Question

Twice the 7th term of an A.P is the same as thrice the sum of 2nd and 11th terms.if the 5th term is 15 , find the A.P

Answer 1

Answer:

The required A.P is 63, 51, 39, . . . . . . . . . . . .

Step-by-step explanation:

Given :  2 a₇ = 3 (a₂ + a₁₁)

              a₅ = 15

Formulae used :

$a_{n}$ = a + (n-1)d

2 a₇ = 3 (a₂ + a₁₁)

2 (a +6d) = 3(a + d + a + 10d)

2a + 12d = 6a + 33d

4a + 21d = 0   . . . . . . . . . (1)

and

a₅ = 15

a + 4d = 15  . . . . . . . (2)

multiply by 4  

∴ 4a + 16d = 60    . . . . . . . . . (3)

(1) - (3)

5d = -60

d = - 12

From (2)  a = 15 - 4(-12)

                 a = 63

∴ The A.P is 63, 51, 39, . . . . . . . . . . .