Question

When the sum of the first ten terms of an a.p is fourtimes the sum of the first five terms. then the K term is ans

Answer 1

sum of first ten terms

$s_{10} = 5(2a + 9d)$

sum of first five terms

$s_{5} = \frac{5}{2} (2a + 4d) = 5(a + 2d)$

now according to question

$s _{10} = 14 s_{5} \\ 5(2a + 9d) = 14 \times 5(a + 2d) \\ 10a + 45d = 70a + 140 d\\ 60a = - 95d \\ 12a = - 19d$

kth term

$a_{k} = a + (k - 1)d \\ a_{k} = \frac{ - 19}{12} d + kd - d \\ a_{k} = d( k- \frac{31}{12} )$

you can also replace d by -12a/19 then answer will be in terms of a and k

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