Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions

Question 1.
Find the seventh term from the end of the series :
√2, 2, 2√2,……32
Solution:
√2, 2, 2√2,……32
Here a = √2
r = \(\frac { 2 }{ \surd 2 } =\surd 2\)
and l =32

Question 2.
Find the third term from the end of the GP.
\(\frac { 2 }{ 27 } ,\frac { 2 }{ 9 } ,\frac { 2 }{ 3 } ,….162\)
Solution:
G.P is \(\frac { 2 }{ 27 } ,\frac { 2 }{ 9 } ,\frac { 2 }{ 3 } ,….162\)
a = \(\\ \frac { 2 }{ 27 } \)
r = \(\frac { 2 }{ 9 } \div \frac { 2 }{ 27 } \)
= \(\frac { 2 }{ 9 } \times \frac { 27 }{ 2 } \)
= 3
l = 162

Question 3.
For the G.P. \(\frac { 1 }{ 27 } ,\frac { 1 }{ 9 } ,\frac { 1 }{ 3 } …..81\)
find the product of fourth term from the beginning and the fourth term from the end.
Solution:
\(\frac { 1 }{ 27 } ,\frac { 1 }{ 9 } ,\frac { 1 }{ 3 } …..81\)
a = \(\\ \frac { 2 }{ 27 } \)
r = \(\frac { 1 }{ 9 } \div \frac { 1 }{ 27 } \)
= \(\frac { 1 }{ 9 } \times \frac { 27 }{ 1 } \)
= 3

Question 4.
If for a G.P., pth, qth and rth terms are a, b and c respectively ;
prove that :
{q – r) log a + (r – p) log b + (p – q) log c = 0
Solution:
In a G.P
T_p = a,
T_q = b,
T_r = c

Question 5.
If a, b and c in G.P., prove that : log aⁿ, log bⁿ and log cⁿ are in A.P.
Solution:
a, b, c are in G.P.
Let A and R be the first term and common ratio respectively.
Therefore,
a = A
b = AR
c = AR²
log a = log A
log b = log AR = log A + log R
log c = log AR² = log A + 2log R
log a, log b and log c are in A.P.
If 2log b = log a + log c
If 2[logA + logR] = log A + log A + 2log R
If 2log A + 2log R = 2log A + 2log R
which is true.
Hence log a, log b and log c are in A.P.

Question 6.
If each term of a G.P. is raised to the power x, show that the resulting sequence is also a G.P.
Solution:
Let a, b, c are in G.P.
Then b² = ac …(i)
Now a^x, b^x + c^x will be in G.P. if (b^x)² = a^x.c^x
=> (b^x)² = a^x.c^x
=>(b²)^x = (ac)^x
Hence a^x, b^x, c^x are in G.P. (∴ b² = ac)
Hence proved.

Question 7.
If a, b and c are in A.P. a, x, b are in G.P. whereas b, y and c are also in G.P. Show that : x², b², y² are in A.P.
Solution:
2 b = a + c _(i)
a, x, b are in G.P.
x² = ab _(ii)
and b, y, c in G.P.
y² = bc _(iii)
Now x² + y² = ab + bc
= b(a + c)
= b x 2b [from(i)]
= 2 b²
Hence x², b², y² are in G.P.

Question 8.
If a, b, c are in G.P. and a, x, b, y, c are in A.P., prove that :
(i)\(\frac { 1 }{ x } +\frac { 1 }{ y } =\frac { 2 }{ b } \)
(ii)\(\frac { a }{ x } +\frac { c }{ y } =2\)
Solution:
a, b, c are in G.P.
b² = ac
a, x, b, y, c are in A.P.
2x = a + b and 2y = b + c

Question 9.
If a, b and c are in A.P. and also in G.P., show that: a = b = c.
Solution:
a, b, c are in A.R
2 b = a + c ….(i)
Again, a, b, c are in G.P.

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11C are helpful to complete your math homework.

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