Question 1
Question: Show that the sum of the cubes of any number of consecutive integers is divisible by the sum of those integers.
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.
Question 2
Question: For given sequence, suggest possible next three terms and find the general term.
7,11,15,...
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
Question 3
Question: For given sequence, suggest possible next three terms and find the general term.
-3, 1, 5...
Answer: 
Question 4
Question: For given sequence, find the first four terms.
Answer: 
Question 5
Question: For given sequence, find the first four terms.
Answer: 
Question 6
Question: Using the given values of a and d find the first five terms of A.P. and the general term.
a = 4, d = 3
Answer: The first five terms are:
Question 7
Question: Find the first five terms of the sequence defined by tn = 6 - 4n. Which term of the sequence is -50?
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).
Question 8
Question: The tenth term and the common difference of an AP are 22 and 3. Find the first term and the nth term.
Answer: 
To find a:
tn = a + (n-1) d
= -5 + (n-1)3
= -5 + 3n - 3
= 3n - 8
Question 9
Question: 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.
Answer: 
98 = 11 + (n - 1)3
98 - 11 + 3 = 3n
3n = 90
n = 30
Hence the number of terms = 30.
Question 10
Question: 
Then, find k.
Answer: If a,b,c are in AP, then a+c = 2b.
