Math, asked by pranjal1104, 1 year ago

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.


himanshusangshe: yup

Answers

Answered by hhm1991
0

Answer:

Step-by-step explanation:

2.6.18.54.162.486.1458.4374.13122

13122*2=26244


pranjal1104: i think question says product of third yerm frm last and third term fr 1st product should be tht no
pranjal1104: u multiplied first and last term
pranjal1104: why
pranjal1104: u should mutliply ar^3 and ar^7 and tht seven is bec i got n=9 by solving
pranjal1104: how come u mutliplied 1st and last term?
pranjal1104: how?
Answered by pinquancaro
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

Sum to infinite terms of gp 1/2+1/6+1/18+1/54.....

https://brainly.in/question/14059848

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