Question
Question: The \(6^{th}\) term of AP is zero. Prove that its \(31^{st}\) term is 5 times the \(11^{th}\) term....
The 6th term of AP is zero. Prove that its 31st term is 5 times the 11th term.
Solution
We solve this problem by first equating the 6th term to zero and using the formula for the nth term of an A.P, a+(n−1)d. Then we get a relation between a and d. . Then we use the same formula and find the values of the 31st term and 11th term and then use the relation obtained between a and d to find their values in a single variable. Then multiply the 11th term by 5 and compare with 31st term.
Complete step-by-step answer:
First let us consider the formula for the nth term of an A.P with the first term a and with common difference d.
an=a+(n−1)d
We are given that the 6th term of an A.P. is zero, which is a6=0.
⇒a+(6−1)d=0
Subtract the value,
⇒a+5d=0
Move 5d to the right side,
a=−5d..........….. (1)
First, find the value of the 31st term of the A.P. using the formula,
⇒a31=a+(31−1)d
Subtract the value,
⇒a31=a+30d
Substitute the value of a from equation (1),
⇒a31=−5d+30d
Simplify the term,
⇒a31=25d...............….. (2)
Now, find the value of the 11th term of the A.P. using the formula,
⇒a11=a+(11−1)d
Subtract the value,
⇒a11=a+10d
Substitute the value of a from equation (1),
⇒a11=−5d+10d
Simplify the term,
⇒a11=5d
Now multiply a11 by 5,
⇒5a11=5×5d
Multiply the terms,
⇒5a11=25d...........….. (3)
Compare both equation (2) and (3),
∴a31=5a11
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 term as an=2n[2a+(n−1)d]. But it is the formula for the sum of first n terms of an A.P. not for the nth term of the A.P.