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Algebra: Sequence and Series Test-2

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Question 1
The sum (53 + 63 + ……103) is equal to
A
2295
B
2425
C
2495
D
2925
Question 1 Explanation: 
Let p = Sum of cubes of all natural numbers till 10
and q = sum of cubes of all natural numbers till 4 The given expression=p-q
$ \begin{array}{l}{{\left( \frac{10(10+1)}{2} \right)}^{2}}-{{\left( \frac{4(4+1)}{2} \right)}^{2}}\\={{55}^{2}}-{{10}^{2}}\\=65X45\\=2925\end{array}$
Question 2
The next term of the sequence
1, 2, 5, 26……. is:
A
677
B
47
C
50
D
152
Question 2 Explanation: 
The general term is square of previous term +1.
Thus the next term is 262+1= 677.
Question 3
The missing term in the sequence
0, 3, 8, 15, 24, …… 48 is
A
35
B
30
C
36
D
39
Question 3 Explanation: 
The nth term is n2-1.
Thus the 6th term is 62-1=35.
Question 4
If 13 + 23 + 33 + ……..+ 103 = 3025, then find the value of 23 + 43 + 63 + …..+ 203
A
6050
B
9075
C
12100
D
24200
Question 4 Explanation: 
$ \displaystyle \begin{array}{l}{{2}^{3}}+{{4}^{3}}+{{6}^{3}}+.......+{{20}^{3}}\\={{\left( 2\times 1 \right)}^{3}}+{{\left( 2\times 2 \right)}^{3}}+{{\left( 2\times 3 \right)}^{3}}+.......+{{\left( 2\times 10 \right)}^{3}}\\=8\times {{1}^{3}}+8\times {{2}^{3}}+8\times {{3}^{3}}.......+8\times {{10}^{3}}\\=8\times \left[ {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+........+{{10}^{3}} \right]\\=8\times 3025=24200\\\left[ \because \,\,\,{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+.........+{{10}^{3}}=3025\,\,\left( given \right) \right]\end{array}$
Question 5
What is the 507th term of the sequence
1, –1, 2, –2, 1, –1, 2, –2, 1…..?
A
–1
B
1
C
–2
D
2
Question 5 Explanation: 
The sequence 1,-1, 2,-2 repeats.
Thus we just need to find the (507 mod 4) = 3rd term of the sequence.
Thus the 507th term will be 2.
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