13. 4th term of an A.P. is equal to 3 times its first
term and 7th term exceeds twice the 3rd term
by 1. Find the first term and the common
difference.
Answers
The 4th term of an A.P. is three times the first and 7th terms exceeds twice the third terms by 1. Find the first term and the common difference?
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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.
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