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Solved Examples on Geometric Progression 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: 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.
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: 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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