Question

Ratio of 11th and 14th term of ap is 7:9 find ration of 10th and 3th term

Answer 1

$\bold{\underline{\underline{Answer:}}}$

$\bold{\therefore a_{10} : a_{3 } = 19 : 5}$

$\bold{\underline{\underline{Step-by-step\:explantion:}}}$

• In the given question information given about ratio of a11 and a14 is given.

• We have to find ratio of a10 and a3.

$\underline \bold{given :} \\ \implies a_{11} : a_{14} = 7 : 9 \\ \\ \underline \bold{to \: find : } \\ \implies a_{10} : a_{3} = ?$

• According to given question :

$\implies \frac{ a_{11} }{a_{14} } = \frac{7}{9} \\ \\ \implies \frac{a + 10d}{a + 13d} = \frac{7}{9} \\ \\ \bold{cross \: multiply \: them : } \\ \implies 9(a + 10d) = 7(a + 13d) \\ \\ \implies 9a + 90d = 7a + 91d \\ \\ \implies 9a - 7a = 91d - 90d \\ \\ \bold{\implies d = 2a- - - - (1)}\\ \\ \bold{by \: given \: question : } \\ \implies \frac{ a_{10} }{ a_{3} } = \frac{a + 9d}{a + 2d} - - - - - (2) \\ \\ \bold{putting \: value \: of \: d \: in \: (2)} \\ \implies \frac{a_{10}}{a_{3}} = \frac{a + 9 \times 2a}{a + 2 \times 2a} \\ \\ \implies \frac{a_{10}}{a_{3}} = \frac{a + 18a}{a + 4a} \\ \\ \implies \frac{a_{10}}{a_{3}} = \frac{19 \cancel{a}}{5 \cancel{a}} \\ \\ \bold {\therefore a_{10} : a_{3 } = 19 : 5}$

Answer 2

ANSWER:-

Given:

${ }^{</u><u>T</u><u>} 11 \ratio {}^{</u><u>T</u><u>} 14 = 7 \ratio 9$

We know that,

nth term= Tn = a+(n-1)d.

Where,

⚫First term= a

⚫Common difference=d

Therefore,

${}^{T} 11 = a + (11 - 1)d \\ \\ = > {}^{T} 11 = a + 10d ...........(1)$

And,

${}^{T} 14 = a + (14 - 1)d \\ \\ = > {}^{T} 14 = a + 13d.............(2)$

From equation (1) & (2) we get;

$\frac{ {}^{T} 11}{ {}^{T}14 } = \frac{a + 10d}{a + 13d} = \frac{7}{9} \\ [cross \: multiplication] \\ \\ = > 9a + 90d = 7a + 91d \\ \\ = > 9a - 7a = 91d - 90d \\ \\ = > 2a = d\\ \\=) d= 2a$

We know that,

${}^{S} n = \frac{n}{2} [2a + (n - 1)d]$

Now,

$\frac{ {}^{S} 10}{ {}^{S} 3} = \frac{ \frac{10}{2} (2a + 9d)}{ \frac{3}{2} (2a + 2d)} \\ \\ = > \frac{ {}^{S} 10}{ {}^{S} 3} = \frac{10(2a + 9d)}{3(2a + 2d)} \\ From \: equation \: (3) \:, we \: get; \\ = > \frac{ {}^{S} 10}{ {}^{S}3 } = \frac{10[2a + 9(2a)]}{3[2a + 2(2a)]} \\ \\ = > \frac{ {}^{S} 10}{ {}^{S} 3} = \frac{10(20a)}{3(6a)} \\ \\ = > \frac{ {}^{S} 10}{ {}^{S}3 } = \frac{200a}{18a} \\ \\ = > \frac{ {}^{S}10 }{ {}^{S}3 } = \frac{100}{9}$

Hence,

S10: S3= 100:9

Hope it helps ☺️