If for a sequence $(t_{n}), S_{n} = 2 n^{2} + 5 n,$ find $t_{n}$ and show that the sequence is an A.P.

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Solution

$T_{n} = S_{n} - S_{n - 1}$
$= 2 n^{2} + 5 n - [2 (n - 1)^{2} + 5 (n - 1)]$
$= 2 n^{2} + 5 n - 2 n^{2} - 2 + 4 n - 5 n + 5$
$∴ T_{n} = 4 n + 3$
$T_{1} = 4 (1) + 3 = 7$
$T_{2} = 4 (2) + 3 = 11$
$T_{3} = 4 (3) + 3 = 15$

Hence these form an A.P.

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Similar questions

(t_{n}), S_{n} = 4 n^{2} - 3 n

A . P .

(t_{n}), s_{n} = 4 n^{2} - 5 n

t_{n} = \frac{2^{n - 2}}{5^{n - 3}}

T_{n} = 5 n + 1

(a) T_n = n find (i) T_n+1 (ii) T _n−1

(b) If T_n = n² − 1 find (i) T_n−2 (ii) T_n+1

(d) In a sequence T_n= 5 − 3n find (i) T_n+1 (ii) T_n+2

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