Question

In a sequence of 21 terms, the first 11 terms are in A.P. with common difference 2 and the last 11 terms are in G.P with common ratio 2. If the middle term of A.P. be equal to the middle term of G.P., then the middle term of the entire sequence is

(a) \(\frac{-10}{31}\)

(b) \(\frac{10}{31}\)

(c) \(\frac{32}{31}\)

(d) \(\frac{-31}{32}\)

Answer 1

(a) \(\frac{-10}{31}.\)

Let the first term of the A.P. be a and common difference d.

Given d = 2 ⇒ T₁₁ of A.P. = a + 10d = a + 20. 

Let the first term of the G.P. be b and common ratio r. 

Given r = 2. Now, the middle term of A.P = middle term of G.P 

⇒ T₆ of A.P = a + 5d = T₆ of G.P = br⁵ 

⇒ a + 5d = br⁵ 

⇒ a + 10 = 32b              (∵ r = 2)                   ...(i) 

Also the last term of A.P. is the first term of G.P.

∴ b = T₁₁ of A.P = a + 20                    ...(ii)

∴ From (i) and (ii) 

a + 10 = 32.(a + 20) 

⇒ 31a = – 630 ⇒ a = \(\frac{-630}{31}\)

∴ Middle term of the entire sequence of 21 terms = 11th term 

= a + 10d

= \(\frac{-630}{31}\) + 20 = \(\frac{-630+620}{31}\) = \(\frac{-10}{31}.\)