# For a sequence, if tn = 5n-27n-3, verify whether the sequence is a G.P. If it is a G.P., find its first term and the common ratio. - Mathematics and Statistics

Sum

For a sequence, if tn = (5^("n" - 2))/(7^("n" - 3)), verify whether the sequence is a G.P. If it is a G.P., find its  first term and the common ratio.

#### Solution

The sequence (tn) is a G.P. if

("t"_("n" + 1))/("t"_"n") = constant for all n ∈ N.

Now, tn = (5^("n" - 2))/(7^("n" - 3))

∴ tn+1 = (5^("n" + 1- 2))/(7^("n" + 1 - 3)) = (5^("n" - 1))/(7^("n" - 2))

∴ ("t"_("n" + 1))/"t"_"n" = (5^("n" - 1))/(7^("n" - 2)) = (7^("n" - 3))/(5^("n" - 2))

= 5^(("n" - 1) - ("n" - 2)) xx 7^(("n" - 3) - ("n" - 2)

= 5(1) x 7–1 = 5/7

= constant , for all n ∈ N.

∴ the sequence is a G.P. with common ratio (r) = 5/7

and first term = t1 = (5^(1 - 2))/(7^(1 - 3))

= (5^(-1))/(7^(-2)) = 7^2/5 = 49/5.

Concept: Sequence and Series - Geometric Progression (G.P.)
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#### APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 11th Standard Maharashtra State Board
Chapter 4 Sequences and Series
Miscellaneous Exercise 4 | Q 3 | Page 64