Question

17. If the sum of first ten terms of an A.P. is 4 times the
sum of the first five terms, then find the ratio of the
first term to the common difference.​

Answer 1

Answer:

Solution

let the first term = a

common difference = d

$S10= \frac{10}{2} [2a+(10−1)d] \\ =S(2a+9d)$

And

$S5= \frac{5}{2} [2a+(5−1)d] \\ = \frac{5}{2} (2a+4d)$

As per question,

S10 =4(S5 )

$5(2a+9d) \\ =4[ \frac{5}{2} (2a+4d)]  \\ 2a+9d = 2(2a+4d)  \\ 2a+9d=4a+8d \\ ∴d=2a</p><p></p><p></p><p>  \\ or  \: \: \: \frac{a}{d} = \frac{1}{2} </p><p></p><p></p><p>$