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Question: The \(6^{th}\) term of AP is zero. Prove that its \(31^{st}\) term is 5 times the \(11^{th}\) term....

The 6th6^{th} term of AP is zero. Prove that its 31st31^{st} term is 5 times the 11th11^{th} term.

Explanation

Solution

We solve this problem by first equating the 6th6^{th} term to zero and using the formula for the nth{n^{th}} term of an A.P, a+(n1)da + \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 31st31^{st} term and 11th11^{th} term and then use the relation obtained between a and d to find their values in a single variable. Then multiply the 11th11^{th} term by 5 and compare with 31st31^{st} term.

Complete step-by-step answer:
First let us consider the formula for the nth{n^{th}} term of an A.P with the first term a and with common difference d.
an=a+(n1)d{a_n} = a + \left( {n - 1} \right)d
We are given that the 6th term of an A.P. is zero, which is a6=0{a_6} = 0.
a+(61)d=0\Rightarrow a + \left( {6 - 1} \right)d = 0
Subtract the value,
a+5d=0\Rightarrow a + 5d = 0
Move 5d to the right side,
a=5da = - 5d..........….. (1)
First, find the value of the 31st31^{st} term of the A.P. using the formula,
a31=a+(311)d\Rightarrow {a_{31}} = a + \left( {31 - 1} \right)d
Subtract the value,
a31=a+30d\Rightarrow {a_{31}} = a + 30d
Substitute the value of a from equation (1),
a31=5d+30d\Rightarrow {a_{31}} = - 5d + 30d
Simplify the term,
a31=25d\Rightarrow {a_{31}} = 25d...............….. (2)
Now, find the value of the 11th11^{th} term of the A.P. using the formula,
a11=a+(111)d\Rightarrow {a_{11}} = a + \left( {11 - 1} \right)d
Subtract the value,
a11=a+10d\Rightarrow {a_{11}} = a + 10d
Substitute the value of a from equation (1),
a11=5d+10d\Rightarrow {a_{11}} = - 5d + 10d
Simplify the term,
a11=5d\Rightarrow {a_{11}} = 5d
Now multiply a11{a_{11}} by 5,
5a11=5×5d\Rightarrow 5{a_{11}} = 5 \times 5d
Multiply the terms,
5a11=25d\Rightarrow 5{a_{11}} = 25d...........….. (3)
Compare both equation (2) and (3),
a31=5a11\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 nth{n^{th}} term as an=n2[2a+(n1)d]{a_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]. But it is the formula for the sum of first nn terms of an A.P. not for the nth{n^{th}} term of the A.P.