Question

if for a sequence Sn=2(3^n-1) then the third term is?​

Answer 1

Step-by-step explanation:

$s1 = a1 = 2( {3}^{n} - 1) = 2( {3}^{1} - 1) \\ = 4 \\ \\ \\ s2 = 2( {3}^{2} - 1) = 16 \\ \\ \\ s3 = 2( {3}^{3} - 1) = 52$

$a2 = s2 - s1 = 16 - 4 = 12 \\ \\ \\ \\ a3 = s3 - s2 = 52 - 16 = 36$

Answer 2

Solution :-

Basic Method :-

given that, for a sequence ,

→ S(n) = 2(3^n - 1)

Putting n = 1,

→ S(1) = a1 = 2(3¹ - 1) = 2 * (3 - 1) = 2 * 2 = 4

Putting n = 2,

→ S(2) = a1 + a2 = 2(3² - 1) = 2 * (9 - 1) = 2 * 8 = 16

then,

→ (a1 + a2) - a1 = 16 - 4

→ a2 = 12

Putting n = 3,

→ S(3) = a1 + a2 + a3 = 2(3³ - 1) = 2 * (27 - 1) = 2 * 26 = 52

then,

→ (a1 + a2 + a3) - (a1 + a2) = 52 - 16

→ a3 = 36 (Ans.)

Hence, Third term is equal to 36 .

Extra :-

→ The given sequence is in GP with common ratio as 3 .

→ nth term of sequence = 4[3^(n - 1)] { T(3) = 4[3^(3 - 1)] = 4[3^2] = 4 * 9 = 36.}

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