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Question (1):
For the given sequence, suggest possible next three terms and find the general term. 7, 11, 15,----
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Answer:
First term = 7, Common Difference = 11 - 7 = 15 - 11 = 4
Next three terms are 15 + 4 = 19, 19 + 4 = 23, 23 + 4 = 27
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Question (2):
For the given sequence, suggest possible next three terms and find the general term. -3, 1, 5,----
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Answer:
First term = -3, Common difference = 1-(-3) = 4 = 5 - 1
Next three terms are 5 + 4 = 9, 9 + 4 = 13, 13 + 4 = 17
General term is = a + (n - 1)d = -3 + (n - 1)4 = 4n - 7
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Question (3):
For the given sequence, find the first four terms.
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Answer:
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Question (4):
For the given sequence, find the first four terms.
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Answer:
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Question (5):
Find the first four terms of the sequence 4 + 5n. Which term of the sequence is 99?
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Answer:
When n = 1, t1 = 4+5(1) = 9. When n = 2, t2 = 4+5(2) = 14
When n = 3, t3 = 4+5(3) = 19. When n = 4, t4 = 4+5(4) = 24
\ The first four terms are 9, 14, 19, 24
19th term is 99.
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Question (6):
The 50th term of an AP is 90 and 90th term is 50. Find the 140th term.
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Answer:
If tn = m and tm = n then tm+n = 0
OR
a = first term, d = common difference
Subtracting (ii) from (i)
40d = -40
= 90 + 90d
\140th term is zero.
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Question (7):
The first term of an AP is 73, common difference is -7. If the last term of the AP is -95, find the number of terms.
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Answer:
Hence, the number of terms is 25.
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Question (8):
Find three numbers in AP whose sum is 33 and whose product is 1155.
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Answer:
Let a - d, a, a + d be the three numbers in AP.
The numbers are
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Question (9):
Find four numbers in AP whose sum is 26. Sum of the products of the extremes and the middle terms is 62.
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Answer:
Let (a - 3d), (a - d), (a + d), (a + 3d) be the four numbers in AP
The numbers are
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Question (10):
Four numbers are in AP. Sum of the first three is 21 and sum of last three is 30. Find the numbers.
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Answer:
Let the numbers a - 3d, a - d, a + d, a + 3d be in AP.
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Question (11):
in AP.
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Answer:
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Question (12):
Show that if the positive numbers a,b,c are in AP, so are the 
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Answer:
Since a,b,c are in an AP, we have
Now, we shall show that sum of extreme terms is double the middle terms.
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Question (13):
Find the sum of the following series:
i. 8 + 21 + 34 +---- to n terms
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Answer:
i. 8 + 21 + 34 +---- to n terms
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Question (14):
Find the sum of the series
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Answer:
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Question (15):
Find the sum of 32 terms of an AP whose 3rd term is 1 and 6th term is -11.
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Answer:
Let a = first term, d = common difference
Solving for a and d, we have a = 9, d = -4
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Question (16):
Find the sum to first hundred even natural numbers divisible by 5.
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Answer:
S = 10 + 20 + 30 + 40 +-----to 100 terms
 
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Question (17):
The sum of 3rdand 7th term of an AP is 42 and the sum of the 7th and the 11th term of same AP is 82. Find the first term and the common difference.
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Answer:
Let a = first term, d = common difference
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Question (18):
If a1, a2, a3, a4----an are in AP prove that
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Answer:
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Question (19):
The ratio of the sums of m and n terms of an AP is  .
Show that the ratio of mth and nth terms is (2m - 1): (2n - 1).
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Answer:
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Question (20):
are in AP.
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Answer:
(b - c)2, (c - a)2, (a - b)2are in AP, then
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Question (21):
There are 25 trees at equal distances of 5 metres in a line with a well, the distance of the well from the nearest tree being 10 metres. A gardener waters all trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees.
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Answer:
Let O be the position of the well (assuming that the trees and the well are collinear). T1, T2, T3......T25 be the position of trees. The total distance covered by the gardener will be:
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Question (22):
A man saves Rs 32 during the first year, Rs 36 in the next year and Rs 40 in the third year. If he continues his savings in the sequence, in how many years will he save Rs 2000.
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Answer:
Let a = 32, d = 4, Sn = 2000
Since the duration is positive, n = 25 years.
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Question (23):
An insect starts from a point and travels in a straight-line path one mm in the first second and half of the distance covered in the previous second in the succeeding second. In how much time would it reach a point 3mm from its starting point?
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Answer:
Let time taken to cover a distance of 3mm be n seconds. The series of the given distances covered from a G P. where
 insect would never reach its goal.
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Question (24):
For each geometric progression, find the common ratio. Then tell whether the ratio is a growth factor or decay factor or neither. Find also the nth term.
i. 5, 10, 20, 40,....
ii.6, 0.6, 0.06,....
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Answer:
i. 5, 10, 20, 40,....
Since r > 1, r is a growth factor.
ii. 6, 0.6, 0.06,....
Since r = 0.1 < 1, r is a decay factor.
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Question (25):
Use the given values for a and r to find the first four terms of a GP and the nth term.
i. a = 1234, r = 0.1
iv. a = 1, r = 1
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Answer:
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Question (26):
The sum of 3 numbers in GP is 35 and their product is 1000. Find the numbers.
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Answer:
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Question (27):
If a2 + b2, ab + bc and b2 + c2 are in G.P. Prove that a, b, c are also in G.P
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Answer:
a2 + b2, ab + bc, b2 + c2are in G.P
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Question (28):
If a, b, c, d are in GP, prove that
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Answer:
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Question (29):
prove that x, y and z are in GP.
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Answer:
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Question (30):
If one Geometric mean G and two arithmetic means p and q be inserted between two quantities, show that G2=(2p -q)(2q - p).
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Answer:
Let the two numbers be a and b
G is the GM between a and b
Then G2= ab ...(1)
p, q are the AM's between a and b.
Then a, p, q and b are in AP
2p = a+q and 2q = b+p
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Question (31):
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Answer:
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Question (32):
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Answer:
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Question (33):
i. Show that the arithmetic mean of two positive numbers is greater than or equal to their geometric mean.
ii. The arithmetic mean of two positive integers a and b (a>b)
is twice their Geometric mean. Prove that
iii. Construct a quadratic equation in x such that AM of its roots is A and GM is G.
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Answer:
i. Let a and b be the numbers.
iii. Let a and b be the numbers. A be the AM and G be the GM
The quadratic equation in x is
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Question (34):
i. If a, b, c are in GP, prove that log a, log b, log c are in AP.
AP.
in AP.
vi. If a, b, c are in GP and x, y be the AM's of a, b and b, c respectively, prove that
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Answer:
Taking log on both sides
 
