Q1: Find the 20^{th} and n^{th}terms of the G.P.
Ans: The given G.P. is
Here, a = First term =
r = Common ratio =
Q2: Find the 12^{th} term of a G.P. whose 8^{th} term is 192 and the common ratio is 2.
Ans: Common ratio, r = 2
Let a be the first term of the G.P.
∴ a_{8} = ar ^{8–1} = ar^{7}
⇒ ar^{7} = 192
a(2)^{7} = 192
a(2)^{7} = (2)^{6} (3)
Q3: The 5^{th}, 8^{th} and 11^{th} terms of a G.P. are p, q and s, respectively. Show that q^{2} = ps.
Ans: Let a be the first term and r be the common ratio of the G.P.
According to the given condition,
a_{5} = a r^{5–1 }= a r^{4} = p … (1)
a_{8 }= a r^{8–1 }= a r^{7} = q … (2)
a_{11} = a r^{11–1 }= a r^{10} = s … (3)
Dividing equation (2) by (1), we obtain
Dividing equation (3) by (2), we obtain
Equating the values of r^{3} obtained in (4) and (5), we obtain
Thus, the given result is proved.
Q4: The 4^{th} term of a G.P. is square of its second term, and the first term is –3. Determine its 7^{th} term.
Ans: Let a be the first term and r be the common ratio of the G.P.
∴ a = –3
It is known that, a_{n} = ar^{n}^{–1}
∴ a_{4 }= ar^{3} = (–3) r^{3}
a_{2} = a r^{1} = (–3) r
According to the given condition,
(–3) r^{3} = [(–3) r]^{2}
⇒ –3r^{3} = 9 r^{2}
⇒ r = –3
a_{7} = a r ^{7–1 }= a r^{6} = (–3) (–3)^{6} = – (3)^{7} = –2187
Thus, the seventh term of the G.P. is –2187.
Q5: Which term of the following sequences:
(a)
(b)
(c)
Ans: (a) The given sequence is
Here, a = 2 and r =
Let the n^{th} term of the given sequence be 128.
Thus, the 13^{th} term of the given sequence is 128.
(b) The given sequence is
Here,
Let the n^{th} term of the given sequence be 729.
Thus, the 12^{th} term of the given sequence is 729.
(c) The given sequence is
Here,
Let the n^{th} term of the given sequence be .
Thus, the 9^{th} term of the given sequence is .
Q6: For what values of x, the numbers are in G.P?
Ans: The given numbers are .
Common ratio =
Also, common ratio =
Thus, for x = ± 1, the given numbers will be in G.P.
Q7: Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 …
Ans: The given G.P. is 0.15, 0.015, 0.00015, …
Here, a = 0.15 and
Q8: Find the sum to n terms in the geometric progression
Ans: The given G.P. is
Here,
Q9: Find the sum to n terms in the geometric progression
Ans: The given G.P. is
Here, first term = a_{1} = 1
Common ratio = r = – a
Q10: Find the sum to n terms in the geometric progression
Ans: The given G.P. is
Here, a = x^{3} and r = x^{2}
Q11: Evaluate
Ans:
The terms of this sequence 3, 3^{2}, 3^{3}, … forms a G.P.
Substituting this value in equation (1), we obtain
Q12: The sum of first three terms of a G.P. is ^{39}/_{10} & their product is 1. Find the common ratio and the terms.
Ans: Let be the first three terms of the G.P.
From (2), we obtain
a^{3} = 1
⇒ a = 1 (Considering real roots only)
Substituting a = 1 in equation (1), we obtain
Thus, the three terms of G.P. are .
Q13: How many terms of G.P. 3, 3^{2}, 3^{3}, … are needed to give the sum 120?
Ans: The given G.P. is 3, 3^{2}, 3^{3}, …
Let n terms of this G.P. be required to obtain the sum as 120.
Here, a = 3 and r = 3
∴ n = 4
Thus, four terms of the given G.P. are required to obtain the sum as 120.
Q14: The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.
Ans: Let the G.P. be a, ar, ar^{2}, ar^{3}, …
According to the given condition,
a + ar + ar^{2} = 16 and ar^{3 +} ar^{4} + ar^{5 }= 128
⇒ a (1 + r + r^{2}) = 16 … (1)
ar^{3}(1 + r + r^{2}) = 128 … (2)
Dividing equation (2) by (1), we obtain
Substituting r = 2 in (1), we obtain
a (1 +2 +4) = 16
⇒ a (7) = 16
Q15: Given a G.P. with a = 729 and 7^{th} term 64, determine S_{7}.
Ans: a = 729
a_{7} = 64
Let r be the common ratio of the G.P.
