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Parin Sir > Quantitative Aptitude (Maths) > Progression > Examples 5 to 9 of progression > Examples 5 to 9 of progression

Examples 5 to 9 of progression

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(5) Find three arithmetic means between 7 and 47.

Soln. Now, d= \frac{b-a}{n+1}   :  Where,  a  =  7,   b  =  47  &  n  =  3

= \frac{47-7}{3+1}\frac{40}{4}

\therefore   d =10

\therefore  Nos.  are  7 + 10 = 17,   17 + 10 = 27,    27 + 10 = 37,

i.e.  Three arithmetic means are 17, 27 and 37.

(6) Find  8th term of  the G.P.  9, 18, 36, … .

Soln. In  this G.P. we have  a  =  9,  r  =  2.

\therefore  T_8  =   ar n – 1   =   9 ( 2 ) 8 – 1  =  9 \times 2 7  =  ( 9 \times 128 )  =  1152.

 (7) How many terms are there in the G.P.  32, 16, 8, 4, .., \frac{1}{16}?

Soln. In this G.P. we have  a  =  32,  r  =  \frac{8}{16} = \frac{1}{2}.

Let the number of  terms be n.  Then,

32 \times \left(\frac{1}{2}\right)^{n-1}  =  \frac{1}{16}

\therefore  \left(\frac{1}{2}\right)^{n-1}  =  \frac{1}{16\times32} = \frac{1}{2^9}

\therefore \left(\frac{1}{2}\right)^{n-1} = \left(\frac{1}{2}\right)^9

\therefore    n – 1  =  9

\therefore       n  =  10

(8) A boy saves Re 1 on first day of  the month, Rs. 2 on 2nd day of  the month, Rs. 4 on the 4th day, Rs. 16 on 5th day and so on. How much does he save in 10 days ?

Soln. His savings are 1, 2, 4, 8, 16, … up to 10 terms.

This is a G.P. with  a  =  1   and   r  =  \frac{2}{1} = 2.

Savings = Sum of  above G.P. up to 10 terms.

= \frac{a\left(r^n-1\right)}{(r-1)}

=  \frac{1\times(2^{10}-1)}{(2-1)}

=  Rs.  1023.

(9) Find  the geometric mean between   32   and   162.

Soln. G.M. = \sqrt{32\times162}

=  \sqrt{2^5\times2\times3^4}

= \sqrt{2^6\times3^4}

=  2^3 \times 3^2

=  8 \times 9

= 72

 

Best   of   Luck

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