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The 7th Term of an A.P. is 32 and Its 13th Term is 62. Find the A.P. - Mathematics

The 7th term of an A.P. is 32 and its 13th term is 62.  Find the A.P.

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Solution

Here, let us take the first term of the A.P. as a and the common difference of the A.P as d

Now, as we know,

`a_n = a + (n -1 )d`

So for the 7th term (n = 7)

`a_7 = a + (7 - 1)d`

32 = a + 6d .......(1)

Also for 12th term (n = 13

`a_13 = a + (13 - 1)d`

62 = a + 12d ......(2)

Now on substracting (2) from (1) we get

62 - 32 = (a + 12d) - (a + 6d)

30 = a + 12d - a - 6d

30 = 6d

`d = 30/6`

d = 5

Substiting the value of d in (1) we get

32 = a + 6(5)

32 = a + 30

a = 32 - 30

a = 2

So, the first term is 2 and the common difference is 5.

There the A.P is 2, 7, 12, 27, ....

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 10 Maths
Chapter 5 Arithmetic Progression
Exercise 5.4 | Q 25 | Page 25
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