It is known that, a_{n} = a r^{n}^{–1}
a_{7} = ar^{7–1} = (729)r^{6}
⇒ 64 = 729 r^{6}
Also, it is known that,
Q16: Find a G.P. for which sum of the first two terms is –4 and the fifth term is 4 times the third term.
Ans: Let a be the first term and r be the common ratio of the G.P.
According to the given conditions,
a_{5} = 4 × a_{3}
ar^{4} = 4ar^{2}
⇒ r^{2} = 4
∴ r = ± 2
From (1), we obtain
Thus, the required G.P. is
4, –8, 16, –32, …
Q17: If the 4^{th}, 10^{th} and 16^{th} terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.
Ans: Let a be the first term and r be the common ratio of the G.P.
According to the given condition,
a_{4} = a r^{3} = x … (1)
a_{10} = a r^{9} = y … (2)
a_{16} = a r^{15 }= z … (3)
Dividing (2) by (1), we obtain
Dividing (3) by (2), we obtain
∴ Thus, x, y, z are in G. P.
Q18: Find the sum to n terms of the sequence, 8, 88, 888, 8888…
Ans: The given sequence is 8, 88, 888, 8888…
This sequence is not a G.P. However, it can be changed to G.P. by writing the terms as
S_{n} = 8+ 88+ 888+ 8888 …………….. to n terms
Q19: Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, ½ .
Ans Required sum =
Here, 4, 2, 1, is a G.P.
First term, a = 4
Common ratio, r =
It is known that,
∴ Required sum =
Q20: Show that the products of the corresponding terms of the sequences form a G.P, and find the common ratio.
Ans: It has to be proved that the sequence, aA, arAR, ar^{2}AR^{2}, …ar^{n}^{–1}AR^{n}^{–1}, forms a G.P.
Thus, the above sequence forms a G.P. and the common ratio is rR.
Q21: Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4^{th} by 18.
Ans: Let a be the first term and r be the common ratio of the G.P.
a_{1} = a, a_{2} = ar, a_{3} = ar^{2}, a_{4} = ar^{3}
By the given condition,
a_{3} = a_{1}+ 9
⇒ ar^{2} = a + 9 … (1)
a_{2} = a_{4} + 18
⇒ ar = ar^{3} + 18 … (2)
From (1) and (2), we obtain
a(r^{2} – 1) = 9 … (3)
ar (1– r^{2}) = 18 … (4)
Dividing (4) by (3), we obtain
Substituting the value of r in (1), we obtain
4a = a+ 9
⇒ 3a = 9
∴ a = 3
Thus, the first four numbers of the G.P. are 3, 3(– 2), 3(–2)^{2}, and 3(–2)^{3} i.e., 3¸–6, 12, and –24.
Q22: If the terms of a G.P. are a, b and c, respectively. Prove that
Ans: Let A be the first term and R be the common ratio of the G.P.
According to the given information,
AR^{p}^{–1 }= a
AR^{q}^{–1 }= b
AR^{r}^{–1 }= c
a^{q–r} b^{r–p} c^{p–q}
= A^{q}^{–r }× R^{(p–1) (q–r)} × A^{r–p} × R^{(q–1) (rp)} × A^{p}^{–q} × R^{(r –1)(p–q)}
= Aq^{ – r + r – p + p – q} × R ^{(pr – pr – q + r) (rq – r + p – pq) (pr – p – qr + q)}
= A^{0} × R^{0}
= 1
Thus, the given result is proved.
Q23: If the first and the n^{th} term of a G.P. are a ad b, respectively, and if P is the product of n terms, prove that P^{2} = (ab)^{n}.
Ans: The first term of the G.P is a and the last term is b.
Therefore, the G.P. is a, ar, ar^{2}, ar^{3}, … ar^{n}^{–1}, where r is the common ratio.
b = ar^{n}^{–1} … (1)
P = Product of n terms
= (a) (ar) (ar^{2}) … (ar^{n}^{–1})
= (a × a ×…a) (r × r^{2} × …r^{n}^{–1})
= a^{n} r ^{1 +2 …(n–1)} … (2)
Here, 1, 2, …(n – 1) is an A.P.
∴ 1+ 2 ………. +(n – 1)
Q24: Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from
Ans: Let a be the first term and r be the common ratio of the G.P.
Since there are n terms from (n + 1)^{th} to (2n)^{th} term,
Sum of terms from(n + 1)^{th} to (2n)^{th} term
a ^{n+ 1} = ar ^{n+ 1} ^{– 1} = ar^{n}
Thus, required ratio =
Thus, the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)^{th} to (2n)^{th }term is ^{ }^{ .}
Q25: If a, b, c and d are in G.P. show that .
Ans: a, b, c, d are in G.P.
