# Progressions MCQ

## Click on any option to know the CORRECT ANSWERS

 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

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)

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

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