Understanding the Problem

We are given that for an arithmetic progression (AP) with non-zero common difference, 100 times the 100^th term equals 50 times the 50^th term. We need to find the 150^th term.

Step 1: Recall the General Term of an AP

The n^th term of an AP with first term $a$ and common difference $d$ is given by:

$T_n=a+(n-1)d$

Step 2: Write the Given Condition as an Equation

According to the problem:

$100\times T_{100}=50\times T_{50}$

Substituting the formula for the n^th term:

$100[a+(100-1\left)d\right]=50[a+(50-1\left)d\right]$

Simplifying the expressions inside the brackets:

$100(a+99d)=50(a+49d)$

Step 3: Simplify the Equation

Divide both sides by 50 to simplify:

$2(a+99d)=1(a+49d)$

Expanding both sides:

$2a+198d=a+49d$

Bring all terms to one side:

$2a-a+198d-49d=0$

$a+149d=0$

Step 4: Find the 150^th Term

The 150^th term is given by:

$T_{150}=a+(150-1)d=a+149d$

But from the previous step, we have $a+149d=0$. Therefore:

$T_{150}=0$

Final Answer

The 150^th term of this AP is zero.

Related Topics and Formulae

Arithmetic Progression (AP)

An arithmetic progression is a sequence of numbers in which the difference between consecutive terms is constant. This difference is called the common difference ($d$).

Key Formulae

• n^th Term: $T_n=a+(n-1)d$
• Sum of first n terms: $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ or $S_n=\frac{n}{2}(a+l)$, where $l$ is the last term.

Properties of AP

A useful property is that for any three consecutive terms, the middle term is the average of the other two. For example, for terms $T_{n-k},T_n,T_{n+k}$, we have $2\times T_n=T_{n-k}+T_{n+k}$. This problem is a specific application of such a property.