If $a = 8$ , $t_{n} = 62$ , $S_{n} = 20$ , find $n$ .

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Solution

$T_{n} = a + (n - 1) d$

$62 = 8 + (n - 1) d$

$54 = (n - 1) d$ ...(1)

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$20 = \frac{n}{2} [2 (8) + (n - 1) d]$

$40 = n [16 + (n - 1) d]$

From (1)

$40 = n [16 + 54]$

$40 = n (70)$

$∴ n = \frac{4}{7}$

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Similar questions

S_{n} =^{n} C_{1} - \frac{^{n} C_{2}}{2} + \frac{^{n} C_{3}}{3} - . . . . . + \frac{{(- 1)}^{n - 1}^{n} c_{n}}{n}

T_{n} = 1 + \frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{n}

S_{n} = T_{n} \forall n

S_{n + 1} - S_{n} = T_{n + 1} - T_{n}

a = 8, T_{n} = 33, S_{n} = 123

d

n

a = 8, a_{n} = 62, S_{n} = 210

Find n and d where a=8 , a_n=62 ,S_n=210.

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