Question

In an

a.P., the sum of the first n terms bears a constant ratio with the sum to next n terms and if first term of an

a.P. Is 3, then the sum of 12 terms of

a.P., is

Answer 1

Answer:

216

Step-by-step explanation:

let say ap

a , a+d  , a+2d , ........ a +(n-1)d

a = 3

S1 = a = 3    (sum of first n terms where n =1)

S2-S1 = a + a+d -a = a +d = 3 +d    (sum of next n terms where n =1)

S2 = a + a+d = 2a +d = 6 +d    (sum of first n terms where n =2)

S4 - S2 = a + a+d + a+2d + a +3d - (a + a +d) = 2a + 5d = 6 +5d

(sum of next n terms where n =2)

(S2-S1)/S1 = (S4-S2)/S2

(3+d)/3 = (6+5d)/(6+d)

d^2 + 9d + 18 = 18 + 15d

d^2 - 6d = 0

d(d-6) = 0

d = 6 or d = 0

d= means all numbers will be 3 only ( Ap would be 3 , 3 ,3  and so on)

then sum of 12 terms = 12 *3 = 36

d= 6 means AP would be

3 , 9 , 15 , 21 , 27 ..... 3+(n-1)6

Sum of 12 term = (n/2)(a + a + (n-1)d)

= (6/2) (3 + 3 + 11*6)

= 3 (72)

= 216