Question

if sum of 1st 10th term of A.p is equal to 4 times of 1st 5 terms of that A.p find a/d ?​

Answer 1

If sum of 1st ten term of A.P is equal to 4 times of 1st five terms of that A.P. Find a/d

$\mathbb{SOLUTION} :$

Let the first term of AP is a.

Then,

$\tt{S_{10} = \frac{10}{2} \{2a + (n - 1)d\}} \\ \\ \tt \to 5 \{2a + 9d \}$

And,

$\tt{S_{5} = \frac{5}{2} \{2a + (n - 1)d \} } \\ \\ \tt \to{2.5 \{ 2a + 4d\} }$

According To Question :

$\tt{}5(2a + 5d) = 4 \{2.5(2a + 4d) \} \\ \tt{5(2a + 5d)} = 10(2a + 4d) \\ \\ \tt{2a + 5d = \frac{10}{2}(2a + 4d) } \\ \\ \tt2a + 5d = 10a + 20d \\ \tt{5d - 20d = 10a - 2a} \\ \tt{ - 15d = 8a} \\ \\ \sf \red{ \frac{a}{d} } = \orange{ \frac{ - 15}{8} }$

Thus,

• a/d = (-15)/8