Question

In GP. 2, 6, 18, 54, .................. 13122; the product of 3rd term from the beginning
and 3rd term from the last is 26244. Show it.
​

Answer 1

Answer:

Step-by-step explanation:

2.6.18.54.162.486.1458.4374.13122

13122*2=26244

Answer 2

Step-by-step explanation:

Given : In GP. 2, 6, 18, 54, .................. 13122; the product of 3rd term from the beginning  and 3rd term from the last is 26244.

To find : Show it ?

Solution :

In GP. 2, 6, 18, 54, .................. 13122.

First term a=2

Common ratio  $r=\frac{6}{2}=3$

Last term is l=13122

The last term formula is $l=ar^{n-1}$

$13122=(2)(3)^{n-1}$

$6561=(3)^{n-1}$

$3^8=(3)^{n-1}$

On comparing the base,

$n-1=8$

$n=9$

So, total number of terms are 9.

The 3rd term from the beginning is 18.

The 3rd term from the last is 7th term which is

$T_7=ar^{6}$

$T_7=(2)(3)^{6}$

$T_7=1458$

The product of 3rd term from the beginning  and 3rd term from the last is

$P=T_3\times T_7$

$P=18\times 1458$

$P=26244$

Hence showed.

#Learn more

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