Question

If the ratio of the 11th term of an A.P to this 18th term is 2:3, finds the ratio of the sum of the first five terms of the A.P to the sum of its 10 terms

Answer 1

Step-by-step explanation:

Given:

$\frac{a + 10d}{a + 17d} = \frac{2}{3}$

To find:

$\frac{s5}{s10} = ?$

Cross multiply the given:

3(a+10d)=2(a+17d)

3a+30d=2a+34d

3a-2a=34d-30d

a=4d-----(1)

Substituting it in the given back we get:

$\frac{4d + 10d}{4d + 17d} = \frac{2}{3} = > ( \frac{14d}{21d} ) \div 7 = \frac{2}{3}$

$\frac{2d}{3d} = \frac{2}{3} = > \frac{d}{d} = \frac{2 \times 3}{3 \times 2} = \frac{6}{6 } = \frac{1}{1}$

:d=1 and a=4.

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

Now,

$\frac{ \frac{5}{2}(4 + (5 - 1) \times 1) }{ \frac{10}{2}(4 + (10 - 1) \times 1 )} = \frac{1 \times (4 + 4)}{2 \times (4 + 9)} = \frac{16}{26} \div 2$

:The ratio is 8:13.

Answer 2

Answer:

answer attached in photo