Question

If the first term of a parallel series is 4 and the
second term is 7 find the third term.​

Answer 1

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${\Huge{\underline{\underline{\bf{\maltese Question:-}}}}}$

If the first term of a parallel series is 4 and the

second term is 7 find the third term.

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${\Huge{\underline{\underline{\bf{\maltese Answer:-}}}}}$

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.

Hence the answer.

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Answer 2

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.

Hence the answer.