The 4th term of AP is equal to 3 times the 1st term and 7th term exceeds twice the 2nd
term by 1 Find a and d
Answers
Let ‘a’ be the first term & 'd' be the common difference of the AP referred to.
Then, 4th term is 3 × 1st term ( as per the problem ).
That is, a + 3d = 3a Or a + 3d -a = 3a - a.
That is, 3d = 2a ———be eqn.( 1)
7th term is 1 added to twice the third term ( as per the problem ).
That is, a+6d = 2 ( a+2d) + 1
That is, a + 6d = 2a + 4d + 1 or, a + 6d -4d - a = 2a + 4d + 1 -4d - a.
That is, 2d = a + 1 or 2d - 1 = a +1 - 1.
Or, a=2d - 1. Or, 2a = 2 × (2d - 1 )
Or, 2a = 4d - 2.
This is also equal to 3d as per eqn.(1) above. That is, 3d = 4d -2 Or, 3d - 3d = 4d - 2 - 3d.
That is, 0 = d -2
Or, 0 + 2 = d -2 + 2 Or, d= 2.
Then, 4th term is thrice of 1st term gives us (a + 3d ) = 3a gives us a + (3 × 2 )= 3a. Or, a + 6 = 3a.
That is, a + 6 - a = 3a - a.
That is, 6 = 2a Or, a = 6/2 = 3.
So, the first term is 3
And common difference is 2.
Answer Check:
First term = 3 & Common Difference =2.
So, A.P. is 3, 5, 7, 9, 11, 13, 15, ……
4th term ÷ 1st term = 9÷3 =3.
7th term ÷ 3rd term = 15 ÷ 7 = 2 with remainder 1.
Answer:
First term(a)=2/5
Common difference (d)=3/5
