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Question 1
Question: For the 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
Next three terms are 15 + 4 = 19, 19 + 4 = 23, 23 + 4 = 27
Question 2
Question: For the given sequence, suggest possible next three terms and find the general term. -3, 1, 5,----
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
Question 3
Question: For the given sequence, find the first four terms.
Answer: 
Question 4
Question: For the given sequence, find the first four terms.
Answer: 
Question 5
Question: Find the first four terms of the sequence 4 + 5n. Which term of the sequence is 99?
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.
Question 6
Question: The 50th term of an AP is 90 and 90th term is 50. Find the 140th term.
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.
Question 7
Question: 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.
Answer: 
Hence, the number of terms is 25.
Question 8
Question: Find three numbers in AP whose sum is 33 and whose product is 1155.
Answer: Let a - d, a, a + d be the three numbers in AP.
The numbers are
Question 9
Question: Find four numbers in AP whose sum is 26. Sum of the products of the extremes and the middle terms is 62.
Answer: Let (a - 3d), (a - d), (a + d), (a + 3d) be the four numbers in AP
The numbers are
Question 10
Question: Four numbers are in AP. Sum of the first three is 21 and sum of last three is 30. Find the numbers.
Answer: Let the numbers a - 3d, a - d, a + d, a + 3d be in AP.
