Question

if the starting term of an AP is 19 and the 6th& 22nd terms are in the ratio 11:31. Show that the sixth term is 20​

Answer 1

Explanation:

Given: 6th term A6= 20, 11th term A11-30

we have, An= A1+(n-1)d

=> A6-A1+(6-1)d

=> 20-A1+5d - equation (1)

A11-A1+(11-1)d

=> 30-A1+(11-1)d

=> 30 A1+10dequation (2)

equating 1 and 2, we have

=> d=2, substitute in eq(1)

=> 20= A1+5(2)

=>A1-20-10=10

Hence, the first term A1-10 and common difference d=2.

Answer 2

• given : first term = a = 19
• then ,

$\huge \sf \frac{6th \: term}{22 \: term} = \frac{11}{31}$

$\large \sf \frac{19 + (6 - 1)(d)}{19 + (22 - 1)(d)} = \frac{11}{31}$

• cross multiplying ,

• 31( 19 + 5d ) = 11(19+21d)
• 589 + 105d = 209 + 231d
• 589 - 209 = 231d - 105d
• 380 = 126d
• d = 380/126 = 190/63