CLASS_11_MATHS_SOLUTIONS_NCERT

Class XI Chapter 9 – Sequences and Series Maths
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2
a 1
r
A2
4
...... 1
1
r
a
4a
5 3
4 2 2
ar ar r
r
2
4 4
From (1), we obtain
2
a 1
2
4
1
2
a
4
1
4
1
4 a 3
a
Also,
4
3
for
a 1 2
4
1 2
a 1
4
4
1
2
a 3
4
3
a
4
r 2
2
Thus, the required G.P. is
for
r 2
4 8 16
, , ,...
3 3 3
or 4, 8, 16, 32...
Question 17:
If the 4 th , 10 th and 16 th terms of a G.P. are xy, , and z, respectively. Prove that
Solution 17:
Let a be the first term and r be the common ratio of the G.P.
According to the given condition,
a a r x
3
4
...... 1
a a r y
9
10
...... 2
a a r z
15
16
...... 3
Dividing (2) by (1), we obtain
x, y,
z
are in G.P.
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