Question

which term of the arthmetics progression 3,15,27,39,......Will be 132 more than its 54th term?​

Answer 1

Answer

Given A.P. is : 3, 15,27,39.....

Here a1

= 3, a2

= 15

a3

= 27, a4

= 39

d = a2

– a1

= 15 – 3 = 12

a54 = a (54 – 1)d

= 3 + 53 x 12

= 3 + 636 = 639

Let ntn term be 132 more than a54 We know that,

an = a + (n - 1)d

a54 + 132 = a + (n - 1)d

639 + 132 = 3 + (n - 1) x 12

771 - 3 = 12(n - 1)

768 = 12(n - 1)

n-1=768÷12=64

n=64+1=65