Question

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

Answer 1

Hint: In this question, first write down the formula to find (m+1) th term of the A.P. assuming first term as ‘a’ and common difference as ‘d’. Similarly write the (n+1) th term. After this form the equation using given information. This will give a result. Use this result to derive the required result.

Complete step-by-step answer:

We know that the nth term of a A.P. is given as:
${{\text{T}}_n}$ = a + (n-1)d; where
‘a’ is the first term of the A.P.
‘d’ is the common difference.
‘n’ is the nth term of A.P.

Now, Putting the value of n = m+1 in above equation, we get:
${{\text{T}}_{m + 1}}$ = a+ ((m+1)-1)d

Similarly, we get:
${{\text{T}}_{n + 1}}$ = a+ ((n+1)-1)d
In the question, it is given that ${{\text{T}}_{m + 1}}$ = 2${{\text{T}}_{n + 1}}$.

So, we can write:
a+ ((m+1)-1)d = 2(a+ ((n+1)-1)d)
$\Rightarrow$ 2a – a = d(m - 2n)
$\Rightarrow$ a = d(m - 2n). (1)

Now we will calculate the (3m+1) th term and (m+n+1) th of the A.P.
${{\text{T}}_{3m + 1}}$ = a+ ((3m+1)-1)d (2)

And ${{\text{T}}_{(m + n + 1)}}$ = a+ ((m+n+1)-1)d (3)

Putting the value of ‘a’ in equation 2, we get:
${{\text{T}}_{3m + 1}}$= d(m - 2n) + ((3m+1) - 1)d = d(4m –2n) =2d(2m – n) (4)

Similarly, Putting the value of ‘a’ in equation 3, we get:
${{\text{T}}_{m + n + 1}}$= d(m - 2n) + ((m+n+1)-1)d = d(2m –n). (5)

From equation 4 and equation, we get:
${{\text{T}}_{3m + 1}}$= 2d(2m – n) = 2$\times$ ${{\text{T}}_{m + n + 1}}$.

Hence, it is proved that ${{\text{T}}_{3m + 1}}$= 2$\times$ ${{\text{T}}_{m + n + 1}}$.

Note: In this type of question, it is important to remember the formula for the nth term of an A.P. Then you should read the information given in the question carefully and accordingly form the equation. Here we have use a = d(m - 2n) to reduce the other terms, you can also use d = $\dfrac{{\text{a}}}{{({\text{m - 2n)}}}}$ to reduce the later terms but it will a little more time consuming.