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Question (1):
Show that the sum of the cubes of any number of consecutive integers is divisible by the sum of those integers.
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Answer:
Let the consecutive integers be (n + 1), (n + 2), (n + 3),....,(n + m)
Sum of the cubes of integers = S1
(n+m) and (n+m+1) are consecutive integers, one of them must be even.
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Question (2):
For 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
 The three terms form an A.P.
Next three terms are 15 + 4 = 19, 19 + 4 = 23, 23 + 4 = 27
General term is = a + (n-1) d = 7 + (n-1) 4 = 4n + 3
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Question (3):
For given sequence, suggest possible next three terms and find the general term.
-3, 1, 5...
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Answer:
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Question (4):
For given sequence, find the first four terms.
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Answer:
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Question (5):
For given sequence, find the first four terms.
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Answer:
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Question (6):
Using the given values of a and d find the first five terms of A.P. and the general term.
a = 4, d = 3
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Answer:
The first five terms are:
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Question (7):
Find the first five terms of the sequence defined by tn = 6 - 4n. Which term of the sequence is -50?
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Answer:
tn = 6 - 4n
When n = 1, t1 = 6 - 4(1) = 2
When n = 2, t2 = 6 - 4(2) = -2
When n = 3, t3 = 6 - 4(3) = -6
When n = 4, t4 = 6 - 4(4) = -10
When n = 5, t5 = 6 - 4(5) = -14
The first five terms are (2, -2, -6, -10, -14).
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Question (8):
The tenth term and the common difference of an AP are 22 and 3. Find the first term and the nth term.
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Answer:
To find a:
tn = a + (n-1) d
= -5 + (n-1)3
= -5 + 3n - 3
= 3n - 8
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Question (9):
If the first term of an AP is 11, the common difference is 3 and the last term is 98, find the number of terms.
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Answer:
98 = 11 + (n - 1)3
98 - 11 + 3 = 3n
3n = 90
n = 30
Hence the number of terms = 30.
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Question (10):
Then, find k.
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Answer:
If a,b,c are in AP, then a+c = 2b.
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Question (11):
Find three numbers in AP whose sum is 24 and whose product is 440.
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Answer:
Let a-d, a, a+d be three numbers in AP
Then (a - d) + (a) + (a + d) = 24
3a = 24
a = 8
(a - d)(a)(a + d) = 440
(8 - d)(8)(8 + d) = 440
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Question (12):
Find three numbers in AP whose sum is 9 and sum of their cubes is 153.
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Answer:
Let a-d, a, a+d be three numbers in AP.
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Question (13):
Find the four numbers in AP whose sum is -4 and the product is 105.
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Answer:
The numbers are
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Question (14):
If a, b, c are in AP, then prove that a2(b + c), b2(c + a), c2(a + b) are in AP.
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Answer:
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Question (15):
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Answer:
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Question (16):
Find the sum of the following series:
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Answer:
Note that we can find n directly by using the formula  for an A.P.
= 15(7 + 145) = 15 x 152 = 2280
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Question (17):
If Sn = 3n2 + 4n denotes the sum of n terms of a progression, Prove that it is an AP. Find the rth term.
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Answer:
(This is independent of n). Hence the progression is an AP.
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Question (18):
If Sn = 2n2 + 3n denote the sum of a progression, prove that it is an AP. Find (2n)thterm.
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Answer:
d is independent of n. The progression is an AP.
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Question (19):
The sum of certain numbers in AP is 5500. If the first and the last terms are 100 and 1000 respectively, find the number of terms.
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Answer:
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Question (20):
If the first term of an AP is 2 and the sum of first five terms is equal to one fourth of the sum of next five terms, then show that
i) t20 = -112
ii) find the sum of the first 40 terms.
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Answer:
a = first term, d = common difference
S10 - S5 = 10 + 35d
But (10 + 35d) = 4S5
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Question (21):
 (pq)th term and the sum of pq terms.
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Answer:
Let a = first term, d = common difference
Subtract (ii) from (i)
 
