Advertisements
Advertisements
प्रश्न
Find 2 + 22 + 222 + 2222 + … upto n terms.
Advertisements
उत्तर
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
संबंधित प्रश्न
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.
For a sequence, if Sn = 2 (3n – 1), find the nth term, hence show that the sequence is a G.P.
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, …
If for a sequence, `t_n = (5^(n- 3))/(2^(n - 3))`, show that the sequence is a G.P.
Find its first term and the common ratio.
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.
Verify whether the following sequence is 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.
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
Verify whether the following sequence is 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.
