Advertisements
Advertisements
Question
Find 2 + 22 + 222 + 2222 + … upto n terms.
Advertisements
Solution
Sn = 2 + 22 + 222 +… upto n terms
= 2(1 + 11 + 111+ … upto n terms)
= `2/9` (9 + 99 + 999 + upto n terms)
= `2/9`[(10 – 1) + (100 – 1) + (1000 – 1) + ... upto n terms]
= `2/9`[(10 – + 100 + 100 + ... upto n terms) – (1 + 1 + 1 ... n terms)]
Since, 10, 100,1000, … n terms are in G.P. with a= 10, r = `100/100` = 10
∴ Sn = `2/9[10((10^"n" - 1)/(10 - 1)) - "n"]`
= `2/9[10/9(10^"n" - 1) - "n"]`
∴ Sn = `2/81[10(10^"n" - 1) - 9"n"]`
APPEARS IN
RELATED QUESTIONS
Find three numbers in G. P. such that their sum is 21 and sum of their squares is 189.
Find four numbers in G. P. such that sum of the middle two numbers is `10/3` and their product is 1.
The fifth term of a G. P. is x, eighth term of the G. P. is y and eleventh term of the G. P. is z. Verify whether y2 = xz.
In a G.P., the fourth term is 48 and the eighth term is 768. Find the tenth term.
Find the nth term of the sequence 0.6, 0.66, 0.666, 0.6666, …
For the G.P. if a = `2/3` t6 = 162, find r.
Express the following recurring decimals as a rational number.
`4.bar18`
For the G.P. if a = `2/3`, t6 = 162, find r.
For the G.P. if a = `2/3`, t6 = 162, find r.
Verify whether the following sequences are G.P. If so, find tn.
`sqrt5, 1/sqrt5, 1/(5sqrt5), 1/(25sqrt5), ...`
For the G.P. if a = `2/3` , t6 = 162 , find r
For the G.P. If a = `2/3, t_6 = 162,` find r
For the G.P. if a = `2/3`, t6 = 162, find r.
