Progressions MCQ

 

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Progressions

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Question 1
The 7th and 21st terms of an AP are 6 and -22 respectively. Find the 26th term.
A
-32
B
-34
C
-10
D
-16
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Question 1 Explanation: 
7th term = 6;

21st term = -22;

That means, 14 times common difference or -28 is added to 6 to get -22;

Thus, d = -2;

7st term = 6 = a+6d;

Or, a+(6*-2) = 6;

Or, a = 18;

26st term = a+25d = 18-25*2 = -32.

Question 2
Find the nth term of the following sequence :

5 + 55 +555 + . . . . Tn

A
(5/9)*(10n - 1)
B
5n(10n - 1)
C
(5/9)n *10n - 1)
D
5(10n - 1)
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Question 2 Explanation: 
\begin{align} & {\text{We will it through option checking method}}: \cr & \left( {\frac{5}{9}} \right) \times \left( {{{10}^n} - 1} \right) \cr & {\text{We}}\, {\text{put}}\, n = 1, \cr & \left( {\frac{5}{9}} \right) \times \left( {{{10}^1} - 1} \right) = 5 \cr & n = 2\left( {\frac{5}{9}} \right) \times \left( {{{10}^2} - 1} \right) = 55 \cr & n = 3\left( {\frac{5}{9}} \right) \times \left( {{{10}^3} - 1} \right) = 555 \cr & {\text{It means Option C is satisfying the sequence so the}} \cr & {n^{th}}\, {\text{term}}\, {\text{would}}\, {\text{be}}\, \left( {\frac{5}{9}} \right) \times \left( {{{10}^n} - 1} \right) \cr\end{align}
Question 3
How many terms are there in 20, 25, 30..... 140
A
23
B
25
C
22
D
24
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Question 3 Explanation: 
\begin{align} & {\text{Number}}\, {\text{of}}\, {\text{terms}}, \cr & = \left\{ {\frac{{\left( {{1^{st}}\, t{\text{erm - last}}\, {\text{term}}} \right)}}{{{\text{common}}\, {\text{difference}}}}} \right\} + 1 \cr & = \left( {140 - \frac{{20}}{5}} \right) + 1 \cr & = \left( {\frac{{120}}{5}} \right) + 1 \cr & = 24 + 1 \cr & = 25 \cr\end{align}
Question 4
The first term of an Arithmetic Progression is 22 and the last term is -11. If the sum is 66, the number of terms in the sequence are:
A
8
B
9
C
12
D
10
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Question 4 Explanation: 
\begin{align} & {\text{Number}}\, {\text{of}}\, {\text{terms}} = n\, \left( {let} \right) \cr & {\text{First}}\, {\text{term}}\, \left( a \right) = 22 \cr & {\text{Last}}\, {\text{term}}\left( l \right) = - 11 \cr & {\text{Sum}} = 66 \cr & {\text{Sum}}\, {\text{of}}\, {\text{an}}\, {\text{AP}}\, {\text{is}}\, {\text{given}}\, {\text{by}}: \cr & = {\text{Number}}\, {\text{of}}\, {\text{terms}} \times \left\{ {\frac{{\left( {{\text{First}}\, {\text{term}} + {\text{Last}}\, {\text{term}}} \right)}}{2}} \right\} \cr & 66 = n \times \left\{ {\frac{{\left( {a + l} \right)}}{2}} \right\} \cr & 66 = n \times \frac{{\left( {22 - 11} \right)}}{2} \cr & 66 = n \times \left( {\frac{{11}}{2}} \right) \cr & n = \frac{{\left( {66 \times 2} \right)}}{{11}} \cr & n = 12 \cr & {\text{No}}{\text{.}}\, {\text{of}}\, {\text{terms}} = 12 \cr\end{align}
Question 5
If the fifth term of a GP is 81 and first term is 16, what will be the 4th term of the GP?
A
24
B
18
C
27
D
54
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Question 5 Explanation: 
\begin{align} & {5^{th}}\, {\text{term}}\, {\text{of}}\, {\text{GP}} \cr & = a{r^{5 - 1}} \cr & = 16 \times {r^4} \cr & = 81 \cr & {\text{Or}}, \, r = {\left( {\frac{{81}}{{16}}} \right)^{\frac{1}{4}}} = \frac{3}{2} \cr & {4^{th}}\, {\text{term}}\, {\text{of}}\, {\text{GP}} \cr & = a{r^{4 - 1}} \cr & = 16 \times {\left( {\frac{3}{2}} \right)^3} \cr & = 54 \cr\end{align}
There are 5 questions to complete.

 

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