Question

The $6^{th}$ term of AP is zero. Prove that its $31^{st}$ term is 5 times the $11^{th}$ term.

Answer 1

Hint: We solve this problem by first equating the $6^{th}$ term to zero and using the formula for the ${n^{th}}$ term of an A.P, $a + \left( {n - 1} \right)d$. Then we get a relation between a and d. . Then we use the same formula and find the values of the $31^{st}$ term and $11^{th}$ term and then use the relation obtained between a and d to find their values in a single variable. Then multiply the $11^{th}$ term by 5 and compare with $31^{st}$ term.

Complete step-by-step answer:
First let us consider the formula for the ${n^{th}}$ term of an A.P with the first term a and with common difference d.
${a_n} = a + \left( {n - 1} \right)d$
We are given that the 6th term of an A.P. is zero, which is ${a_6} = 0$.
$ \Rightarrow a + \left( {6 - 1} \right)d = 0$
Subtract the value,
$ \Rightarrow a + 5d = 0$
Move 5d to the right side,
$a = - 5d$..........….. (1)
First, find the value of the $31^{st}$ term of the A.P. using the formula,
$ \Rightarrow {a_{31}} = a + \left( {31 - 1} \right)d$
Subtract the value,
$ \Rightarrow {a_{31}} = a + 30d$
Substitute the value of a from equation (1),
$ \Rightarrow {a_{31}} = - 5d + 30d$
Simplify the term,
$ \Rightarrow {a_{31}} = 25d$...............….. (2)
Now, find the value of the $11^{th}$ term of the A.P. using the formula,
$ \Rightarrow {a_{11}} = a + \left( {11 - 1} \right)d$
Subtract the value,
$ \Rightarrow {a_{11}} = a + 10d$
Substitute the value of a from equation (1),
$ \Rightarrow {a_{11}} = - 5d + 10d$
Simplify the term,
$ \Rightarrow {a_{11}} = 5d$
Now multiply ${a_{11}}$ by 5,
$ \Rightarrow 5{a_{11}} = 5 \times 5d$
Multiply the terms,
$ \Rightarrow 5{a_{11}} = 25d$...........….. (3)
Compare both equation (2) and (3),
$\therefore {a_{31}} = 5{a_{11}}$
Hence, it is proved.

Note: There is a possibility of one making a mistake while solving this problem by taking the formula for the ${n^{th}}$ term as ${a_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$. But it is the formula for the sum of first $n$ terms of an A.P. not for the ${n^{th}}$ term of the A.P.