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JEE Main Mock Test Series 2020 & Previous Year Papers

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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∞ = a/ (1-r), if |r| < 1
= ∞,    if |r| ≥ 1
Hence, S = 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 t3 = 4.
Hence, this means ar2 =4
Now product of first five terms =
a.ar.ar2.ar3.ar4
= a5r10
= (ar2)5
= 45

Illustration: Consider an infinite geometric series with first term ‘a’ and common ratio ‘r’. Find the values of ‘a’ and ‘r’ if its second term is ¾ and its sum is 4.
Solution: It is given in the question that the second term is ¾ and the sum is 4.
Further, the first term is ‘a’ and the common ratio is ‘r’.
Hence, we have a/ (1-r) = 4 and ar = 3/4.
This gives the value of r as 3/4a.
So, 4a2/(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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