Advertisements
Advertisements
प्रश्न
For a sequence Sn = 4(7n – 1), verify whether the sequence is a G.P.
उत्तर
Sn = 4(7n – 1)
∴ Sn–1 = 4(7n–1 – 1)
But, tn = Sn – Sn–1
= 4(7n – 1) – 4(7n–1 – 1)
= 4(7n – 1– 7n– 1 + 1)
= 4(7n–1+1 – 7n–1)
= 4.7n–1 (7 – 1)
∴ tn = 24.7n–1
∴ tn+1 = 24(7)n+1–1
= 24(7)n
The sequence (tn) is a G.P., if `("t"_("n" + 1))/"t"_"n"` = constant for all n ∈ N.
∴ `("t"_("n" + 1))/"t"_"n" = (24(7)^"n")/(24(7)^("n" - 1)`
= 7 = constant, for all n ∈ N
∴ the sequence is a G.P.
APPEARS IN
संबंधित प्रश्न
Verify whether the following sequence is G.P. If so, write tn:
2, 6, 18, 54, ...
Verify whether the following sequence is G.P. If so, write tn:
1, – 5, 25, – 125, ...
Verify whether the following sequence is G.P. If so, write tn:
3, 4, 5, 6, ...
For the G.P., if r = `1/3`, a = 9, find t7.
For the G.P., if a = `7/243, "r" = 1/3`, find t3.
For the G.P., if a = `2/3`, t6 = 162, find r.
If p, q, r, s are in G. P., show that p + q, q + r, r + s are also in G. P.
Insert two numbers between 1 and – 27 so that the resulting sequence is a G.P.
For a G.P. a = `4/3 and "t"_7 = 243/1024`, find the value of r.
Find k so that k – 1, k, k + 2 are consecutive terms of a G.P.
If pth, qth and rth terms of a G.P. are x, y, z respectively, find the value of xq – r .yr – p .zp – q.
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.
Verify whether the following sequences are G.P. If so, find tn.
`sqrt5, 1/sqrt5, 1/(5sqrt5), 1/(25sqrt5), ...`
Verify whether the following sequence are G.P. If so, find tn
`sqrt5, 1/sqrt5, 1/(5sqrt5), 1/(25sqrt5),.......`
Verify whether the following sequence is G.P. If so, find tn.
`sqrt(5), 1/sqrt(5), 1/(5sqrt(5)), 1/(25sqrt(5)), ...`
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.
Verify whether the following sequence is G.P. If so, find tn.
`sqrt5,1/(sqrt5),1/(5sqrt5), 1/(25sqrt5)`, ......