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Question (22):
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Answer:
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Question (23):
The sum of n terms of three APs are S1,S2 and S3. The first term of each is 1 and common differences are 1, 2 and 3 respectively. Prove that S1 + S3 = 2S2.
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Answer:
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Question (24):
The first and the last term of an AP are a and l respectively. If S denotes the sum of terms show that the common 
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Answer:
a = first term, l = last term, Sn = S|, d = common difference.
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Question (25):
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 (26):
If the sums of the first p, q, r terms of AP are a, b, c respectively prove that
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Answer:
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Question (27):
If a, b, c are in AP, show that (a - c)2 = 4(b2 - ac).
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Answer:
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Question (28):
If a2 + 2bc, b2+ 2ca, c2 + 2ab are in AP, show that 
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Answer:
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Question (29):
Two cars start together in the same direction at the same place. The first goes with uniform speed of 15 km/hr, the second goes at the speed of 10 km/hr in the first hour and increases the speed by 0.1 km each succeeding hour. After how many hours will the second car over take the first if both cars go non stop?
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Answer:
Speed of second car = 10 km/hr
Uniform Speed of the first car = 15 km/hr
The second car increases its speed by 1 km every hour.
Since both cars are travelling in the same direction, the distance covered will be same.
They are together initially or after 11 hours. Thus the second car over takes the first car after 11 hours of journey.
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Question (30):
Find the indicated terms in the following G.P.:
iv. Find the 12th term if a GP of which the 5th and 9th terms 
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Answer:
iv. Find the 12th term if a GP of which the 5th and 9th terms are 
Let a = first term, r = common ratio
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Question (31):
In a finite G.P., the product of the terms equidistant from the beginning and the end is always same and equal to the product of the first and last term.
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Answer:
Let a1, a2, a3,- - - -,an be a finite GP with common ratio = r.
Then ak = kth term from the beginning = ark-1
And an-k+1 is the kth term from the other end.
Hence the problem.
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Question (32):
The first term of a GP is 1. The sum of the third and fifth terms is 90. Find the common ratio of the GP.
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Answer:
Let r = common ratio
The common ratio is either +3 or -3.
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Question (33):
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.
ii. 1, 1.1, 1.21, 1.331,....
iii. -1, 1, -1, 1, -1,....
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Answer:
Since r = 1.1 > 1, r is a growth factor.
Here the progression is oscillating.
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Question (34):
Use the given values for a and r to find the first four terms of a GP and the nth term.
1. a = 2, r = 5
2. a = -2, r =3
4. a = x, r = y
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Answer:
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Question (35):
What is the decay factor for the value of a car that depreciates 20% per year. Give the value at the end of 0, 1, 2, 3 and 4 years for a car that costs Rs. 300000/- as a sequence.
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Answer:
The value of car at the 0th year = Rs. 300000
The value of car at the end of 1st year
= (100 - 20)% of Rs. 300000 = Rs. 240000
The value of car at the end of 2nd year = Rs. 192000
The value of car at the end of 3rd year = Rs. 153600
The value of car at the end of 4th year = Rs. 122880
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Question (36):
Find three numbers in GP such that their sum is 38 and their product is 1728.
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Answer:
 
 
 
The numbers are
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Question (37):
Find four numbers in GP in which the third term is greater than the first by 9 and the second term is greater than the fourth by 18.
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Answer:
Let the four numbers in GP be a, ar, ar2, ar3
Third term-first term = 9
ar2 - a = 9... (i)
Second term-fourth term = 18
ar - ar3 = 18... (ii)
Dividing (ii) by (i) we get
r = -2... (iii)
Now putting r = -2 in (i)
The numbers are
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Question (38):
If a, b, c, d are in GP, Prove that a+b, b+c, c+d are also in GP.
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Answer:
From the problem we have a, b, c, d are in GP
Now a+b, b+c, c+d will be in GP if,
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Question (39):
If a, b, c are in GP, prove that the following are also in GP.
 
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Answer:
a, b, c, d are in GP
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Question (40):
If a, b, c are three consecutive terms of an AP, then prove that ka, kb and kc are three consecutive terms in GP, where k is a non zero real number.
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Answer:
Since a, b, c are in AP
b - a = c - b...(i)
If ka, kb, kcwill be in G.P.,
then kb-a = kc-b. This is true because of (i).
Hence ka, kb and kc are in GP.
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Question (41):
If the pth, qth, rth, sth terms of an AP are in GP, show that
p - q, q - r, r - s are in GP.
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Answer:
Let a = first term and d = common difference of the AP.
Since ap, aq, ar, as are in GP, then
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Question (42):
If p, q, r are in AP and x, y, z are in GP, then prove that 
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Answer:
Also q = p + d, r = p + 2d, where d is the common difference of the AP.
x, y, z are in GP. Let R be the common ratio
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Question (43):
Find geometrically the GM between a and b?
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Answer:
Let G be the Geometric mean between a and b.
Geometrically, if AB = a, BC = b.
On a line ABC, draw a semi circle on AC, with AC as diameter. Draw BD perpendicular to AC at B to cut the semicircle at D.
Draw AD and CD.
BD is the GM between AB and BC.
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Question (44):
Insert 3 GM's between 8 and 648.
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Answer:
a = 8, b = 648 R = Common ratio
Let the three GM be aR, aR2, aR3. then b = aR4
\ The three GM's are 8(3), 8(3)2, 8(3)3
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Question (45):
If one arithmetic mean A and two geometric means g1, g2 be 
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Answer:
Let the two numbers be a and b.
If A is the AM between a and b then a, A, b are in AP.
If g1, g2 are the GM's between a and b then a, g1, g2, b are in GP.
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Question (46):
If S1, S2, S3 be respectively the sum of n, 2n, 3n terms of GP, prove that:
i) S12+S22 = S1(S2 + S3)
ii) S1(S3 - S2) = (S2 - S1)2
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Answer:
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Question (47):
Find the sum of n terms of the series 5 + 55+ 555 + ...
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Answer:
Here a = 10, r = 10, n = n
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Question (48):
If y = x + x2 + x3+ ... to infinite number of terms, then 
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Answer:
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Question (49):
Insert
i. 2 Geometric means between 2 and 16
iii. 4 Geometric means between 5 and 0.00005
iv. 5 Geometric means between 3 and 192
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Answer:
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Question (50):
If the AM between two numbers exceeds their GM by 2 and the ratio of two numbers is 4, find the numbers.
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Answer:
Let the numbers be a and b (a > b)
A - G = 2 (by data)
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