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Home » 7. If term of an A.P. is zero, prove its term is double the term.
7. If {{9}^{th}} term of an A.P. is zero, prove its {{29}^{th}} term is double the {{19}^{th}} term.
Arithmetic Progression | Class 10 | ICSE | Maths | RD Sharma
7. If {{9}^{th}} term of an A.P. is zero, prove its {{29}^{th}} term is double the {{19}^{th}} term.

An arithmetic progression or arithmetic sequence is a number’s sequence such that the difference between the consecutive terms is constant.

Solutions:

Given,

{{a}_{9}}=0

We know that, {{n}^{th}} term {{a}_{n}}=a+\left( n-1 \right)d

So, a+\left( 9-1 \right)d=0 ⇒ a+8d=0 ……(i)

Now,

{{29}^{th}} term is given by {{a}_{29}}=a+\left( 29-1 \right)d

⇒ {{a}_{29}}=a+28d

And, {{a}_{29}}=\left( a+8d \right)+20d [using (i)]

⇒ {{a}_{29}}=20d ….. (ii)

Similarly, {{19}^{th}} term is given by {{a}_{19}}=a+\left( 19-1 \right)d

⇒ {{a}_{19}}=a+18d

And, {{a}_{19}}=\left( a+8d \right)+10d [using (i)]

⇒ {{a}_{19}}=10d …..(iii)

On comparing (ii) and (iii), it’s clearly seen that

{{a}_{29}}=2\left( {{a}_{19}} \right)

Therefore, {{29}^{th}} term is double the {{19}^{th}} term.

i

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