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**Illustration 1: An infinite G.P. has its first term as x and sum as 5. Then what is the range of x? (2004)****Solution: **we know that the sum of an infinite G.P. is S_{âˆž }= a/ (1-r), if |r| < 1

= âˆž, if |r| â‰¥ 1

Hence, S_{âˆž} = x/(1-r) = 5

Or 1-r = x/5

Hence, r = (5-x)/5 exists only when |r| < 1

Hence, -1 < (5-x)/5 <1

-10 < -x < 0

So this gives 0 < x < 10.

**Illustration 2: The third term of a geometric progression is 4. What is the product of the first five terms?****Solution:** here it is given that t_{3} = 4.

Hence, this means ar^{2} =4

Now product of first five terms =

a.ar.ar^{2}.ar^{3}.ar^{4}

= a^{5}r^{10}

= (ar^{2})^{5}

= 4^{5}

**Illustration:** Consider an infinite geometric series with first term â€˜aâ€™ and common ratio â€˜râ€™. Find the values of â€˜aâ€™ and â€˜râ€™ if its second term is Â¾ and its sum is 4.**Solution:** It is given in the question that the second term is Â¾ and the sum is 4.

Further, the first term is â€˜aâ€™ and the common ratio is â€˜râ€™.

Hence, we have a/ (1-r) = 4 and ar = 3/4.

This gives the value of r as 3/4a.

So, 4a^{2}/(4a-3) = 4

This gives (a-1)(a-3) = 0

Hence, a = 1 or 3.

When a = 1 then r =3/4 and when a = 3 then r = 1/4.

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