Question

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

Answer 1

Step-by-step explanation:

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

3a + 30d = 2a + 34d

3a - 2a = 34 - 30 

a = 4d

a₅ = a + 4d

= 4d + 4d 

a₅ = 8d

a₂₁ = a + 20d

= 4d + 20d

a₂₁ = 24d

So, a₅ : a₂₁ = 8 : 24 = 1 : 3

The ratio of the sum of the first 5 terms to the sum of the first 21 terms.

S₅ = 5/2(8d + 4d)

= 60d/2

S₅ = 30d

S₂₁ = 21/2(8d + 20d)

= 588d/2

S₂₁ = 294d

So, 

S₅ : S₂₁ = 30d : 294d

 = 5 : 49

Answer 2

Correct Question:

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

Solution :

Given , t11 : t18 = 2 :3

Therefore,

$\implies\sf{ \frac{a + 10d \: }{a + 17d \: } }$

Therefore, 3 a + 30 d = 2a + 34d

Therefore, a = 4d

Consider S5 : S10

Therefore ,

$\implies\sf{ \frac{S5}{S10} }$

$\implies\sf{ \frac{ \frac{5}{2} (2a + 4d)}{ \frac{10}{2}(2a + 9d) } }$

$\implies\sf{\dfrac{\cancel{5}(a+2d)}{\cancel{5}(2a+9d)}}$

$\implies\sf{ \frac{a + 2d}{2a + 9d} }$

Therefore ,

$\implies\sf{\dfrac{6 \cancel{d} }{ 17 \cancel{d} }}$

$\implies\sf{ \frac{6}{17} }$

= S5 :S10 = 6:17

Hence , 6 : 17 is the correct Answer !!