Question

Find 7th term of an ap is 32 and it's 13 term is 62 find the ap

Answer 1

Given: 7th term = 32 and 13th term = 62.

→ 7th term = a + 6d = 32 __(1)

→ 13th term = a + 12d = 62 __(2)

On subtracting (2) by (1),

→ a + 12d - a - 6d = 62 - 32

→ 6d = 30

→ d = 5

By substituting value in eq(1),

→ a + 6(5) = 32

→ a = 32 - 30 = 2

Hence existing Arithmetic Progression is

→ 2, 7, 12, 17, 22, 27, 32,.....

Answer 2

Given:

• 7th term of AP=32

• 13th term of AP=62

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To find:

• AP

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Solution:

7th term of AP can be expressed as:

» a+6d=32   ...(1.)

13th term of AP can be expressed as:

» a+12d=62  ...(2.)

Subtract equation (1.) From equation (2.)

» (a+12d)-(a+6d)=62-32

» a+12d-a-6d=30

» a-a+12d-6d=30

» 6d=30

» d=30/6

» d=5

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Since, we have obtained the common difference of AP, but we also need value of first term (a) so as to complete our AP.

Substitute value of d in equation (1.)

» a+6d=32

» a+6(5)=32

» a+30=32

» a=32-30

» a=2

So the first term of AP is 2.

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Now we know that any Arithmetic Progression (AP) can be expressed as::

» a, a+d, a+2d, a+3d...

But putting values of a and d, we get::

» (2), (2)+(5), (2)+2(5), (2)+3(5)...

» 2, 7, 2+10, 2+15...

» 2, 7, 12, 17...

This is the required AP.

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Extra:-

• How to obtain common difference of any AP?

To find common difference of any AP, we have to subtract any preceeding term from succeeding term.

Let's understand with an example::

Let's assume an AP, 2,5,8,11,...

Here common difference can be obtained by subtracting any of preceeding and succeeding term.

Common difference=5-2=8-5=11-8

Common difference=3

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• Short terms usually used in AP::

» AP=Arithmetic Progression

» a= first term

» d=common difference

» an=nth term

» Sn=sum of nth term

» n=number of terms

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