1)^{th}} term
⇒a3m+1=a+(3m+1−1)d\Rightarrow {a_{3m + 1}} = a + (3m + 1 - 1)d
⇒a3m+1=a+3md\Rightarrow {a_{3m + 1}} = a + 3md … (4)
Substitute the value of ‘a’ from equation (3)
⇒a3m+1=md−2nd+3md\Rightarrow {a_{3m + 1}} = md - 2nd + 3md
⇒a3m+1=4md−2nd\Rightarrow {a_{3m + 1}} = 4md - 2nd
⇒a3m+1=2(2md−nd)\Rightarrow {a_{3m + 1}} = 2(2md - nd) … (5)
Also we have to calculate the value of (m+n+1)th{(m + n + 1)^{th}}term
⇒am+n+1=a+(m+n+1−1)d\Rightarrow {a_{m + n + 1}} = a + (m + n + 1 - 1)d
⇒am+n+1=a+(m+n)d\Rightarrow {a_{m + n + 1}} = a + (m + n)d
Substitute the value of ‘a’ from equation (3)
⇒am+n+1=md−2nd+md+nd\Rightarrow {a_{m + n + 1}} = md - 2nd + md + nd
⇒am+n+1=2md−nd\Rightarrow {a_{m + n + 1}} = 2md - nd … (6)
Substitute the value of (2md−nd)(2md - nd) from equation (6) in equation (5)
⇒a3m+1=2am+n+1\Rightarrow {a_{3m + 1}} = 2{a_{m + n + 1}}

Hence proved

Note:
Many students make mistake of writing the value of the terms wrong as they only change the subscript from n to m but keep in mind we have to change the complete value that exists in subscript, so if subscript has m+1m + 1 in place of n then the value of n is replaced by m+1m + 1 even in the formula.