Q. 73.7( 6 Votes )

# The n^{th} term of a sequence is given by a_{n} = 2n + 7. Show that it is an A.P. Also, find its 7^{th} term.

Answer :

Given,

a_{n} = 2n + 7

We can find first five terms of this sequence by putting values of n from 1 to 5.

When n = 1:

a_{1} = 2(1) + 7

⇒ a_{1} = 2 + 7

⇒ a_{1} = 9

When n = 2:

a_{2} = 2(2) + 7

⇒ a_{2} = 4 + 7

⇒ a_{2} = 11

When n = 3:

a_{3} = 2(3) + 7

⇒ a_{3} = 6 + 7

⇒ a_{3} = 13

When n = 4:

a_{4} = 2(4) + 7

⇒ a_{4} = 8 + 7

⇒ a_{4} = 15

When n = 5:

a_{5} = 2(5) + 7

⇒ a_{5} = 10 + 7

⇒ a_{5} = 17

∴ First five terms of the sequence are 9, 11, 13, 15, 17.

A.P is known for Arithmetic Progression whose common difference = a_{n} – a_{n-1} where n > 0

a_{1} = 9, a_{2} = 11, a_{3} = 13, a_{4} = 15, a_{5} = 17

Now, a_{2} – a_{1} = 11 – 9 = 2

a_{3} – a_{2} = 13 – 11 = 2

a_{4} – a_{3} = 15 – 13 = 2

a_{5} – a_{4} = 17 – 15 = 2

As, a_{2} – a_{1} = a_{3} – a_{2} = a_{4} – a_{3} = a_{5} – a_{4}

The given sequence is A.P

Common difference, d = a_{2} – a_{1} = 2

To find the seventh term of A.P, firstly find a_{n}

We know, a_{n} = a + (n-1) d where a is first term or a_{1} and d is common difference

∴ a_{n} = 3 + (n-1) 2

⇒ a_{n} = 3 + 2n – 2

⇒ a_{n} = 2n + 1

When n = 7:

a_{7} = 2(7) + 1

⇒ a_{7} = 14 + 1

⇒ a_{7} = 15

Hence, the 7^{th} term of A.P. is 15

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