Question

answer to this question plz

Answer 1

aloha user!!
----------------

thanks for asking this question.
here is the answer:


the ratio of the 11th term to the 18th term of an AP is given to be 2 : 3.

the 11th term would be a11= a + (n -1)d
                                              = a + (11- 1)d
                                              = a + 10d-----------------(1)

the 18th term would be a18= a + (n-1)d
                                               = a + (18-1)d
                                               = a + 17d--------------------(2)

now their ratio is given to be:

[tex] \frac{a+10d }{a+17d } = \frac{2}{3} [/tex]

-----------------------------------------------------------

3( a + 10d ) = 2( a + 17 d )
3a + 30d = 2a + 34d
3a- 2a = 34d - 30d
a= 4d

----------------------------------------------------------

now,

we have got to find the ratio of 5th term to 21st term.

$\frac{a + 4d}{a + 20d} = ?$

[tex] \frac{ 4d + 4d}{4d + 20d} = \frac{8d}{24d} = \frac{1}{3} [/tex]

hence the ratio of 5th term to 21st term is 1:3

--------------------------------------------------------

now,

we have got to find the ratio of sum of first 5 terms to the sum of first 21 terms.

sn= n/2 { 2a + (n- 1)d }

s5= 5/2 { 2(4d) + 4d }
s5= 5/2 { 8d + 4d }
s5= 5/2 { 12d }
s5=30d

s21= 21/2 { 2(4d) + 20d }
s21= 21/2 { 8d + 20d }
s21= 21/2 { 28d }
s21= 21× 14d
s21= 294d

so,

the ratio would be:


$\frac{s5}{s21} = \frac{30d}{294d} = \frac{5}{49}$


-------------------------------------------------------
hope it helps :^)