Page 23 - 11-Math-6 Sequences and Series

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Using an = a1rn-1 we have,

= a5 a1=r5-1 i.e., a5 a1r 4 (i)
Now substituting the values of a5 and a1 in (i) we have

=1 2=r4 or r4 1 (ii)
2 4

Taking square root of (ii), we get,

r2 = ± 1
2

So, we have, r2 =1 or r2 =- 1 =i2 ( )-1 =i2
2 22

⇒ r =± 1 or r =± 1 i
22

when=r 1, then =G1 2 =1 2,=G2 2 12=2 1,=G3 2  12=3 1
2 2  2

when r =-1 , then G1 =2 -21  =- 2,G2 =2 -21 2 =1,G3 =2 -21 3 =- 1
2 2

when r =i , then G1 =2 × i =2 i, G2 =2 i 2 =-1, G3 =2 i 3 =- i
2 2 2  2  2

when r =-i , then G1 =2 -2i  =- 2 i, G2 =2 -2i 2 =-1, G3 =2 -2i 3 =i
2    2

Note: The real values of r are usually taken but here other cases are considered to widen
the out-look of the students.

Example 3: If a, b, c and d are in G.P. show that a + b, b + c, c + d are in G.P.

Solution: Since a, b, c are in G.P therefore, version: 1.1
ac = b2 (i)
Also b, c, d are in G.P., so we have
bd = c2 (ii)

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