Let A = first term, R = Common ratio
 (dividing both sides by xy)
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Question (35):
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Answer:
Let g1, g2, g3,----gn be the GM's between a and b.
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Question (36):
The vibration of a group are damped so that the amplitude of successive deflection are in GP. If the amplitude is first 12cm and then 8cm. Find the amplitude of 6th Vibration.
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Answer:
Here the successive amplitudes are in GP
ar = 8
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Question (37):
If S1,S2,S3------Sn are the sums of n terms of n GP's, whose common ratio are respectively 1,2,3,--------- Show that,
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Answer:
 
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Question (38):
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Answer:
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Question (39):
If S1, S2, S3,....., Sn are the sums of infinite GP's whose first
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Answer:
This is an AP with a = 2, d = 1 and n = n
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Question (40):
Find the first term.
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Answer:
\ The first term is 16.
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Question (41):
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Answer:
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Question (42):
One side of a square is 8 cm. The midpoints of its sides are joined to form another square, whose midpoints are again joined to form one more square and this process is continued indefinitely. Find the sum of areas of all the squares formed.
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Answer:
First square is ABCD. Second square is A1 B1 C1 D1. Third square is
A2 B2 C2 D2.
Let each side of square be a units.
Then the length of side of the second square is
The side if the third square is
Area of first square = a2
If the length of each side of the square is 8 cm, then the area is
2a2 = 2(8)2 = 128 sq cm.
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Question (43):
The sum of the first n term of a certain series  show that the terms of the series form a GP and find the first term and the common ratio.
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Answer:
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Question (44):
Find the sum of the squares of first n odd natural numbers.
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Answer:
S= 12 + 32 + 52+.... to n terms
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Question (45):
Sum the series
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Answer:
Let tn = nth term of the given series
(Note that sum of first n odd natural numbers is n2)
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Question (46):
The sequence N of natural numbers is divided into groups as follows:
Show that the sum of the numbers in the nth row is n(2n2+1).
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Answer:
We observe that the number of terms in 1st row = 2, 2nd row = 4, 3rd row=6 and so on. The number of terms in nth row = 2n.
\ The number of numbers upto nth row (inclusive of the nth row).
If Sn= sum of (1+2+3+...n), then the required sum = Sn(n+1) - Sn(n-1)
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Question (47):
Find the sum of the following series 23 + 43 + 63+.... to n terms.
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Answer:
23 + 43 + 63+.... to n terms
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Question (48):
Find the sum of the following series 23 + 53 + 83+....
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Answer:
23 + 53 + 83+.... to n terms
Sn = 23 + 53 + 83+.... to n terms
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Question (49):
Find the sum of the following series 1.2 + 2.3 + 3.4 +....to n
terms.
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Answer:
Sn = 1.2 + 2.3 + 3.4 +....to n terms.
Sn = Stn = S(n2 + n)
= Sn2 + Sn
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Question (50):
Find the sum of the following series 1.22 + 2.32 + 3.42 +....
to n terms.
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Answer:
Sn = 1.22 + 2.32 + 3.42 +.... to n terms.
Sn = Stn = S(n3 + 2n2 + n)
= Sn3 + 2Sn2 + Sn
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