Therefore,
bc = ad … (1)
b^{2} = ac … (2)
c^{2} = bd … (3)
It has to be proved that,
(a^{2} + b^{2} + c^{2}) (b^{2}+ c^{2} + d^{2}) = (ab + bc – cd)^{2}
R.H.S.
= (ab + bc+ cd)^{2}
= (ab + ad + cd)^{2} [Using (1)]
= [ab + d (a + c)]^{2}
= a^{2}b^{2} + 2abd (a + c) + d^{2} (a + c)^{2}
= a^{2}b^{2} + 2a^{2}bd+ 2acbd + d^{2}(a^{2} + 2ac + c^{2})
= a^{2}b^{2} + 2a^{2}c^{2}+ 2b^{2}c^{2} + d^{2}a^{2} + 2d^{2}b^{2} + d^{2}c^{2} [Using (1) and (2)]
= a^{2}b^{2} + a^{2}c^{2} + a^{2}c^{2} + b^{2}c^{2 } + b^{2}c^{2} + d^{2}a^{2} + d^{2}b^{2} + d^{2}b^{2} + d^{2}c^{2}
= a^{2}b^{2} + a^{2}c^{2} + a^{2}d^{2 }+ b^{2 }× b^{2} + b^{2}c^{2+} b^{2}d^{2} + c^{2}b^{2} + c^{2 }× c^{2} + c^{2}d^{2}
[Using (2) and (3) and rearranging terms]
= a^{2}(b^{2} + c^{2} + d^{2}) +b^{2} (b^{2} + c^{2} + d^{2}) + c^{2} (b^{2} + c^{2} + d^{2})
= (a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2})
= L.H.S.
∴ L.H.S. = R.H.S.
∴
Q26: Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Ans: Let G_{1} and G_{2} be two numbers between 3 and 81 such that the series, 3, G_{1}, G_{2}, 81, forms a G.P.
Let a be the first term and r be the common ratio of the G.P.
∴ 81 = (3) (r)^{3}
⇒ r^{3} = 27
∴ r = 3 (Taking real roots only)
For r = 3,
G_{1} = ar = (3) (3) = 9
G_{2} = ar^{2} = (3) (3)^{2} = 27
Thus, the required two numbers are 9 and 27.
Q27: Find the value of n so that may be the geometric mean between a and b.
Ans: G. M. of a and b is .
By the given condition,
Squaring both sides, we obtain
Q28: The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio .
Ans: Let the two numbers be a and b.
G.M. =
According to the given condition,
Also,
Adding (1) and (2), we obtain
Substituting the value of a in (1), we obtain
Thus, the required ratio is .
Q29: If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are .
Ans: It is given that A and G are A.M. and G.M. between two positive numbers. Let these two positive numbers be a and b.
From (1) and (2), we obtain
a + b = 2A … (3)
ab = G^{2} … (4)
Substituting the value of a and b from (3) and (4) in the identity (a – b)^{2} = (a + b)^{2} – 4ab, we obtain
(a – b)^{2} = 4A^{2} – 4G^{2} = 4 (A^{2}–G^{2})
(a – b)^{2} = 4 (A + G) (A – G)
From (3) and (5), we obtain
Substituting the value of a in (3), we obtain
Thus, the two numbers are .
Q30: The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2^{nd} hour, 4^{th} hour and n^{th} hour?
Ans: It is given that the number of bacteria doubles every hour. Therefore, the number of bacteria after every hour will form a G.P.
Here, a = 30 and r = 2
∴ a_{3} = ar^{2} = (30) (2)^{2} = 120
Therefore, the number of bacteria at the end of 2^{nd} hour will be 120.
a_{5} = ar^{4} = (30) (2)^{4} = 480
The number of bacteria at the end of 4^{th} hour will be 480.
a_{n}_{ + 1 }= ar^{n} = (30) 2^{n}
Thus, number of bacteria at the end of n^{th} hour will be 30(2)^{n}.
Q31: What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?Ans: The amount deposited in the bank is Rs 500.
At the end of first year, amount = = Rs 500 (1.1)
At the end of 2^{nd} year, amount = Rs 500 (1.1) (1.1)
At the end of 3^{rd} year, amount = Rs 500 (1.1) (1.1) (1.1) and so on
∴ Amount at the end of 10 years = Rs 500 (1.1) (1.1) … (10 times)
= Rs 500(1.1)^{10}
Q32: If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
Ans: Let the root of the quadratic equation be a and b.
According to the given condition,
The quadratic equation is given by,
x^{2}– x (Sum of roots) + (Product of roots) = 0
x^{2} – x (a + b) + (ab) = 0
x^{2} – 16x + 25 = 0 [Using (1) and (2)]
Thus, the required quadratic equation is x^{2} – 16x + 25 = 0
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1. What is the difference between a sequence and a series in mathematics? 
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3. What is the formula for the sum of an arithmetic series? 
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