Class 10 Math All topics Arithmetic Progressions

The $19_{th}$ term of an A.P. is equal to three times its $6_{th}$ term. If its $9_{th}$ term is $19,$ find the A.P.

Solution:

From the question it is given that, $a_{19}=19_{th}$ term of an A.P. is equal to three times its $6_{th}$ term $=3a_{6}$

$a_{9}=19$

As we know, $a_{n}=a+(n−1)d$

$a_{9}=a+8d=19$

Then, $a_{19}=3(a+5d)$

$a+18d=3a+15d3a−a=18d−15d2a=3d $

$a=(3/2)d$

Now substitute the value of a in equation (i) we get,

$(3/2)d+8d=19(3d+16d)/2=19(19/2)d=19d=(19×2)/19d=2 $

To find out the value of a substitute the value of $d$ in equation (i)

$a+8d=19a+(8×2)=19a+16=19a=19−16a=3 $

Therefore, A.P. is $3,5,7,9,…$

Similar topics

introduction to trigonometry

functions

some applications of trigonometry

quadratic equations

polynomials

introduction to trigonometry

functions

some applications of trigonometry

quadratic equations

polynomials

Related Questions

Related Questions