Question

The sum of first ten terms of an A.P. is four times the sum of its first five terms, then ratio of the first term and common difference is

• A. 1 : 3
• B. 1 : 2
• C. 3 : 4
• D. 1 : 4

Answer 1

The correct option is B 1 : 2

Let the A.P. be:
$a+(a+d)+(a+2d)+....$

$S_n=\frac{n}{2}\times (2a+(n-1)d$)

$\text{Hence,}{S}_{10}=\frac{10}{2}\times (2a+(10-1)d$)

$=5(2a+9d)$

Similarly we have,
$S_5=\frac{5}{2}\times (2a+(5-1)d$)
$=\frac{5}{2}\times (2a+4d$)

Given: $S_{10}=4S_5$

$\Rightarrow 5(2a+9d)=4\times \frac{5}{2}\times (2a+4d)$
$\Rightarrow 5(2a+9d)=2(2a+4d)$
$\Rightarrow 2a+9d=4a+8d$
$\Rightarrow \frac{a}{d}=\frac{1}{2}$